Chemistry Part I Chapter 3 Classification Of Elements And Periodicity In Properties
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    NCERT Solution For Class 11 Chemistry Chemistry Part I

    Classification Of Elements And Periodicity In Properties Here is the CBSE Chemistry Chapter 3 for Class 11 students. Summary and detailed explanation of the lesson, including the definitions of difficult words. All of the exercises and questions and answers from the lesson's back end have been completed. NCERT Solutions for Class 11 Chemistry Classification Of Elements And Periodicity In Properties Chapter 3 NCERT Solutions for Class 11 Chemistry Classification Of Elements And Periodicity In Properties Chapter 3 The following is a summary in Hindi and English for the academic year 2021-2022. You can save these solutions to your computer or use the Class 11 Chemistry.

    Question 1
    CBSEENCH11005372
    Question 2
    CBSEENCH11005373

    What are typical elements?

    Solution
    These are the elements of the third short period in the periodic table.
    Question 3
    CBSEENCH11005374

    Give the ‘liquid elements’ in the periodic table.

    Solution
    Mercury and Bromine.
    Question 4
    CBSEENCH11005375

    What is valency of an element?

    Solution
    Valency of an element is defined as the combining power of element. As, the number of electrons gained or lost or shared with other atoms in the formation of compounds.
    Question 5
    CBSEENCH11005376

    Gallium was named by Mendeleev as

    Solution
    Eka-aluminium.
    Question 6
    CBSEENCH11005377

    What are magic numbers?

    Solution
    In the periodic table, elements with similar properties are repeated after intervals of 2, 8, 18, 32 etc. These numbers are called magic numbers.
    Question 7
    CBSEENCH11005378

    Which property is the basis of long form of periodic table?

    Solution
    The long form of periodic table is based upon the property that ‘Physical and chemical properties of the elements are a periodic function of their atomic number.’
    Question 8
    CBSEENCH11005379

    How many elements are placed in each period of the p-block?

    Solution
    Six elements are placed in each period of the p-block.
    Question 9
    CBSEENCH11005380

    What is the number of groups in: (i) p-block  (ii) d-block?

    Solution

    (i) p-block: Six groups
    (ii) d-block: Ten groups.

    Question 10
    CBSEENCH11005381

    How many elements are present in:
    (i) second period    
    (ii) fourth period    
    (iii) sixth period?

    Solution
    The number of elements present in the given periods are;
    (i) 8 elements
    (ii) 18 elements
    (iii) 32 elements.
    Question 11
    CBSEENCH11005382

    What is the name given to the s-block elements?

    Solution

    These are called alkali elements.

    Question 13
    CBSEENCH11005384

    Why are there 10 elements in each series of d-block?

    Solution

    As the d-subshell can have a maximum of ten electrons.

    Question 14
    CBSEENCH11005385

    In terms of electronic configuration,
    what is common in a given period and group?

    Solution
    For elements in a period, the number of shells is equal.
    For elements in a group, the number of electrons in the outermost shell is the same.
    Question 15
    CBSEENCH11005386

    Give the general electronic configuration of
    (i) Transition elements     (ii) Inner transition elements.

    Solution

    (i) Transition elements:
         (n - 1)d1-10  ns1-2.
    (ii) Inner transition elements:
          (n - 2)f1-14   (n - 1)d0-1  ns2.

    Tips: -

     
    Question 16
    CBSEENCH11005387

    Name the group and period to which an element with Z = 15 belongs ?

    Solution
    Element (Z = 15) = 1s22s22p63s23p3.  It belongs to 3rd period and group 15.
    Question 17
    CBSEENCH11005388

    What are the advantages of periodic classification of elements?

    Solution

    (i) The classification simplifies the learning of chemistry.

    (ii) Periodic table explains clearly why elements in a group display similar properties. The periodic table serves as a guide for scientific investigations.

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    Question 18
    CBSEENCH11005389

    In the modern periodic table, the period indicates the value of:
    (a) atomic number
    (b) atomic mass
    (c) principal quantum number
    (d) azimuthal quantum number.

    Solution
    (c) Principal quantum number. This is because each period begins with the tilling of new shell having a particular value of principal quantum number. 
    Question 21
    CBSEENCH11005392

    Write the atomic number of the element present in the third period and seventeenth group of the periodic table. 

    Solution

    In the third period the filling up of only 3s and 3p-orbitals occur. Therefore, in this period there are only two s- and six p-biock elements. Since the third period (n = 3) starts with Z = 11 and ends at Z = 18, hence the element is chlorine (CI) with atomic number (Z) = 17 which is present in the third period and seventeenth group of the periodic table.

     
    Question 23
    CBSEENCH11005394

    How is the atomic radius of an element actually expressed?

    Solution
    Atomic radius of an element is generally expressed as covalent radius because most of the elements form covalent bonds in their atoms.
    Question 24
    CBSEENCH11005395
    Question 25
    CBSEENCH11005396

    Which radius, vander Waal's or covalent, is larger for an atom?

    Solution
    Vander Waal’s radius is larger by 40%. 
    Question 26
    CBSEENCH11005397

    What is the correlation between atomic size and ionisation enthalpy?

    Solution
    Larger the size of atom, lesser is the force of attraction from the valence electrons. Thus, lower value of ionisation enthalpy.
    Question 27
    CBSEENCH11005398
    Question 28
    CBSEENCH11005399

    Arrange,  l, l-, l+ in increasing order of size.

    Solution
    Increasing size order of given series is,
     l+< l < l-
    Question 29
    CBSEENCH11005400

    Arrange the following in decreasing order of ionic size:

    Mg2+,  O2-,  Na+, F-,  N3-.

    Solution
    Decreasing order of ionic size
    N3- > O2- > F- > Na+ > Mg2+.
    Question 30
    CBSEENCH11005401

    Name two ions which are iso-electronic with Ar (Z = 18).

    Solution
    K+, Ca2+ are isoelectronic ions.
    Question 31
    CBSEENCH11005402

    Which one out of O and Se is bigger in size and why?

    Solution
    Selenium has bigger atomic size because it comes after oxygen in the group (O, Se) and down the group size of atom increase.
    Question 32
    CBSEENCH11005403

    Which one out of Na+ and Cl- is bigger in size and why?

    Solution
    The size of Cl- is greater as it has three shells (2, 8, 8) while Na+ ion has two shells only (2, 8).
    Question 33
    CBSEENCH11005404

    Out of P3-,  S2- and Cl- ions, which has the smallest value of ionic radius and why?

    Solution
    All these elements are isoelectronic ions i.e. having same number of electrons (18) but their atomic numbers are different. Cl- ion (Atomic number 17) has the maximum effective nuclear charge, therefore, the size of Cl- ion must be the smallest out of the three ions. 
    Question 34
    CBSEENCH11005405

    Out of the following which has the smallest ionic 'Size and why: Li+, Na+, K+?

    Solution
    All the three ions belong to alkali earth metal group (group 1). Ionic radius increases down the group, therefore Li+ ion has the smallest ionic size out of the three ions.
    Question 35
    CBSEENCH11005406

    What is the trend of ionisation enthalpy in a period?

    Solution

    It increases from left to right in a period.

    Question 36
    CBSEENCH11005407

    What is the trend of ionisation enthalpy in a group?

    Solution
    It decreases as we move from top to bottom down a group.
    Question 37
    CBSEENCH11005408

    Among the elements Li, K, Ca, S and Kr, which one has the highest first ionisation enthalpy and which one has the lowest first ionisation enthalpy?

    Solution

    The element with highest IE1 value: Kr (Noble gas).
    The element with lowest IE1 value: K (Potassium)

    Question 38
    CBSEENCH11005409

    For each of the following pairs state which one has greater ionisation enthalpy:
    (i) I, I- (ii) Br, K.

    Solution

    (i) I-  (ii) Br has the greater ionisation enthaply.

    Question 39
    CBSEENCH11005410

    From amongst the following, choose the element with the lowest IE: Be, B and C.

    Solution

    Boron (B) has lowest Ionization energy.

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    Question 40
    CBSEENCH11005411

    What is the relationship between the first ionisation enthaplies and metallic properties?

    Solution

    Higher the value of ionisation enthalpy of an atom, greater will be its electronegative and non-metallic character. 

    Question 41
    CBSEENCH11005412

    What is the relationship between the first ionisation enthalpy and reducing power and basic nature of element?

    Solution

    Lower the Ionisation enthalpy, greater will be the reducing power and basic nature of the element.

    Question 42
    CBSEENCH11005413

    Arrange the following in order of increasing first ionisation enthalpy:
    Mg, Al, Si, Na.

    Solution
    Increasing order of first ionisation enthalpy of the given series.
    Na < Al < Mg < Si.
    Question 43
    CBSEENCH11005414
    Question 44
    CBSEENCH11005415

    Out of Na and Na+, which has higher ionisation enthalpy?

    Solution
    Na+ has higher ionisation enthalpy.
    Question 46
    CBSEENCH11005417

    Which one of the following statements is incorrect in relation to ionization enthalpy?
    (a) Ionization enthalpy increases for each successive electron.
    (b) The greatest increase in ionization enthalpy is experienced on removal of electron from core noble gas configuration.
    (c) End of valence electrons is marked by a big jump in ionization enthalpy.
    (d) Removal of electron from orbitals bearing lower n value is easier than from orbital having higher n value.

    Solution
    (d) is incorrect. Ionization potential or enthalpy is the minimum amount of energy which is needed to remove the most loosely bound electron from a neutral isolated gaseous atom to form a cation. Electrons in orbitals bearing a lower n value are more attracted to the nucleus than electrons in orbitals bearing a higher n value. Hence, the removal of electrons from orbitals bearing a higher n value is easier than the removal of electrons from orbitals having a lower n value.
    Question 47
    CBSEENCH11005418

    Considering the elements B, Al, Mg, and K, the correct order of their metallic character is:
    (a) B > Al > Mg > K                         (b) Al> Mg> B > k
    (c) Mg > Al > K > B                         (d) K > Mg > Al > B 

    Solution
    (d) is correct because metallic character decreases from left to right in a period and increases from top to bottom in a group. K is expected to be the most metallic and B the least. 
    Question 48
    CBSEENCH11005419

    Considering the elements B, C, N, F and Si, the correct order of their non-metallic character is:
    (a) B> C > Si > N > F                  (b)  Si > C > B > N > F
    (c)  F > N > C > B > Si                (d) F > N > C > Si > B

    Solution
    (d) is correct because non-metallic character increases from left to right in a peiod and decreases from top to bottom in group. F is expected to be most non-metallic and B the least.
    Question 49
    CBSEENCH11005420

    Considering the elements F, Cl, O and N, the correct order of their chemical reactivity in terms of oxidizing property is:
    (a) F > Cl > O > N                               (b) F > O > Cl > N
    (c) Cl> F >O > N                                  (d) O > F > N > Cl

    Solution
    (b) is correct because oxidising character increases from left to right. Therefore oxidising power decreases in the order F > O > N. But within a group, oxidising power decreases from top to bottom. Thus F is stronger oxidising agent than Cl. Also O is more electronegative than Cl, therefore O is stronger oxidising agent than Cl. 
    Question 50
    CBSEENCH11005421

    The second ionisation enthalpy of an atom is greater than its first ionisation enthalpy because __________.

    Solution

    More enthalpy is requried to remove an electron held more firmly by the unipositive ion.

    Question 51
    CBSEENCH11005422
    Question 52
    CBSEENCH11005423

    Which elements occupy the maxima of ionisation potential curve?

    Solution

    Noble gases elements occupy the maxima of ionisation potential curve.

    Question 53
    CBSEENCH11005424

    Do elements have negative ionisation enthalpy?

    Solution
    No, since enthalpy is always needed to remove electrons.
    Question 55
    CBSEENCH11005426
    Question 57
    CBSEENCH11005428

    What is the electron gain enthalpy of neon?

    Solution

    +116 kJ mol-1.

    Question 58
    CBSEENCH11005429

    The size of isoelectronic species: F-, Ne and Na+ is affected by:
    (a) nuclear charge (Z)
    (b) valence principal quantum number (n)
    (c) electron-electron interaction in the outer orbitals.
    (d) none of the factors because their size in the same.

    Solution
    (a) nuclear charge (Z). Isoelectronic species are the species belonging to different atoms or ions which have same number of electrons but different magnitudes of nuclear charges.

    The size of an isoelectronic species increases with a decrease in the nuclear charge (Z). For example, the order of the increasing nuclear charge of F, Ne, and Na+ is as follows:

           F < Ne < Na+

    Z     9     10     11

    Therefore, the order of the increasing size of F, Ne and Na+ is as follows:

    Na+ < Ne < F

    Question 59
    CBSEENCH11005430

    From the electronic configuration of the elements A, B and C given below:
    space space straight A space colon space 1 straight s squared space 2 straight s squared space 2 straight p subscript straight x superscript 1 space space 2 straight p subscript straight y superscript 1
space space straight B thin space colon space 1 straight s squared space 2 straight s squared space 2 straight p subscript straight x superscript 2 space 2 straight p subscript straight y superscript 1 space 2 straight p subscript straight z superscript 1
space space straight C thin space colon space 1 straight s squared space 2 straight s squared space 2 straight p subscript straight y superscript 1 space 2 straight p subscript straight z superscript 1
    (i) Which one has least negative electron gain enthalpy and
    (ii) Which one has highest ionisation enthalpy value?

    Solution

    (i) Element having lowest negative electron gain enthalpy is C.
    (ii) Element having highest I.E1 value is C.

    Question 60
    CBSEENCH11005431

    Are the oxidation state and covalency of Al in open square brackets AlCl left parenthesis straight H subscript 2 straight O right parenthesis subscript 6 close square brackets to the power of 2 plus end exponent same?

    Solution

    Valency of chlorine = (-1)
    Valency of water = 0
    Let the oxidation state of Al be x.
               x -1         0
           [Al Cl   (H2O)6]2+
    therefore        x - 1 + 6(0) = +2
    or                         x = +3
    Thus the oxidation state of Al is +3. Since six legends (which can denote a pair of electrons) i.e. H2O molecules are attached to the alumninum atom, thereofre, its covalency is 6.

    Question 61
    CBSEENCH11005432

    What is periodic table? What do you mean by classification of elements?

    Solution
    Periodic table is defined as the arrangement of all the known elements according to their properties in a tabular form. The method of arranging similar elements together and separating them from dissimilar elements is called classification of elements. The aim of classification is to make the study of the elements and their compounds easier.
    Question 62
    CBSEENCH11005433

    What was the need for the classification of elements?

    Solution

    In chemistry 114 element are known. The most of elements take a part in variety of chemical transformations to number of substances and it is also very difficult to study the properties of all these element and their compound individually. Thus in order to solve this problem it is necessary to classify the element.

    Question 63
    CBSEENCH11005434

    Discuss the earlier attempts to classify elements.

    Solution

    Different scientists tried to group elements with similar elements.
    (i) Dobereiner Triads (1822). Dobereiner arranged similar elements in a group of three so that the atomic mass of the central element was nearly equal to arithmetic mean of the atomic mass of the other two elements. The group of three elements was named as triad. For example,
    Triad                    Lithium     Sodium          Potassium           Mean atomic mass
    Atomic masses       7                23                    39                     space space fraction numerator 7 plus 39 over denominator 2 end fraction space equals space 23
    Triad                  Chlorine        Bromine              Iodine             Mean atomic mass
    Atomic masses       35.5            80                      127                    space space fraction numerator 35.6 plus 127 over denominator 2 end fraction space equals space 81.25
    But this concept of triad could be applied to a limited number of elements. It was rejected soon.
    (ii) Newland's law of octaves (1864). Newland observed that if the known elements were arranged in order of their increasing atomic masses, similar properties recurred in every eighth element like nodes in a musical scale. This generalisation was known as Law of octaves. This system worked very well for lighter elements.

     

    But this law failed in case of heavier elements and was rejected.
    (iii) Lother Meyer’s work (1870). Lother Meyer plotted a graph between atomic volumes of elements in the solid state and their atomic masses. He found that the elements with similar physical properties occupied similar positions on the curve. By taking into account this fact, he drew up a periodic table in which elements were arranged in order of increasing atomic masses.

    Question 64
    CBSEENCH11005435

    Which important property did Mendeleev use to classify the elements in his Periodic Table? and did he stick to that?

    Solution
    Mendeleev arranged  all the known element in horizontal rows and vertical columns in the order of their  increasing atomic weights. While doing so, he left certain gaps in his table and predicted that new elements would be discovered to fill these gaps.
    Question 65
    CBSEENCH11005436

    What is the basis of Mendeleev's Periodic table?

    Solution
    Mendeleev’s periodic table is based on the Mendeleev’s periodic law which sates that the physical and chemical properties of elements are a periodic function of their atomic masses. It means that when the elements are arranged in order of increasing atomic masses, the elements with similar properties recur at regular intervals.
    Question 66
    CBSEENCH11005437

    Give a brief discussion of the Mendeleev's Periodic table. 

    Solution

    Mendeleev arranged all the known elements (63 at that time) in horizontal rows and vertical columns in order of increasing atomic masses in the form of table known as Periodic table. Mendeleev’s table has few vertical columns called groups and a few horizontal rows called periods. Mendeleev left certain gaps in his table and predicted that new elements would be discovered to fill up the gaps.
    The Mendeleev's modified periodic table consists of:
    (i) Nine vertical columns called groups. These are numbered from I to VIII and zero (The members of the zero group were not discovered at the time of Mendeleev). Each group from I to VII, is further sub-divided into two sub-groups-A and B. Group VIII consists of three sets, each containing three elements. Group zero consists of noble gases.

    (ii) Seven horizontal rows are called periods. These are numbered from I to VII. First period contains two elements, second and third periods contain eight elements each. These periods are called short periods. Fourth and fifth periods contain eighteen elements each. These periods are called long periods. Sixth period contains thirty two elements and is called longest period. Seventh period is incomplete and contains nineteen elements (At. No. 87-105).

    Question 67
    CBSEENCH11005438

    What are the merits and applications of Mendeleev's Periodic table?

    Solution

    The periodic table has following merits and applications:
    (i) Prediction of new elements: Mendeleev left various vacant places in his table which provided a clue for the discovery of new elements.
    (ii) Determination of correct atomic masses. This table has also been useful in correcting the doubtful atomic masses of some elements.
    (iii) In research work. Periodic table provided useful information which helped in stimulating research work.
    (iv) Prediction of properties. The properties of elements and their compounds can be predicted from their positions in the table.

    Question 68
    CBSEENCH11005439

    What are the defects or drawbacks of Mendeleev’s table? 

    Solution

    Drawbacks of Mendeleev's table:
    (i) Position of hydrogen: Hydrogen is placed in group I-A of the periodic table. However, it resembles the elements of both group I A (alkali metals) and group VII A (halogens). Therefore, the position of hydrogen in the periodic table is not clear.
    (ii) Anomalous pairs of elements: In some cases, elements of higher atomic masses are placed before those having lower atomic masses. Examples:
      Ar        and         K;
    39.95                     39.1
    Co          and        Ni;         Te       and    I
    58.93                  58.71    127.6           126.9
    (iii) Metals and non-metals: No attempt has been made to place metals and non-metals separately in it.
    (iv) Lanthanoides and Actinodes: Fifteen elements are placed in one position.
    Group III B
    6th period:  La + 14 lathanoide elements. (58Ce to 71Lu).
    7th period:   Ac + 14 actinoide elements. (90Th to 103Lr)
    Thus, lanthanoids and actinoides have not been provided separate and proper places in the Mendeleev's periodic table.
    (v) Position of isotopes:  Isotopes of elements are placed in the same position in the table though according to their atomic masses they should have been placed in different positions.
    (vi) Similar elements separated in the table: Certain elements such as copper and mercury or gold and platinum which possess similar chemical properties are placed in different groups.
    (vii) Dissimilar elements placed together in the same group: Alkali metals such as Li, Na, K etc. (group IA) are grouped together with coinage metals such as Cu, Ag and Au (group IB) though their properties are quite different.
    (viii) Number of elements in the periods. The presence of only 2 elements in the first period, 8 in the second and third etc. can not be explained.

    Question 69
    CBSEENCH11005440

    State and explain Modern Periodic Law.

    Solution
    In 1913, Moseley stated that the physical and chemical properties of the elements are determined by their atomic numbers instead of their atomic masses. The acceptance of atomic number as the fundamental property of an element led to the development of modern periodic law. It states that “Physical and chemical properties of the elements are a periodic function of their atomic numbers.” It means that if the elements are arranged in order of their increasing atomic numbers, the elements with similar properties recur after certain regular intervals.

    Question 70
    CBSEENCH11005441

    What is periodicity? What is its cause?

    Solution

    The repetition of the properties of elements after certain intervals when the elements are arranged in order of increasing atomic numbers is known as periodicity.

    Cause of periodicity: We know that the physical and chemical properties of the elements depend upon the number of electrons present in their outermost shell. Elements having similar outer electronic configurations have similar properties. When the elements are arranged in the periodic table on the basis of their increasing atomic numbers, similar valence shell electronic configurations are repeated after certain regular intervals of atomic numbers i.e. 2, 8, 8, 18, 18, 32 (magic numbers). This repetition of similar valence shell electronic configurations of the elements in a group is the cause of periodicity of the elements in the periodic table. For example, all the members of alkali metal family, which have definite gaps of atomic numbers (8, 8, 18, 18, 32), have similar physical and chemical properties because they have similar valence shell electronic configuration i.e. ns1 as shown below:

    Element Atomic No. Electronic configuration
    Lithium (Li) 3 [He]221
    Sodium (Na) 11 [Ne]103s1
    Potassium (K) 19 [Ar]18 4s1
    Rubedium (Rb) 37 [Kr]36 5s1
    Cesium (Cs) 55 [Xe]54 6s1
    Francium (Fr) 87 [Rn]86 7s1

    In a similar manner, all the halogens i.e. elements of group 17 have similar outer electronic configuration (ns2 np5) and as such possess similar properties.

     

    Question 71
    CBSEENCH11005442

    What is long form of the periodic table? Discuss in brief its salient structural features.

    Solution

    Long form of the periodic table (also called Bohr's table) is constructed on the basis of repeating configuration of the atoms when they are arranged in order of increasing atomic numbers.

    Salient structural features:  This table has the following features:
    (i) All the elements have been arranged in the increasing order of atomic numbers.
    (ii) Elements having similar electronic configurations have been placed together at one place and those having different electronic configurations have been placed at different places in the periodic table.
    (iii) Table consists of seven horizontal rows called periods and each period starts with different principal quantum number. The first, second and third periods are called short periods. They contain 2.8, 8 elements respectively. Fourth, fifth and sixth periods are called long periods. They contain 18, 18 and 32 elements respectively. Seventh period is incomplete and contains 19 elements (2+14+3).
    (iv) The series of 14 elements just after lanthanum (57La) i.e. from cerium (58Ce) to lutetium (71Lu) is collectively called lanthanoides and are placed at the bottom of the periodic table to avoid undue side-wise expansion of the periodic table.
    (v) The series of 14 elements just after actinium (89Ac) i.e. from thorium (90Th) to lawrencium (103Lr) is collectively called actinoides and are placed at the bottom of the periodic table to avoid undue side-wise expansion of the periodic table.
    (vi) Lanthanoids and actinoids are also called Inner Transition elements.
    (vii) The elements beyond uranium (atomic number 93 to 105) which have been prepared artificially by nuclear reactions are also called Transuranic elements or man made elements.
    (viii) Table consits of eighteen vertical columns called groups or families.
    (ix) Elements in each column have the greatest electronic similarities on the basis of their electronic configuration, the elements have been divided into s-block, p-block, d-block and f-block.
    (x) According to 1UPAC system, groups are numbered as 1-18:
    (i) Alkali metals (Group 1)
    (ii) Alkaline earth metals (Group 2)
    (iii) In between groups 2 to 13 there are ten columns which constitute group 3 to 12. The elements present in these groups are known as Transition elements.
    (iv) Pincogens (group 15) (v) Chalcogens (Group 16) (vii) Halogens (group 17) (viii) Aerogens, Noble gases, inert or rare gases (group 18).
    (ix) The elements constituting groups 1, 2 and group 13 to 18 are called normal or representative elements.

    Question 72
    CBSEENCH11005443

    Why atomic number is a better basis for classifying the elements than atomic mass?

    Solution
    The atomic mass of the element relates to the nucleus which is present in the centre of the atom. But the properties of the elements depend upon the electronic configuration which is linked with atomic number. Therefore, atomic number is a better basis for the classification of the elements than the atomic mass.
    Question 73
    CBSEENCH11005444

    Why is long form periodic table regarded better than the Mendeleev's periodic table?

    Solution

    Long form periodic table is regarded as better than the Mendeleev’s table due to the following reasons:

    Long form periodic table Mendeleev's 
    1. All elements have been arranged in the increasing order of atomic numbers. 1. Elements have been arranged in the increasing order of atomic masses.
    2. The position of hydrogen has been justified. 2. There is no justification for the position of hydrogen.
    3. It takes into consideration the filling of the electrons in s, p, d and f subshell. 3. It takes into consideration only the atomic numbers of the element.
    4. The table has been divided into four blocks i.e. s, p, d and f block. 4. There are no blocks in the table.
    5. Groups are not further sub-divided into subgroups. 5. There are sub-groups (A and B) in each group.
    6. This is simple and easy to reproduce. 6. This is quite difficult to reproduce.

     

     

    Question 74
    CBSEENCH11005445

    What are the defects of long form of the periodic table?

    Solution

    1. Position of hydrogen. Hydrogen shows similarities with both metals(alkali metals) and non-metals (halogens) but it is placed with alkali metals.
    2. Position of helium. It is kept along with p-block elements. 
    3. Position of lanthanoids and actinoids. They are kept outside the main body of the periodic table. 

    Question 75
    CBSEENCH11005446

    Discuss the nomenclature of the element with atomic number greater than 100.

    Solution

    In 1997, IUPAC recommended the approved official names for elements with atomic numbers 104 to 109. The names for elements with atomic number 110 and beyond have yet to be officially announced. This nomenclature is based on the Latin words for the atomic number of the element.
    (i) The names are derived directly from the atomic number using the numerical roots for 0 and numbers 1to 9.

    Digit Name Abbreviation
    0 nil n
    1 un u
    2 bi b
    3 tri t
    4 qued q
    5 pent p
    6 hex h
    7 sept s
    8 oct o
    9 enn e

    (ii) The roots are strung together in the order of digits which make up the atomic number and “ium” is added at the end.
    For example let us write the IUPAC name and symbol for the elements having atomic number 112. The roots for 1, 1, 2 are un, un and bi respectively. Hence the name and the symbol respectively are ununbium and Hub.

    Question 76
    CBSEENCH11005447

    What would be the IUPAC name and symbol for the element with atomic number 120?

    Solution

    Element with atomic number 120:
    The roots for 1 is un
    The roots for 2 is bi
    The roots for 0 is nil
    Hence the IUPAC name for the element with atomic number 120 is unbinilium and its symbol is Ubn.

    Question 77
    CBSEENCH11005448

    How many blocks are present in the long form of periodic table? How will you classify the elements of the periodic table into s, p, d and f blocks? 

    Solution

    The long form of periodic table has been divided into four blocks. These are called:
    (i) s-block (ii) p-block
    (iii) d-block , (iv) f-block.
    Division of elements into s, p, d and f-blocks: This division is based upon the name of the orbital which receives the last electron.
    1. s-block elements. The elements in which the last electron enters the s-orbital of the outermost enthalpy level are called .s-block elements. The general electronic configuration of s-block elements is ns1-2 where n stands for the outermost shell. In s-block, the alkali metals of group 1 and alkaline earth metals of group 2 are included.
    2. p-block elements: The elements in which the last electron enters the p-orbital of the outermost enthalpy level are called p-block elements. The general electronic configuration of p-block elements in ns2np1-6 where n stands for the outermost shell. The elements of Groups 13, 14, 15, 16, 17, 18 having 3, 4, 5, 6, 7 and 8 electrons respectively in the outermost enthalpy levels constitute p-block elements.

    3. d-block elements:The elements in which the last electron enters the d-orbital of the penultimate enthalpy level (last but one shell) are called d-block elements. General electronic configuration of d-block elements is (n - 1) d1-10 ns1-2 where n represents the valence shell, d-block elements have three complete series of ten elements in each whereas the fourth series is incomplete.
    Note, Exception is 46Pd whose configuration is 4d10 550:
    4. f-block elements:The elements in which the last electron enters the f-orbital of the anti-penultimate (third to the outermost shell) enthalpy level are called f-block elements. The general electronic configuration of f-block elements is (n -2) f1-14 (n-1) d0-1 ns2where n represents the outermost shell. Such type of elements have two series. First is 4f series which is also called lanthanoides series whereas second is 5f series which is also described as actinoide series. These two series are placed separately at the bottom of the periodic table. The elements included in these two series are called inner transition elements.

    Question 78
    CBSEENCH11005449

    What are the advantages of classifying the elements into s, p, d and f-block elements?

    Solution

    These are:
    (i) By classifying the elements into s, p, d and f blocks, we can study their physical and chemical properties in a very organised and systematic manner.
    (ii) By knowing the block of the element, it is very easy to study the general behaviour and important properties of the elements. For example if the element belongs to s-block, it means it must have 1 or 2 electrons in outermost shell. Hence it must have low value of IE, strongly metallic and electropositive character and form cations easily.
    (iii) This classification of elements also explains why properties of d-block elements vary slightly within the same period. This is because inner (n -1) d orbital is being progressively filled up.
    (iv) In f-block elements there is progressively filling of (n - 2) f orbitals and this classification also explains why f-block elements are nearly identical in their chemical properties.

    Question 79
    CBSEENCH11005450

    Write the general outer electronic configuration of s-, p-, d- and f- block elements.

    Solution

    General outer electronic configuration of s-, p-,d-and f- block elements:
    (i) s- Block elements:
    ns1-2 where n = 2 to 7
    (ii) p-Block elements: ns2 np1-6 where n = 2 to 6
    (iii) d-Block elements:
    (n - 1)d1-10 ns0-2 where n = 4 to 7
    (iv) f-block elements:
    (n - 2)f0-14 (n -1)d0-1 ns2 where n = 6 to 7.

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    Question 80
    CBSEENCH11005451

    Which element do you think would have been named by:
    (i)  Lawrence Berkeley Laboratory
    (ii) Seaborg's group?

    Solution

    i) Lawrencium (Lr) with Z = 103  is an actinide which involves the filling of 5f orbital andBerkelium (Bk) with Z = 97  is an actinide which involves the filling of 5f orbital (both of them are f block elements).

    ii) Seaborgium (Sg) with Z = 106 is a d- block element present in 7th period &  6th column of periodic table.

    Question 81
    CBSEENCH11005452

    Why do elements in the same group have similar physical and chemical properties?

    Solution
    Since elements in the same group have similar electronic configuration and therefore have similar physical and chemical properties. 
    Question 82
    CBSEENCH11005453

    In terms of period and group were would you locate the element with Z = 114?

    Solution

    We know that the gaps of atomic numbers in a group are 8, 8, 18, 18, 32. Hence the element which proceeds the element with Z = 114 in the same group must have an atomic number equal to 114 - 32 = 82. This represents lead (Pb) which is present in the 6th period and belongs to group 14 of the p-block. This means the element with Z = 114 must also belong to group 14 (second p-block element) of 7th period. Thus the location of the element with Z = 114 in the periodic table is:
    period = 7th, Block = p-Block, Group = 14.

    Question 83
    CBSEENCH11005454

    Distinguish between s and p-block elements.

    Solution
    The important points of distinctions are:
    s-block elements p-block elements
    1. General electronic configuration is ns1-2 1. General electronic configuration:
    ns2 np-1-6 ns2 np -1-6
    2. There are two groups in s-block 2. There are six groups in p-block.
    3. The elements present in this block are all metals. 3. The elements present in this block are mostly non-metals.
    4. They show fixed oxidation state. For example, elements in group 1 show oxidation state of + 1 while those in group 2 show oxidation state of +2. 4. Most of the members show variable oxidation states i.e. exhibit more than one oxidation state.
    5. The compounds are mostly ionic in nature. 5. The compounds are mostly ionic in nature.
    6. They are good reducing agents. 6. They are generally strong oxidising agents.
    7. Most of the elements impart characteristic colours to the flame. 7. Most of these do not impart colours to the flame.

    Question 84
    CBSEENCH11005455

    Define transition elements. Name different transition series.

    Solution

    The transition elements are those elements which represent a change of state from the highly metallic elements of s-block to the non-metals of p-block i.e. they lie between the elements of s-and p-blocks. There are four transition series each marking the filling of 3d, 4d, 5d and 6d orbitals respectively as:
    (i) First transition series (3d). Scandium (21Sc) to Zinc (30Zn).
    This series lies in the fourth period of the periodic table and consists of 10 elements.
    General electronic configuration is 3d1-104s1-2.
    (ii) Second transition series (4d). Ytterium (39Y) to Cadmium (48Cd).
    This series lies in the fifth period of the periodic table and consists of 10 elements.
    General electronic configuration is 4d1-10 5s0-2
    (iii) Third transition series (5d). Lanthanum (57La), Hafinium 72Hf) to Mercury (80Hg).
    This series lies in the sixth period of the periodic table and consists of 10 elements.
    General electronic configuration is 5d1-10 6s1-2.
    (iv) Fourth transition series (6d). Actinium (89Ac), Kurchatovium (104Ku) to atoic number 112 have been synthesised. This series lies in the seventh period of the periodic table and consits of 10 elements.
    Hence d-block elements consist of 40 elements.

    Question 85
    CBSEENCH11005456

    Enlist the important characteristics of transition elements (or d-block elements) ?

    Solution

    The important charateristics of transition elements:
    (i) These elements are divided into four transition series each marking the filling of 3d, 4d, 5d and 6d orbitals respectively.
    (ii) They are hard and brittle metals having high melting and boiling points.
    (iii) They are good conductors of heat and electricity.
    (iv) They show variable oxidation states.
    (v) They are generally paramagnetic due to the presence of unpaired electrons.
    (vi) They form coloured ions.
    (vii) These elements have a strong tendency to form complexes.
    (viii) Most of the transition elements (Mn, Co, Ni, Cr) and their compounds are used as catalysts.

    Question 86
    CBSEENCH11005457

    Enlist the number of elements in each block.

    Solution

    The number of elements in each block are:
    (i) s-block: There are 13 elements in s-block. Except the first period (having only one element), each other period of the block has 2 elements.
    (ii) p-block: There are 31 elements. With the exception of first period (having only one element), each other period has got 6 elements.
    (iii) d-block: This block has 40 elements. Each period i.e. fourth, fifth, sixth and seventh of the block has got 10 elements.
    (iv) f-block: This block has 28 elements in the form of 2 horizontal rows of 14 elements each.
    Lanthanoides = 14 elements.
    Actinoides = 14 elements

    Question 87
    CBSEENCH11005458

    Account for the fact that the 4th period has eighteen and not eight elements.

    Solution

    In the long form of the periodic table, each period starts with the filling up of a new principal shell. Clearly, the 4th period will begin with filling of 4th shell i.e. N shell. Thus, fourth period starts with the filling up of 4s orbtial (n = 4). After filling 4s orbital, the filling of 3rd and then 4p takes place. It is so because enthalpy of 3d sub level in between 4s and 4p sub levels. As,
    4s = 2 electrons
    3d = 10 electrons             open square brackets because space 4 straight s space less than space 3 straight d space less than space 4 straight p close square brackets
    4p = 6 electrons
            _______________
            18 electrons
    Therefore, 18 elements (not eight) are present in the 4th period. The 4d and 4f sublevels are higher in enthalpy than 5s and hence are filled up in the next periods.

     

    Question 88
    CBSEENCH11005459

    Lanthanoides and actinodes are placed in separate rows at the bottom of the periodic table. Explain the reason for this arrangement.

    Solution

    In lanthanoids (58Ce to 71Lu) 4f-subshell is being filled and in actinides (90Th to103Lr), 5f-subshell is being filled. All the 14 lanthanoids and all the 14 actinoids have similar properties. They have been placed at the bottom of the periodic table in two separate horizontal rows because of the following reasons:
    (i) to save space,
    (ii) to keep the elements having similar properties in a single column,
    (iii) to avoid undue sidewise expansion of the periodic table.

    Question 89
    CBSEENCH11005460

    How will you predict the position of an element in the periodic table?

    Solution

    The period group or block of an element can be predicted from the electronic configuration of the elements.
    (i) Period of the element corresponds to the principal qauantum number of the valence shell.
    (ii) Block of the element corresponds to the sub-shell which receives the last electron.
    (iii) Group is predicted from the number of electrons in the outermost (n) or penultimate (n -1) shell as follows:
    For s-block elements, group number is equal to the number of valence electrons (ns electrons).
    For p-block elements, group number is equal to 10+ number of valence electrons (ns and np).
    For d-block elements, group number is equal to the sum of the numbers of (n - 1)d and ns electrons.
    For f-block elements group number is 3.
     

    Question 90
    CBSEENCH11005461

    Name the block to which the elements with following valence shell electronic configurations belong:
    (i) 3s2 3p5       
    (ii) 3d
    104s2
    (iii) 3p64s2       
    (iv) 6s2 4f0
    (v) 4s1 3d5

    Solution

    (i) 3s2 3p5 : p-block
    (ii) 3d10 4s2 : d- block
    (iii) 3p6 4s2 : s-block
    (iv) 6s2 4f0 : s-block
    (v) 4s1 3d5 : d-block

    Question 91
    CBSEENCH11005462

    Assign the position of the element having outer electronic configuration:
    (i) ns2np4 for n = 3
    (ii) (n-1)d2ns2 for n = 4, and
    (iii) (n-2) f7 (n-1)d1ns2 for n = 6, in the periodic table.

    Solution


    (i) The electronic configuration of the element is 3s23p4. Thus the element belongs to p-block, third period ( ∵ n = 3) and group 16.
    (ii) The electronic configuration of the element is 3d2As2. Thus the element belongs to d-block, fourth period (∵ n = 4) and group 4.
    (iii) The electronic configuration of the element is 4f7 5d16s2. Thus the element belongs to f-block, sixth period (n = 6) and group 3.

    Question 92
    CBSEENCH11005463

    Give the atomic number and electronic configuration of the following elements:

    (i) The third alkali metal

    (ii) The second transition element

    (iii) The fourth noble gas

    (iv) The first halogen.

    Solution

    (i) The third alkali metal :The atomic number is 19 and the electronic configuration is, 1s2 2s2 2p6 3s2 3p6 4s1.
    (ii) The second transition element: The atomic number is 22 and the electronic configuration is, 1s2 2s2 2p6 3s2 3p6 4s2 3d2.
    (iii) The fourth noble gas: The atomic number is 36 and electronic configuration is, 1s22s2 2p3s2 3p6 4s2 3d10 4p6.
    (iv) The first halogen: The atomic number is 9 and electronic configuration is, 1s2 2s22p5.

    Question 93
    CBSEENCH11005464

    Elements A, B, C and D have atomic numbers 7, 19, 29 and 36 respectively. Write the electronic configuration and also predict their period, group and block.

    Solution

    (i) A(Atomic number 7):
    Electronic configuration: 1s2 2s2 2p3
    Period = 2nd
    Group = 15
    Block  = p-block
    (ii) B(Atomic number 19):
    Electronic configuration: 1s2 2s2 2p6 3s2 3p6 4s1
    Period = 4th
    Group = one (1)
    Block = s-block

    (iii) C (Atomic number 29):
    Electronic configuration:
    1s2 2s2 2p6 3s2 3p6 4s1 3d10
    Period = 4th
    Group = Eleven (11)
    Block = J-block

    (iv) D (Atomic number 36)
    Electronic configuration:
    1s2 2s2 2p6 3s2 3p6 4s2 3d10 4p6
    Period = 4th
    Group = Eighteen (18)
    Block = p-block

    Question 94
    CBSEENCH11005465

    The elements Z = 107 and Z = 109 have been made recently; element Z = 108 has not been made. Indicate the groups in which you will have the above elements.

    Solution
    The last element of the actinoid series is lawrencium (103Lr) in which 5f subshell is complete.
    Lr presubscript 103 space open square brackets Rn close square brackets to the power of 86 space 5 straight f to the power of 14 space 6 straight d to the power of 1 space 7 straight s squared
    Thereafter, the filling of 6d subshell which started at actinium (89Ac) will continue. Therefore, ten elements i.e. Ac (Z = 89), and elements with atomic number 104 to atomic number 112 should belong to the 4th transition series. In other words, all the three elements with Z = 107, Z = 108 and Z = 109 are d-block elements. Their respective groups in the periodic table will be,
    (i) Element (Z = 107) = open square brackets Rn close square brackets to the power of 86
         5 straight f to the power of 14 space 6 straight d to the power of 5 space 7 straight s squared semicolon space space Group space 7.
    (ii) Element (Z = 108) = open square brackets Rn close square brackets to the power of 86
    5 straight f to the power of 14 space 6 straight d to the power of 6 space 7 straight s squared semicolon space space Group space 8
    (iii) Element (Z = 109) = open square brackets Rn close square brackets to the power of 86
     5 straight f to the power of 14 space 6 straight d to the power of 7 space 7 straight s squared semicolon space space Group space 9
    Question 95
    CBSEENCH11005466

    An element belongs to third period of the p-block. It has four valence electrons. Predict its group. How many unpaired electrons does it has?

    Solution

    The element belonging to 3rd period and with four valence electrons:
    (i) belongs to Group 14
    (ii) is silicon (Si) with atomic number 14.
    (iii) The electronic configuration of 14Si is
    1 straight s squared 2 straight s squared 2 straight p to the power of 6 3 straight s squared 3 straight p subscript straight x superscript 1 3 straight p subscript straight y superscript 1
    therefore space space Number space of space unpaired space electrons space equals space 2.

    Question 96
    CBSEENCH11005467

    The elements with Z = 117 and 120 have not yet been discovered. In which family/group would you place these elements. Also give the electronic configuration in each case.

    Solution

    (i) The electronic configuration of element with Z = 117 can be represented as
    open square brackets straight R subscript straight n close square brackets space 4 straight f to the power of 14 space 5 straight d to the power of 10 space 7 straight s squared space 7 straight p to the power of 5
    This element belongs to Group 17 (halogen family).
    (ii) The electronic configuration of element with Z = 120 can be represented as
    open square brackets Uuo close square brackets 8 straight s squared
    This element belongs to Group 2 (alkaline earth metals).

    Question 97
    CBSEENCH11005468

    What are metals, non-metals and metalloids?

    Solution

    Metals: Metals are generally solids at room temperature (mercury is an exception), lustrous, malleable, ductile, high melting and boiling points and good conductors of heat and electricity. Metals comprise more than 75% of all known elements and appear on the left side of the perodic table. For example iron, copper, aluminium, lead etc. The elements become more metallic as we go down a group.
    Non-metals: Non-metals are usually solids or gases at room temperature, non-lustrous, low melting and boiling points and poor conductors of heat and electricity. They are located at the top right hand side of the periodic table. For example oxygen, hydrogen, carbon, sulphur, etc.

    The non-metallic character increases as we go from left to right across the periodic table.

    Metalloids or semimetals: These are the elements which show common characteristic of metals and non-metals. For example, germanium, silicon, arsenic, antimony and tellurium.

    Question 98
    CBSEENCH11005469

    What are the major differences between metals and non-metals?

    Solution
    Metal Non-metal
    Metals have a strong tendency to lose electrons to form cations. Thus, metal is strong reducing agent. Non-metals have a strong tendency to accept electrons to form the anion. Thus, non-metal are the strong oxidising agent.
    Metal have low ionization enthalpies. Non-metal have high ionization enthalpies.
    Less negative electron gain enthalpies. High negative electron gain enthalpies.
    Metal have low electronegativity. Non-metal have high electronegativity.
    Metal have less negative electron gain enthalpies. Non-metal high negative electron gain enthalpies.
    Metal forms a basic oxide and ionic compounds. Non-metal forms acidic oxides and covalent compounds.
    Question 99
    CBSEENCH11005470

    Arrange the following elements in the increasing order of their metallic character: Si, Be, Mg, Na, P.

    Solution
    Metallic character increases down a group and decreases along a period as we move from left to right. Hence the order of increasing metallic character is P < Si < Be < Mg < Na.
    Question 100
    CBSEENCH11005471

    What are periodic properties of the elements? Name a few important periodic properties.

    Solution

    Periodic properties are those properties which get repeated after a regular interval in a regular manner. The important periodic properties are:
    1. Atomic and ionic radius
    2. Ionization enthalpy
    3. Electron gain enthalpy
    4. Electronegativity
    5. Melting points
    6. Density
    7. Valency.

    Question 101
    CBSEENCH11005472

    Define atomic size or atomic radius of an element. Why is it not possible to determine its absolute value?

    Solution

    Assuming atom to be spherical, the distance between the nucleus and the outermost orbit of the atom is called the atomic radius of the element. However, it can not be determined precisely due to the following reasons:
    (i) Atom is too small to be isolated.
    (ii) According to the probability picture of electrons, an atom does not have a well-defined boundary.
    (iii) The probability distribution of an atom is also affected by the presence of other atoms in its neighbourhood.
    (iv)The size of the atom also changes from one bonding state to another.
    Thus, it is not possible to determine the absolute value of atomic radius practically. In other words, atomic radius has no meaning.

    Question 102
    CBSEENCH11005473

    What do you mean by covalent radius?

    Solution
    The covalent radius (or covalent atomic radius) is defined as half of the distance between the nuclei of two like atoms forming a single covalent bond.
    rcov = 1/2 [Internuclear distance between two bonded atoms]
    For example, for hydrogen molecule, the internuclear distance between two hydrogen atoms is 74 pm.

    where 198 pm = Internuclear distance between two chlorine atoms (Cl—Cl) in a molecule of Cl2.
    In case of heteronuclear molecules (containing different atoms) such as HCl, the covalent radius may be defined as: “The difference of the distance between the nuclei of two bonded atoms and covalent radius of one of the atoms in a heteronuclear molecule.”
    Covalent radius = [Internuclear distance of two bonded atoms] -[Covalent radius of one of the two atoms in heteronuclear molecules]
    For example,
    Covalent radius of H-atom (in HCl)
    = Internuclear distance of HCl - covalent radius of Cl atom
    = 136 pm - 99 pm = 37 pm.


    Question 103
    CBSEENCH11005474

    How do atomic radius vary in a period and in a group? How do you explain the variation?

    Solution

    Variation of atomic radii in a group. As we move down a group in the periodic table, the atomic radii go on increasing.
    Reasons :
    (i) Down a group, the nuclear charge goes on increasing. As a result, atomic radii must decrease.

    (ii) As we move down a group, a new enthalpy shell is added at each succeeding element though the number of electrons in the valence shell the remain the same. As a result, atomic radii must increase.
    The effect of the progressive addition of a new shell outweighs the effect of increased nuclear charge. Hence atomic radii of elements increase with the increase in atomic number as we move from top to bottom down a group.


    Variation of atomic radii in a period. As we move from left to right in a period, atomic radii go on decreasing.

    Reason: As we move from left to right in a particular period, the atomic number i.e. nuclear charge increases by one unit in each succeeding element but the corresponding addition of electron takes place in the same enthalpy shell. As a result, electrons are pulled little closer to the nucleus thereby making each individual shell smaller and smaller.

    Question 104
    CBSEENCH11005475

    Why do the noble gas elements have exceptionally high atomic radii?
    Or
    In a period, generally the size of the atom decreases with rise in atomic number, but at the end of each period, atomic size of the noble gas increases abruptly. Why is it so?

    Solution
    In the case of noble gases, we measure Vander Waal’s radii instead of covalent radii since these elements do not form covalent bonds. As by definition, Vander Waal’s radii are larger than covalent radii, the atomic size of an inert gas in a period is much higher than that of preceding halogen.
    Question 105
    CBSEENCH11005476

    Define vander Waal’s radius and compare it with covalent radius.

    Solution

    Vander Waal's radius is defined as one half of the distance between the nuclei of two adjacent atoms belonging to two neighbouring

    molecules of an element in the solid state, vander Waal’s radii are determined by X-ray diffraction method.
    Comparison of covalent radius and vander Waal’s radius.
    vander Waal’s radii are always larger than covalent radii. This is because for the formation of covalent bond, atoms move to come closer to each other due to the overlapping of orbitals.
    On the other hand, vander Waal’s forces operating between atoms are weak, therefore, adjacent atoms belonging to two neighbouring molecules are at relatively larger distances. Clearly, one half of internuclear distance between adjacent atoms belonging to two neighbouring molecules will be more than one half of the distance between the nuclei of two like atoms forming a single covalent bond. Consequently, vander Waals’ radii are always larger than covalent radii. For example, covalent radius of chlorine is 99 pm whereas vander Waal’s radius of chlorine is 180 pm.

    Question 106
    CBSEENCH11005477

    Atomic radii of noble gases are essentially the vander Waal's radii. Comment.

    Solution
    Since noble gases, ordinarily, do not form covalent bonds, therefore, only weak Vander Waal’s forces of attraction; are operative in crystals of noble gases. Therefore, atomic radii of noble gases are essentially the Vander Waal’s radii.

    Question 107
    CBSEENCH11005478

    What is ionic radius?

    Solution
    Ionic radius may be defined as the effective distance from the centre of the nucleus of the ion to which it exerts its influence on its electron cloud, The internuclear distance in any ionic compound is determined from X-ray measurements. This distance is then taken equal to the sum of the radii of positive and negative ions involved. Thus, if we know the radius of one ion, that of the other ion can be calculated. 
    Question 108
    CBSEENCH11005479

    Why is the radius of cation less than that of parent atom?

    Solution
    A cation is formed by the loss of one or more electrons from the neutral gaseous atom. With the removal of one or more electrons from an atom, a magnitude of nuclear charge remains the same as that of the parent atom while the number of electron decreases. As a result, the same nuclear charge acts on the lesser number of electrons. In other words, the effective nuclear charge per electron increases and therefore, electrons are more strongly attracted and pulled towards the nucleus. This causes the decrease in the size of positive ion.
    For example.
    Atomic radius of Na = 154 pm (Nuclear charge = + 11 ; e = 11)
    Ionic radius of Na++ = 95 pm (Nuclear charge = + 11 ; e = 10)
    Question 109
    CBSEENCH11005480

    Why is the radius of anion more than that of the parent atom?

    Solution

    Anion (or negative ion) is formed by the gain of one or more electrons by the gaseous atom. In the anion, nuclear charge is the same as that in parent atom but the number of electrons has increased. As a result, the same nuclear charge acts on relatively larger number of electrons. Thus, effective nuclear charge per electron is decreased which causes the electron cloud to expand. Therefore, anion is always larger than its parent neutral atom.
    For example,
    Atomic radius of Cl = 99 pm (nuclear charge = - 17 ;e = 17)
    Atomic radius of Cl= 131 pm (nuclear charge = 17 ; e = 18)

    Question 110
    CBSEENCH11005481

    What does atomic radius and ionic radius really mean to you?

    Solution

    Atomic radius means the size of the atom i.e. the distance from the centre of the nucleus of the atom to the outermost shell of electrons. In the case of non-metals, atomic radius is called covalent radius. It is defined as one-half of the distance between the centres of the nuclei of two similar atoms bonded by a single covalent bond. In the case of metals, atomic radius is called metallic radius. It is defined as one-half of the distance between the two adjacent atoms in the crystal lattice.
    Ionic radius means the size of the ions. It may be defined as the effective distance from the nucleus of the ion to the point to which it has an influence in the ionic bond. The size of the cation is always smaller as compared to that of the parent atom. The size of the anion is always larger as compared to that of the parent atom.

    Question 111
    CBSEENCH11005482

    Describe the theory associated with the radius of an atom as it gains on electron.

    Solution
    Gain of  the electron: When a neutral atom gains one electron to form an anion, its radius increases. In the anion, the nuclear charge is the same as that of the parent atom but the number of electrons has increased. Since same nuclear charge now attracts the greater number of electrons, therefore, a force of attraction of the nucleus on the electrons of all the shells decreases i.e. effective nuclear charge per electron decreases in the anion and hence electron cloud expands. The relative sizes of the chlorine atom and chloride ion have been shown.


    Question 112
    CBSEENCH11005483

    Describe the theory associated with the radius of an atom as it loses on electron.

    Solution
    Loss of electron: When a neutral atom loses one electron to form a cation, its radius decreases. In the cation, the nuclear charge remains the same as that of the parent atom but the number of electrons becomes less. Since the same nuclear charge now attracts the lesser number of electrons, therefore, the force of attraction of the nucleus on the electrons of all the shells increases. i.e. effective nuclear charge per electron increases in the cation and hence the size of the cation decreases. Thus




    Question 113
    CBSEENCH11005484

    What do you understand by iso-electronic ions? In iso-electronic ions, the value of ionic radius decreases with the increase in atomic number. Explain.

    Solution

    Isoelectronic ions are those which have a same number of electrons and hence same electronic configuration but the different magnitude of nuclear charge. A set of species (cations or anions) having the same number of electrons is known as iso-electronic series, e.g. N3-, O2-, F-, Na+, Mg+, Al3+ are isoelectronic species.
    Within the series of isoelectronic ions, as the atomic number or nuclear charge increases, the attractive force between the nucleus and a same number of electrons also increases. This results in the decrease of ionic radius. 

    For example,


     Ion                          =N3-    O2-    F-     Na+    Mg+       Al3+


    No. of electrons   =  10      10     10      10       10      10


    At. No                      =  7        8      9       11       12      13

    Since, the size of the iso-electronic ion decreases with the increase in atomic number, the order of decreasing size is

    N3- > O2- > F- > Na+ > Mg2+ > Al3+

    Question 114
    CBSEENCH11005485

    What do you understand by isoelectronic species? Name a species that will be isoelectronic with each of the following atoms or ions.
    (i) F-  (ii) Ar    (iii) Mg2+   (iv) Rb+ 

    Solution

    Ions of different elements which have the same number of electrons but different magnitude of the nuclear charge are called isoelectronic ions.


    (i) Since F- has 10 electrons, therefore N3- (7 + 3), O2- (8 + 2), Ne (10 + 0),  Na+ (11 - 1), Mg2+ (12 - 2): Al3+ (13 - 3) are isoelectronic, as each one of these contains 10 electrons.


    (ii) As Ar has 18 electrons, therefore P3- (15 + 3), S2- (16 + 2), Cl- (17 + 1), K+ (19 - 1), Ca2+ (20 - 2) are isoelectronic with it as cach one of each contains 18 electrons.


    (iii) As Mg2+ has 10 electrons, therefore N3+ F , Ne, Na+, Al3+ are isoelectric with it as each one of these contains 10 electrons.


    (iv) As Rb+ has 36 electron, therefore Br- (35 + 1), Kr (36 + 0) and Sr2+ (38 -2) are isoelectronic with it as each one of these contains 36 electrons.

    Question 115
    CBSEENCH11005486

    Consider the following species : N3-, O2-, F-, Na+, Mg2+ and Al3+

    (a) What is common in them?

    (b) Arrange them in the order of increasing ionic radii. 

    Solution

    (a) Each one of these ions contains 10 electrons and hence all are isoelectronic ions.
    (b The ionic radii of isoelectronic species increase with a decrease in the magnitudes of nuclear charge. The arrangement of the given species in order of their increasing nuclear charge is as follows:

    N3-< O-2<F-<Na+<Mg2+<Al3+

    Question 116
    CBSEENCH11005487

    For each of the following pairs, state which one is larger in size and why?
    (i) Na, Na+     (ii) Br, Br-    (iii) Br, I

    Solution

    (i) Na, Na+ pair. The size of Na atom is greater because the cation i.e. Na+ is formed by the loss of one electron from the atom.
    (ii) Br, Br- pair. The size of Br- ion is greater as anion is formed by the gain of one electron by bromine atom.
    (iii) Br, I pair. The size of 1 is greater because it comes after Br in the halogen family and atomic size increases down the group.

    Question 117
    CBSEENCH11005488

    Give the formula of a species which will be isoelectronic with the following atoms or ions:
    (i) Na2+ 
    (ii) Cl

    (iii) Ca2+ 
    (iv) Rb
    +.

    Solution

    (i) Na+: Mg2+ (10 electrons)
    (ii) Cl-: Ar  (18 electrons)
    (iii) Ca2+: K(18 electrons)
    (iv) Rb+: Kr (36 electrons)

     
    Question 118
    CBSEENCH11005489

    Which of the following species will have the largest and the smallest size?
    Mg, Mg2+, Al, Al3+.

    Solution

    Atomic radii decrease across a period, cations are smaller than their parent atoms. Among isoelectronic ions, the one with the larger positive nuclear charge will have a smaller radius.
    Hence the largest species is Mg and the smallest one is Al3+ .

    Question 119
    CBSEENCH11005490

    Arrange the following ions in order of increasing size:
    Be2+, Cl-, S2-, Na+, Mg2+, Br-.

    Solution
    Order of increasing size,
    Be2+ < Mg2+ < Na+ < Cl- < S2- < Br-.

    Sponsor Area

    Question 120
    CBSEENCH11005491

    In each of the following pairs,  which species has a larger size? Explain.
    (i) K or K+ 
    (ii) Br or Br
    - 
    (iii) O
    2- or F
    (iv) Li+ or Na+ 
    (v) P or As
    (vi) Na
    + or Mg2+

    Solution

    (i) K or K+: K atom has large size than K+ ion because the size of cation (or radius) is always smaller as compared to the parent atom.
    (ii) Br or Br- : Br- ion has large size because it is an anion and it has always large size as compared to the parent atom.
    (iii) O2- or F-: Both the anions are isoelectronic species. Since the nuclear charge of O2-ion is small, its size is, therefore, more than that of F-ion.
    (iv) Li+ or Na+: Both the cations are formed from the atoms which are present in group 1. As Na atom is larger than Li atom, therefore, Naion is also bigger in size than Li+ion.
    (v) P or As: Both these atoms belong to group 15 of the periodic table. Since As is placed after P, therefore its size is large because atomic size increases down the group.
    (vi) Na+ or Mg2+: Both the cations are isoelectronic in nature with 10 electrons each. Since the nuclear charge of Na+ ion is more than that of Mg2+ ion, its size is, therefore, large.

    Question 121
    CBSEENCH11005492

    Define ionsiation enthalpy. State its units. 

    Solution
    Ionisation enthalpy is the minimum amount of enthalpy required to remove most loosely held electron from an isolated gaseous atom in its ground state.
    Atom (g)  + Ionisation enthalpy rightwards arrow Positive ion (g) + electron (g) Na (g) + IE(496 kJ) rightwards arrow Na+(g) + e-
    It is expressed either in terms of electron volts per atom (eV/atom) or kilo joules per mole of atoms (kJ mol-1). The two units are related as 1eV per atom = 96.3 kJ mol-1
    Question 122
    CBSEENCH11005493

    What are successive ionisation enthalpies? Explain why the second ionisation enthalpy is higher than the first ionisation enthalpy?

    Solution
    Successive ionisation enthalpies : The enthalpy required to remove the first loosely bound electron from gaseous isolated atom is called the first ionisation enthalpy (IE1). The enthalpies required to remove second, third and fourth electrons are called second (IE2), third (IE3) or fourth (IE4) ionisation enthalpies respectively. The amounts of enthalpies required to remove first, second and subsequent electrons from the gaseous atom one after the other are collectively called successive ionisation enthalpies.
    straight M left parenthesis straight g right parenthesis space rightwards arrow with I E subscript 1 on top space straight M to the power of plus space left parenthesis straight g right parenthesis space plus space straight e to the power of minus
straight M to the power of plus left parenthesis straight g right parenthesis space rightwards arrow with I E subscript 2 on top space straight M to the power of 2 plus end exponent left parenthesis straight g right parenthesis space plus space straight e to the power of minus
straight M to the power of 2 plus end exponent left parenthesis straight g right parenthesis space rightwards arrow with I E subscript 3 on top straight M to the power of 3 plus end exponent left parenthesis straight g right parenthesis space plus space straight e to the power of minus
    Theoretically speaking, there are as many ionisation enthalpies for an atom as there are electrons in it. In the above example
    IE3 > IE2 > IE1.
    Second ionisation enthalpy is always greater than the first ionisation enthalpy. After the removal of first electron, the atom changes into monovalent positive ion (M+). In the ion (M+), the number of electrons decreases but the nuclear charge remains the same as the parent atom. As a result the attraction of the nuclear charge (protons) increases over the remaining electrons. Hence, more enthalpy is required to remove the second electron. In other words, the second ionisation enthalpy is greater than the first ionisation enthalpy (IE2 > IE1).
    Similarly, the removal of second electron results in the formation of divalent positive ion (M++) and the attraction between the nucleus and remaining electrons increases further. This accounts for the progressive increase in the value of ionisation enthalpies. For example, for aluminium,
    IE, = 577.4 kJ mol-1, IE2 = 1816 kJ mol-1 and IE3 = 2744 kJ mol-1
     
    Question 123
    CBSEENCH11005494

    Discuss in brief the factors which influence the magnitude of ionsiation enthalpy.

    Solution

    The main factors are:
    (i) Size of the atom: As the size of the atom increases, the distance between the nucleus and outermost electrons increases. So the attraction between the nucleus and the outermost electron decreases and the enthalpy required to remove the electron also decreases. Thus, with an increase in atomic size, the ionisation enthalpy decreases.

    (ii) Nuclear charge: With the increase in nuclear charge, the force of attraction between the nucleus and the valence electrons increases. Consequently, more enthalpy is required to remove a valence electron. Hence ionisation enthalpy increases with the increase in nuclear charge.

    (iii) Screening effect of inner shells: The electron shells present in between the outermost shell and the nucleus act as screens. These shells reduce the attraction of nuclear charge over the outermost electrons. This reduction in force of attraction by the shells present in between the nucleus and valence electrons is called screening or shielding effect. As the number of inner shells increases, the screening effect also increases. As the screening effect increases, the ionisation enthalpy decreases.

    (iv) Penetration effect of electrons: It is well known that s-electrons are more penetrating towards the nucleus than p-electrons and the penetration power decreases in a given shell (same value of n) in the order s >p >d>f.
    Now if the penetration of the electron is more, it will be closer to the nucleus and held firmly. Hence ionisation enthalpy will be high. Thus, for the same shell, it is easier to remove p-electron in comparison to s-electron. Therefore, for the same value of n, ionisation enthalpy decreases in the order s > p > d > f

    (v) Stability of the electronic configuration: More enthalpy is required to remove an electron from an element having configuration in which all the orbitals of the same sublevel are exactly half filled (p3 or d5 : stable configuration) or completely filled (p6or d10: stable configuration). Thus, more stable the electronic arrangement, greater is the ionisation enthalpy.

    डायफ़्राम
    Question 124
    CBSEENCH11005495

    How does ionisation enthalpy vary along a group?

    Solution

    As we move down a group. the ionisation enthalpy goes on decreasing.
    Reason. As we move down the group.
    (i) nuclear charge regularly increases.
    (ii) The size of the atom regularly increases due to the addition of a new shell every time and
    (iii) shielding effect also go on increasing due to the increase in the number of electrons in the inner shells. 
    The effect of the increase in atomic size and shielding effect is much more than the effect of the increase in nuclear charge. As a result, an electron becomes less and less firmly held to the nucleus as we move down the group. Hence there is the gradual decrease in the value of ionisation enthalpy on moving from top to bottom in a group.

    Question 125
    CBSEENCH11005496

    What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?

    Solution

    Within the main group elements, the ionization enthalpy decreases regularly as we move down the group due to the following two factors:
    (i) Atomic size: As we move down the group, the atomic size of the atom regularly increases due to the addition of one new principal energy shell at each succeeding element.
    (ii) Screening effect: With the addition of new shells, the number of inner electron shells, which shield the valence electron increases. In other words, shielding effect or screening effect increases.
    As a result of both these factors, the force of attraction of the nucleus for the valence electrons decreases. Thus, the electron becomes less and less firmly held to the nucleus as we move down the group. Hence there is a regular decrease in the value of ionization enthalpy on moving down the group.

    Question 126
    CBSEENCH11005497

    Why does the first ionsiation enthalpy increase as we go from left to right through a period of the periodic table?
    Or
    How does ionisation enthalpy vary along a period? 

    Solution

    In general, ionisation enthalpy increases as we move from left to right across a period.
    Reason:
    We know that on moving across the period from left to right.


    (i) the nuclear charge increases,
    (ii) progressive addition of electrons occurs in the same enthalpy level and
    (iii) atomic size decreases.


    Thus, due to the gradual increase in nuclear charge and the simultaneous decrease in atomic size, the valence electrons are more and more tightly held by the nucleus. Therefore, more and more enthalpy is needed to remove the electron and hence ionisation enthalpy keeps on increasing.

    Question 127
    CBSEENCH11005498

    Why is ionisation enthalpy of Be more than that of B?

    Solution

    The electronic configuration of Be (Z = 4) and B(Z = 5) are,

    (i) The electronic configuration of Be in which all the occupied orbitals are fully filled is more stable than boron.
    (ii) It is relatively easier to remove electron from 2p subshell in boron than removal of electrons from 2s subshell in Be as 2p orbital is less penetrating than 2s orbital. Therefore, 2p electron of B is less firmly held by the nucleus and so, it is relatively easily removed.

    Question 128
    CBSEENCH11005499

    Why ionsiation enthalpy of N is more than that of O even though oxygen has higher nuclear charge than nitrogen?

    Solution

    This is due to the fact that electronic configuration of nitrogen in which all 2p orbitals,

    are half filled, is much more stable than that of oxygen,

    in which all the 2p orbitals are neither half filled nor fully filled. Therefore. IE1 of N is more than that of oxygen.

    Question 129
    CBSEENCH11005500

    Ionsiation enthalpy of fluorine is greater than that of oxygen. Explain.

    Solution
    Ionsiation enthalpy of fluorine is greater than oxygen, it is primarily due to:
    (i) the increase in nuclear charge and,
    (ii) decrease in atomic size of fluorine. Due to the combined effect of these two factors, the force of attraction between the nucleus and electron to be removed increases. Hence ionisation enthalpy increases.
    Question 130
    CBSEENCH11005501

    Why do noble gases have very high value of ionisation enthalpy?

    Solution
    The very high value of the ionisation enthalpies of noble gases are due to their high values of effective nuclear charge and stable electronic configuration (ns2 np6) i.e. fully filled s and p-orbitals).
    Question 131
    CBSEENCH11005502

    Nitrogen and phosphorus have high value of ionisation enthalpies. Explain.

    Solution
    The electronic configurations of N (Z = 7) and P(Z = 15) are:


    Both these elements have the configurations in which all the orbitals of the same sub-level are exactly half-filled. Such an arrangement gives extra stability to the atom. As a result, the removal of electron becomes difficult and hence they have high value of ionisation enthalpies.
    Question 132
    CBSEENCH11005503

    Be and Mg have high value of ionisation enthalpies. Explain.

    Solution
    The electronic configurations of Be (Z=4) and Mg (Z=12) are:

    Both these elements have stable electronic arrangement in which all the occupied orbitals are completely filled. Hence they have high value of ionisation enthalpies.
    Question 133
    CBSEENCH11005504

    Would you expect the first ionization enthalpies for two isotopes of the same element to be the same or different? Justify your answer. 

    Solution

    The ionization enthalpy of an atom depends on the number of electrons and protons (Nuclear charge) of that atom. Since the two isotopes of the same element have the same atomic number i.e. have the same number of electrons, same nuclear charge, same size and therefore they are expected to have same ionization enthalpy.

    Question 134
    CBSEENCH11005505

    Which of the following pairs of elements would you expect to have lower first ionisation enthalpy? Explain your answer:

    (a) Cl or F

    (b) Cl or S

    (c) K or Ar

    (d) Kr or Xe.

    Solution

    (a) Cl or F: Chlorine is expected to have lower IE1 as compared to fluorine since ionisation enthalpy decreases down a group. Chlorine comes after fluorine in group 17 of the periodic table.
    (b) Cl or S: Sulphur is expected to have lower IE1 as compared to chlorine since ionisation enthalpy increases along a period. Chlorine having smaller size than sulphur comes after sulphur in the third period.
    (c) K or Ar: Potassium is expected to have lesser IE1 than argon (noble gas-element) because the configuration of argon is very much symmetrical stable. Hence argon has a very high IE, value.
    (d) Kr or Xe: Both the elements belong to group 18 (noble gas family). As we know ionisation enthalpy decreases down the group, thus xenon comes after krypton has lesser IE1 value.

    Question 135
    CBSEENCH11005506

    Among the elements of second period i.e. from Li to Ne, pick out the elements:

    (a) with the highest first ionisation enthalpy
    (b) with the highest electronegativity
    (c) with the largest atomic radius
    (d) which is the most reactive non-metal
    (e) which is the most reactive metal ?

    Solution

    The elements of the second period are:
    Li, Be, B, C, N, O, F, Ne
    (a) The element with highest first ionisation enthalpy: Neon (Ne).
    (b) The element with highest electronegativity: Fluorine (F).
    (c) The element with largest atomic radius: Lithium (Li).
    (d) The most reactive non-metal: Fluorine (F).
    (e) The most reactive metal: Lithium (Li).

     
    Question 137
    CBSEENCH11005508

    The IE1,  of carbon, is more than that of boron while its IE2 value is smaller. Explain.

    Solution

    IE1 of carbon and boron:
    It is primarily due to:
    (i) the increase in nuclear charge.
    (ii) The decrease in the atomic size of carbon. Due to the combined effect of these two factors, the force of attraction between the nucleus and the electron to be removed increases. Hence IE, value of carbon is more than that of boron.


    IE2 of carbon and boron:
    The electronic configuration of B+ and C+ ions are:



    The electronic configuration of B+ ion is more symmetrical than that of C+ ion since in the former case, both the orbitals are filled. Therefore, the configuration of B+ ion is more stable and its IE2 value is more than that of Cion.

    Question 138
    CBSEENCH11005509

    Arrange the following in order of increasing ionisation enthalpies and assign reason:

    (i) K+, Ar, Cl-    
    (ii) Fe, Fe2+, Fe3+

    Solution

    (i) K+, Ar, Cl-:
    The correct order is Cl- < Ar < K+.
    All the three species are isoelectronic, each containing 18 electrons. The nuclear charge increases in the order Cl- (+17) < Ar (+18) < K(+19). As the nuclear charge increases, the attractive force between the nucleus and  the same number of electrons also increases. This results in the decrease of size and hence ionisation enthalpy increases.
    (ii) Fe, Fe2+, Fe3+:
    The correct order is Fe < Fe2+ < Fe3+.
    In the formation of the cation from the atom, electrons are removed. As a result, the effective nuclear charge increases and so is the ionisation enthalpy. Greater the magnitude of positive charge on the ion, more will be its ionisation enthalpy. Hence the order is justified.

    Question 139
    CBSEENCH11005510

    How would you explain the fact that the first ionization enthalpy of sodium is lower than that of magnesium but its second ionization enthalpy is higher than that of magnesium? 

    Solution

    The electronic configurations of Na and Mg are:
    11Na ; 1s22s22p63s1.
    12Mg : 1s22s22p63s2
    The first ionization enthalpy of sodium is lower than that of magnesium because the nuclear charge of Na (+11) is lower than that of Mg (+12) and size of Na is lower than Mg. After the loss of first electron, the electronic configuration of Na+ is 1s22s12p6 (has noble gas configuration) and Mgis 1s22s22p63s1. Hence removal of the second electron from Na+ is very difficult as compared to Mg(electron to be removed from 3s). Thus second ionisation enthalpy of sodium is higher than that of magnesium.

    Question 140
    CBSEENCH11005511

    The first ionization enthalpy values (in kJ mol-1) of group 13 elements are:
    B       Al      Ga      In        Tl
    801    577   579    558    589
    How would you explain this deviation from the general trend?

    Solution
    The decrease in first ionization enthalpy (∆IH1) value from B to Al is due to the bigger size of Al atom. But the element Ga has ten 3d electrons present in the 3d sub-shell which does not screen as much as is done by s and p electrons. Therefore there is an unexpected increase in the magnitude of effective nuclear charge resulting in increased ∆iH1, value. The same explanation can be given in moving from In to TT. The latter has fourteen 4/ electrons with very poor shielding effect. This also results in an unexpected increase in the effective nuclear charge resulting in inflated ∆IH1 value.
    Question 141
    CBSEENCH11005512

    Explain why difference in IE, and IE2 values of sodium is more than that of magnesium.

    Solution

    The (IE1) and IE2 are as follows:
    Na (IE1) = 496 kJ mol-1
    Na (IE2) = 4562 kJ mol-1
    Mg (IE1 = 737 kJ mol-1
    Mg (IE2) = 1450 kJ mol-1
    In the case of sodium, the second electron is to be removed from Na+ ion which has a stable electronic configuration.
    Na+ (Z = 11) = 1s22s22p6 (all occupied orbitals are fully filled, noble gas configuration).
    In the case of magnesium, the second electron is to be removed from Mg+ ion which is yet to achieve a noble gas configuration.
    Mg+ (Z = 12) = 1s22s2 2p63s1
    Therefore, the removal of the second electron from sodium ion is more difficult than from magnesium atom. As a result, the difference is more in sodium than in magnesium atom.

    Question 142
    CBSEENCH11005513

    Energy of an electron in the ground state of the hydrogen atom is -2.18 X 10-18J. Calculate the ionization enthalpy of atomic hydrogen in terms of J mol-1.

    Solution

    The enthalpy required removing an electron in the ground state of hydrogen atom
    = - (its energy in the ground state)
    = -(-2.18 x 10-18 J)
    = 2.18 x 10-18 J
    ∴ Ionization enthalpy per mole of hydrogen atoms
    equals space 2.18 space cross times space 10 to the power of negative 18 end exponent space cross times space 6.022 space cross times space 10 to the power of 23 kJ
equals space 13.127 space straight x 10 to the power of 5 space straight J space mol to the power of negative 1 end exponent

    Question 143
    CBSEENCH11005514

    Would you expect the second electron gain enthalpy of O as positive, more negative or less negative than the first? Justify your answer.
    Or
    What are successive electron gain enthalpies? Explain why second or higher electron gain enthalpy will have positive value.

    Solution

    When a single electron is added to a gaseous atom, it forms a monovalent anion and the enthalpy released is called the first electron gain enthalpy. However, when we add the second electron to the negatively charged ion, this addition of electron will be strongly opposed due to the coulomb forces of repulsion. Hence enthalpy must be supplied for the addition of the second electron. For these reasons, the value of second electron affinity or higher electron affinity will have a positive value. For example

    First electron gain enthalpy:

    straight O left parenthesis straight g right parenthesis space plus space straight e to the power of minus space space space rightwards arrow space space space straight O to the power of minus left parenthesis straight g right parenthesis semicolon space space
space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis increment subscript eg straight H right parenthesis subscript 1 space equals space 141 space straight k to the power of 1 space mol to the power of negative 1 end exponent space
    Second electron gain enthalpy:
    straight O to the power of minus left parenthesis straight g right parenthesis space plus space straight e to the power of minus space space rightwards arrow space space space straight O to the power of 2 minus end exponent left parenthesis straight g right parenthesis semicolon
space space space space space space space space space space space space space space space space space space space space space space space space space space space left parenthesis increment subscript eg straight H right parenthesis subscript 2 space equals space plus 780 space kJ space mol to the power of negative 1 end exponent

    Question 144
    CBSEENCH11005515

    Discuss in brief the factors which influence the magnitude of electron gain enthalpy.

    Solution

    The main factors are:
    1. Nuclear charge. Greater the magnitude of the nuclear charge, greater will be the attraction for the incoming electron. As a result, electron gain enthalpy becomes more negative.
    2. Size of the atom. Larger the size of the atom, more will be the distance between the nucleus and the additional electron and this results in lesser attraction. Consequently, electron gain enthalpy becomes less negative.
    3. Electronic configuration. An atom with stable configuration of half filled or completely filled orbitals has no tendency to gain an electron. Such atoms have zero or almost zero electron gain enthalpy.

    Question 145
    CBSEENCH11005516

    How does electron gain enthalpy vary along a group?

    Solution
    The electron gain enthalpy of elements become less negative as we move from top to bottom down a group. This is because on moving down a group, there is the simultaneous increase in atomic size and nuclear charge. However, the effect of the increase in size is greater than the increase in nuclear charge. As a result, the incoming electron feels lesser attraction by the large atom. Consequently, electron gain enthalpy becomes less negative down a group.
    Question 146
    CBSEENCH11005517

    How does electron gain enthalpy vary along a period?

    Solution
    The electron gain enthalpies of elements become more negative along a period. This is because as we move across a period, the atomic size decreases and effective nuclear charge increases. Both these factors result in greater attraction for the incoming electron. Hence electron gain enthalpy increases in a period from left to right.
    Question 147
    CBSEENCH11005518

    Why do halogens have very higher value of electron gain enthalpy?

    Solution
    This is due to the fact that halogens have the general electronic configuration of ns2np5 and have only one electron less than the stable noble gas (ns2np6) configuration. In order to acquire the noble gas configuration, halogens have a very strong tendency to accept an additional electron and their electron gain enthalpies become more negative.
    Question 148
    CBSEENCH11005519

    Why do noble gases have large positive electron gain enthalpies?

    Solution
    Noble gases have large positive electron gain enthalpies because the electron has to enter the next-higher principal quantum level leading to a very unstable electronic configuration.
    Question 149
    CBSEENCH11005520

    The electron gain enthalpy of fluorine has less negative value than that of chlorine while it is expected to be more. Explain.

    Solution
    The electronic configuration of F and Cl can be represented as:
    straight F presubscript 9 space space space space space space space space space space space space space space space space space space 1 straight s squared 2 straight s squared 2 straight p to the power of 5
Cl presubscript 17 space space space space space space space space space space space space space space space space 1 straight s squared 2 straight s squared 2 straight p to the power of 6 3 straight s squared 3 straight p to the power of 5
    Within a group, electron gain enthalpy becomes less negative down a group. However, adding an electron to the 2p orbital in fluorine leads to greater repulsion than adding an electron to the larger 3p orbital of chlorine. Hence chlorine has more negative electron gain enthalpy than that of fluorine.
    Question 150
    CBSEENCH11005521

    Explain which of the following will have the most negative electron gain enthalpy and which the least negative: P, S, CI, F.

    Solution
    The electronic configuration of P, S, E and CI can be represented as,
    straight P presubscript 15 space space space space space space space space space space space space space space space space space space 1 straight s squared 2 straight s squared 2 straight p to the power of 6 3 straight s squared 3 straight p cubed
straight S presubscript 16 space space space space space space space space space space space space space space space space space space 1 straight s squared 2 straight s squared 2 straight p to the power of 6 3 straight s squared 3 straight p to the power of 4
straight F presubscript 9 space space space space space space space space space space space space space space space space space space space space space 1 straight s squared space 2 straight s squared space 2 straight p to the power of 5
Cl presubscript 17 space space space space space space space space space space space space space space space space space space 1 straight s squared 2 straight s squared 2 straight p to the power of 6 3 straight s squared 3 straight p to the power of 5
    Electron gain enthalpy generally becomes more negative across a period as we move from left to right. Within a group, electron gain enthalpy becomes less negative down a group. However, adding an electron to the 2p orbital leads to greater repulsion than adding an electron to the larger 3p orbital. Hence the element with most negative electron gain enthalpy is chlorine ; the one with the least negative electron gain enthalpy is phosphorus.
    Question 151
    CBSEENCH11005522

    Which of the following elements has more negative electron gain enthalpy?
    (i) [Ne] 3s2sp3
    (ii) [Ne] 3s23p4
    (iii) [Ne] 3s2 3p5

    Solution
    The element (chlorine) corresponding to the electronic configuration (iii) will have more negative electron gain enthalpy. This is due to the reason that it contains one electron less than the stable noble gas configuration i.e. [Ne] 3s2 3pand hence, has a strong tendency to accept one electron to acquire the stable gas configuration.
    Question 152
    CBSEENCH11005523

    Which of the following pairs of elements would have a more negative electron gain enthalpy?
    (i) O or F    (ii) F or Cl

    Solution

    (i) O or F. Electron gain enthalpy of F is much more negative than that of O. This is because as we from O to F, the atomic size decreases and nuclear charge increases, thereby increase the attraction of the nucleus for the incoming electron and hence electron gain enthalpy becomes more negative. Moreover gain of one electron by F gives F- ion which has stable inert gas configuration (1s22s2, p6).

    (ii) F or Cl: In general, the electron gain enthalpy becomes less negative on moving down the group. But the electron gain enthalpy of chlorine is more negative than that of flourine. This is due to the small size of F atom. Due to its small size, the electron-electron repulsion in the relatively compact 2p-subshell are comparatively larger than those present in the bigger 3p-subshell of Cl. Therefore, chlorine has higher tendency to accept an addition electron than F.

    Question 153
    CBSEENCH11005524

    What is the significance of the terms - 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy?

    Solution

    (a) The significance of term ‘isolated gaseous atom:’ When an atom is isolated in the gaseous state, its electron releasing tendency and electron accepting tendency are both absolute in nature. It means that their values of ionization enthalpy and of electron gain enthalpy are not influenced by the presence of the other atoms. It is not possible to express these when the atoms are in the liquid or solid state.
    (b) The significance of ground state: As we know that both ionization enthalpy and electron gain enthalpy are related to the force binding the electrons with the nucleus of the atom, therefore these are related to the energy states of the atoms whether ground states or excited states.

    Question 154
    CBSEENCH11005525

    Explain, which of the following pairs of elements would have a more negative electron gain enthalpy:
    (i) N or O     (ii) O or S

    Solution

    (i) O has more negative electron gain enthalpy than N. This is because O atom has higher nuclear charge and lower atomic size than N.
    (ii) S has more negative electron gain enthalpy than O. The size of O is much smaller than that of S., As a result, the electron-electron repulsions in the smaller 2p-subshell of O are comparatively larger than those present in the bigger 3p-subshell of S. Therefore S has a higher tendency to accept an addition electron than O.

    Question 155
    CBSEENCH11005526

    Using the periodic table predicts the formulae of compounds which might be formed by the following pairs of elements:

    (a) Silicon and bromine.

    (b) Aluminium and sulphur.

    Solution

    (a) Silicon belongs to group 14 with a valence of 4. Bromine being halogen belongs to group 17 with a valence of 1. Hence the formula of the compound formed would be SiBr4.
    (b) Aluminium is group 13 element with a valence of 3. Sulphur belongs to group 16 elements with a valence of 1. Hence the formula of the compound formed would be Al2S3.

    Question 157
    CBSEENCH11005528

    Differentiate between ionisation enthalpy and electron gain enthalpy ?

    Solution
    Ionisation enthalpy Electron gain enthalpy
    1. It is the amount of enthalpy required to remove the most loosely held electron from an isolated gaseous atom of an element in the ground state. 1. It is enthalpy change when an electron is added to the neutral gaseous atom of an element to form the monovalent negative ion.
    2. When the electron is removed, cation is formed. 2. When the electron is added, anion is formed.
    3. Noble gases have a  high value of ionisation enthalpy. 3. Noble gases have a  positive value of electron gain enthalpy.
    Question 158
    CBSEENCH11005529

    Write the electronic configuration of the elements with Z = 9, 11 and 36. Predict the following from the configuration:

    (i) Which of them has the large negative value of electron gain enthalpy?

    (ii) Which of them has the lowest ionisation enthalpy?

    (iii) Which has positive electron gain enthalpy?

    Solution
    The electronic configuration of the elements are: 
    straight Z left parenthesis 9 right parenthesis space equals space 1 straight s squared space 2 straight s squared space 2 straight p subscript straight x superscript 2 space 2 straight p subscript straight y superscript 2 space space 2 straight p to the power of 1
straight Z left parenthesis 11 right parenthesis space equals space 1 straight s squared 2 straight s squared 2 straight p subscript straight x superscript 2 space 2 straight p subscript straight y superscript 2 space 2 straight p squared space 3 straight s to the power of 1
straight Z left parenthesis 36 right parenthesis space equals space 1 straight s squared space 2 straight s squared space 2 straight p to the power of 6 space 3 straight s squared space 3 straight p to the power of 6 space 4 straight s squared space 3 straight d to the power of 10 space 4 straight p to the power of 6
    (i) The element with Z=9 i.e. fluorine has the large negative electron gain enthalpy.
    (ii) The element with (z=11) i.e. sodium has the lowest ionisation enthalpy.
    (iii) The element with Z=36 i.e krypton (noble gas) has positive electron gain enthalpy.
    Question 159
    CBSEENCH11005530

    Use the periodic table to answer the following questions:

    (a) Identify an element with five electrons in the outer subshell.

    (b) Identify an element that would tend to lose two electrons.

    (c) Identify an element that would tend to gain two electrons.

    (d) Identify the group having metal, non-metal, liquid as well as gas at the room temperature. 

    Solution

     (a) The general electronic configuration of the elements having five electrons in the outermost subshell is ns2np3. This electronic configuration is characteristic of elements of group 15. Hence element belongs to nitrogen family (group 15).
    (b) The elements which have a tendency to lose two electrons must have two electrons in the valence shell. Therefore their general electronic configuration should be ns2. This electronic configuration is characteristic of elements of group 2. Hence element belongs to alkaline earth family (Group 2).
    (c) The elements which have a tendency to gain two electrons must have six electrons in the valence shell. Therefore their general electronic configuration is ns2np4. This electronic configuration is characteristic of group 16 elements. Hence element belongs to oxygen family (Group 16).
    (d) A metal which is liquid at room temperature is mercury and belongs to group 12. A non-metal which is gas at room temperature is nitrogen (Group 15), oxygen (Group 16), flourine and chlorine (Group 17) and inert gases (Group 18). A non-metal which is liquid at room temperature is bromine (Group 17).

    Question 160
    CBSEENCH11005531

    The increasing order of reactivity among group 1 elements is Li < Na < K < Rb is F > CI > Br > I. Explain.

    Solution

    In group 1 elements (alkali metals) the reactivity of the metals is mainly due to the electron releasing tendency of their atoms, which is related to ionisation enthalpy. As we know, the ionisation enthalpy decreases down the group, the reactivity of alkali metals increases.
    In group 17 elements (halogens), the reactivity is mainly due to electron gain tendency of the members of the family. Thus reactivity is linked with electronegativity and electron gain enthalpy. As we know, both of these decrease down the group, the reactivity therefore decreases.

    Question 161
    CBSEENCH11005532

    A, B, C are three elements, B is an inert gas element other than helium; with this information, complete the table given below:

    Element Atomic No. No. of electrons in the valence shell Group to which element belongs
    A Z - 1    
    B Z    
    C Z+1    

    Also, explain the following:
    (i) Electron gain enthalpy of element A has large negative value than that of C.

    (ii) Ionisation enthalpy of element C is less than that of A.

    (iii) Electron gain enthalpy of B is positive.

    Solution

    Let B be an inert gas element neon with atomic number Z=10. Therefore, atomic number of A=9 (halogen) and of C=11 (alkali metal).
    The electronic configuration of elements are:
    straight A presubscript 9 space equals space 1 straight s squared space 2 straight s squared space 2 straight p to the power of 5 space left parenthesis Number space of space electrons space in space the space valence space shell space equals space 7 right parenthesis
straight B presubscript 10 space equals space 1 straight s squared space 2 straight s squared space 2 straight p to the power of 6 space left parenthesis Number space of space electrons space in space the space valence space shell space equals space 8 right parenthesis
straight C presubscript 11 space equals space 1 straight s squared space 2 straight s squared space 2 straight p to the power of 6 space 3 straight s to the power of 1 space left parenthesis Number space of space electrons space in space the space valence space shell space equals space 1 right parenthesis

    Groups:
    Element A belongs to Group 17 (Halogens).
    Element B belongs to Group 18 (Noble gases).
    Element C belongs to Group 11 (Alkali metals).

    (i) Electron gain enthalpy of element A has large negative value than that of C. The element A (fluorine) requires only one electron to have a noble gas configuration. Thus, electron gain enthalpy of A is much more than that of C (sodium).
    (ii) Ionisation enthalpy of C is less than that of A. The atomic size of A is smaller (atomic radius = 71 pm) than that of C (atomic radius = 154 pm).Thus, it is easier to remove an electron from C as compared to A. Hence ionisation enthalpy of C is less than that of B.
    (iii) Electron gain enthalpy of B is positive. B being a noble gas element has a symmetrical electronic configuration ( 1s-2s22p6). Added electron has to enter the next higher principal quantum level leading to the positive value of electron gain enthalpy.

    Question 162
    CBSEENCH11005533

    Define electronegativity. Explain the factors influencing electronegativity. How does it vary along periods and groups? what are its units?

    Solution

    Electronegativity: It may be defined as the relative tendency of an atom to attract the shared pair of electrons towards itself in a covalent bond. Greater the tendency of the atom to attract the shared pair of electrons, more will be its electronegativity. For example, a covalent bond A-B can be represented as:
    A : B (If A is more electronegative than B)
    A : B (If B is more electronegative than A)

    Factors influencing electronegativity:
    (i) The size of the atom: The smaller the size of the atom more will be the electronegativity.
    (ii) Nuclear charge: Electronegativity of an element increases with the increase in the nuclear charge of its atom.
    Variation of electronegativity along periods and groups : In general, electronegativity increases along a period due to the decrease in atomic size and decreases down the group due to increase in atomic size.
    Units: Electronegativity represents only the comparative values and it has no specific units. F (halogen) is the most electronegative element (4 . 0) while Cs alkali metal is the least electronegative (0.70).

     
    Question 163
    CBSEENCH11005534

    What is the importance of electronegativity?

    Solution

    1. Electronegativity can be used to predict the nature of bonds between two atoms.
    (i) If the two atoms are identical and have high electronegativity then the bond between them is called non-polar covalent bond or purely covalent bond. For example, H2 (or H-H), Cl2 (or Cl - Cl).
    (ii) If the two atoms are dissimilar and have high electronegativity, then the bond between them is called a polar covalent bond. For example H -Br. Li—Cl etc.

    2. Electronegativity helps us in predicting the polarity of covalent bonds.
    (i) The greater the electronegativity difference between the two bonded atoms, the greater is the polarity of the covalent bond. For example, a bond between H and Cl is more polar (electronegativity difference = 0.9) than the bond between Br and CI (electronegativity difference = 0.2).


    (ii) If the two atoms are dissimilar and have a large difference in their electronegativity,then the bond between them is called an electrovalent bond (or ionic bond).

    Question 164
    CBSEENCH11005535

    Explain giving various reasons for the variation of electronegativity along a period and along a group.

    Solution

    (i) Variation of electronegativity in a period: In a period electronegativity increases in moving from left to right. This is due to the reason that nuclear charge increases whereas atomic radius decreases as we move from left to right in a period. Halogens have the highest value of electronegativity in their respective period.
    (ii) Variation of electronegativity in a group: In a group, electronegativity decreases on moving down the group. This is due to the effect of increased atomic radius. For example among halogens, fluorine has the highest electronegativity.

    Question 165
    CBSEENCH11005536

    What is the basic difference between the terms electronegativity and electron gain enthalpy?

    Solution

     The main points of difference are as:

    Electronegativity Electrogain enthalpy
    1. It is the relative tendency of an atom in a molecule to attract the share pair fo electron towards itself in a covalent bond.  1. It is the tendency of an isolated gaseous atom to attract an electron to form a negative ion.
    2. It is a number and has no units. 2. It is measured in electron volts/atom or kJ mol-1.
    3. It is the property of a bonded atom. 3. It is a property of an isolated atom.
    4. Electronegativity values are only the comparative values.  4. Electron affinity values are the actual values.
    5. It changes regularly in a period or group. 5.  It does not change regularly in a period or a group.



     

    Question 166
    CBSEENCH11005537

    How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds? 

    Solution
    The above statement is wrong because electronegativity of any given atom is not constant but varies with the state of hybridization and the oxidation state of the element. The electronegativity increases as the percentage of s-character of a hybrid orbital increases or oxidation state of the element increases. For example, the electronegativity of N increases as we move from sp3-sp2-sp hybrid orbitals. Similarly,electronegativity of N is different in NO3 and NH3.
    Question 167
    CBSEENCH11005538

    Give a brief account of valence ?

    Solution

    Valence. The electrons present in the outermost shell (or valence shell) of the elements are called valence electrons and they normally represent the valency of the element.
    In the case of representative elements (s-and p-block elements), valency is either equal to the number of valence electrons or eight minus the number of valence electrons. For example
    In HCl, the valency of H =1
    In CaCI2. the valency of Ca = 2 (valence electrons = 2)
    In NH3, the valency of N = 3 {8-5} (valence electrons = 5)
    Variation along a period: The number of electrons in the valence shell increases from one to eight as we move from left to right in the periodic table. The valency with respect to hydrogen initially increases from 1 to 4 and then decreases from 4 to 0.
    LiH  BeH2  BH3  CH4   NH3   H2O  HF   Ne
    1      2       3      4      3      2      1     0

    In a similar manner, the valency with respect to chlorine also initially increases from 1 to 4 and then decreases from 4 to 0.
    Variation in a group: The valency of the element depends on upon its valence shell configuration. Since in a group, the valence shell configuration of the elements remains almost the same, they exhibit a common valency. For example, all the elements of group 1 have valency one and those of group 2 have valency two.

     

    Question 168
    CBSEENCH11005539

    The valence of representative elements is either equal to the number of valence electrons or eight minus the number. What is the basis of their rule?

    Solution
    The electrons present in the valence shell (outermost shell) are called valence electrons and their number determines the valency of the element. It has been found that in the case of representative elements (s-and p-block elements), the valency of an element is either equal to the number of valence electrons or eight minus the number of valence electrons, e.g. consider the hydrides of 2nd period i.e. groups 1,2, 13, 14, 15, 16, 17.
    Group 1 2 13 14 15 16 17
    Compounds LiH BeH2 BH3 CH4 NH3 H2O HF
    No. of valence electrons  1 2 3 4 5 6 7
    Valency of element 1 2 3 4 8-5 =3 8-6 =2 8-7 =1

    From the above table, it is clear that valency of the representative elements of groups 1, 2, 13 and 14 is equal to the number of valence electrons. But the valency of representative elements of group 15, 16 and 17 is equal to eight minus number of valence electrons.

    The valency of any element is the number of electrons which it loses, joins or shares with other atoms to attain the nearest inert gas configuration. The elements of groups 1, 2 and 13 lose 1, 2 and 3 electrons respectively to attain the nearest noble gas configuration. Thus, the valency of the elements of groups 1, 2 and 13 is equal to the number of their valence electrons. The elements of group 14 have four electrons in their respective valence shells and thus attain noble gas configuration only by sharing with four more electrons of other atoms. Therefore, the valency of elements of group 14 is also equal to the number of their valence electrons.

    The elements of group 15, 16 and 17 have 5, 6 and 7 electrons in their respective valence shells and hence they acquire 3 (8-5), 2 (8-6) and 1 (8-7) electrons either by gaining or sharing electrons from other atoms to attain the nearest inert gas configuration. Thus, the valency of the elements of groups 15, 16 and 17 is equal to eight minus the number of electrons in their respective valence shells.

    Question 169
    CBSEENCH11005540

    Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:

    silicon and oxygen


    Solution

    Silicon belongs to group 14 with a valence of 4. Oxygen belongs to group 16 elements with a valence of 2. Hence the formula of the compound formed would be SiO2.

    Question 170
    CBSEENCH11005541

    Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:
    aluminium and bromine

    Solution
    Aluminium belongs to group 13 with a valence of 3. Bromine being halogen belongs to group 17 with a valence of 1. Hence the formula of the compound formed would be AlBr3.
    Question 171
    CBSEENCH11005542

    Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:
    calcium and iodine

    Solution
    Calcium belongs to group 2 with a valence of 2. Iodine being halogen belongs to group 17 with a valence of 1. Hence the formula of the compound formed would be CaI2.
    Question 172
    CBSEENCH11005543

    Predict the formula of the stable binary compounds which would be formed by the following pairs of elements:

    (i) element 114 and fluorine

    (ii) element 120 and oxygen.

    Solution

    (i) Element 114 belongs to group 14 with a valence of 4. Fluorine belongs to group 17 with a valence of 1. Hence the formula of the compound formed would be Uuq. I4.
    (ii) Element 120 belongs to group 2 with a valence of 2. Oxygen belongs to group 16 with a valence of 2. Hence the formula of the compound formed would be Ubn.O.

    Question 173
    CBSEENCH11005544

    Discuss briefly the anomalous behviour of elements of second period.
    Or
    The first element of each group differs significantly from rest of the elements of the same group. Give reasons.

    Solution

    The first element of each of the groups 1 (lithium) and 2 (beryllium) and groups 13-17 (boron to fluorine) differs in many respects from the other members of their respective group. The difference in the behaviour of the first member of a group in the s- and p-block compared to the other members in the same group can be due to the following factors:
    (i) Small atomic size of the first element.
    (ii) Large charge/radius ratio
    (iii) High electronegativity
    (iv) The absence of d-orbitals in the valence shell of the first element.
    The first member of s- and p-block has only four, valence orbitals (2s and 2p) available for bonding. As a result, the maximum covalency of the first member of each group is 4. On the other hand, the second member of the groups has nine valence orbitals (3s, 3p, 3d). Hence these members can expand their covalency beyond 4, through the participation of d-orbital in bond formation.
    (v) Ability to form pπ-pπ multiple bonds. The first member of each group of p-block elements has great tendency to form pπ-pπ multiple bonds to itself (C = C. C ≡ N, N = N, N ≡ N and to the other second period N, N ≡ N) elements (C = 0, C = N, C ≡ N, N = 0).

    Question 174
    CBSEENCH11005545

    What do you understand by diagonal relationship?

    Solution
    Certain elements resemble closely one-step down in the next higher group in the periodic table. The resemblance in properties of a member of the group and a diagonally situated element of the adjacent group is called diagonal relationship. For example, the first three elements of the second period (Li. Be. B) show diagonal similarity with elements (Mg, AI, Si) of the third period placed on the right-hand side.


    Cause of diagonal relationship. The cause of the diagonal relationship is due to opposing trends in periodic properties along a period and down the group. For example,
    (i) The electropositive character increases down a group and it decreases from left to right in a period. Consequently, diagonally opposite elements in the periodic table have more or less similar electropositive character and hence, in turn, have many similar properties also.
    (ii) In case of metallic elements, the diagonally opposite metals possess nearly similar polarizing power [i.e. ionic charge/(ionic radius)2] because:
    (a) ionic charge increases from left to right in a period and
    (b) ionic radius decreases from left to right in a period but increases down a group.
    Question 175
    CBSEENCH11005546

    Show by a chemical reaction with water that Na2O is a basic oxide and Cl2O7 is an acidic oxide. 

    Solution
    Na2O reacts with water to form sodium oxide which turns red litmus blue.
    space space stack Na subscript 2 straight O with Sodium space oxide below space plus space straight H subscript 2 straight O space space space rightwards arrow space space space space NaOH
    Therefore, Na2O is a basic oxide.
    In contrast, Cl2O7 reacts with water to form perchloric acid which turns blue litmus red.
    space space Cl subscript 2 straight O subscript 7 plus straight H subscript 2 straight O space rightwards arrow space stack 2 HClO subscript 4 with Perchloric space acid below
    Hence, Cl2O7 is an acidic oxide. 
    Question 176
    CBSEENCH11005547

    What is the basic theme of organisation in the periodic table?

    Solution

    Before the 18th century, only a 31 elements are known & it was quite east to study their chemical & physical properties. But with the passage of time, more & more elements were discovered. By 1865, the number of identified element had more than doubled to 63. At present 114 elements are known. It was not possible to study & remember each of them individually . Because many of them have different properties.So, there arise the need of developing a table in which all the elements can be classified according to similar properties so that less confusion can be generated & true information can be generated by the different elements present in the periodic table. 

     

     

    Question 177
    CBSEENCH11005548

    What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?

    Solution

    Mendeleev periodic law states that the physical & chemical properties of the elements are a periodic function of their atomic masses. In his periodic table, the elements are arranged in vertical rows called groups & horizontal rows are known as periods.There are nine groups & seven periods. He arranged only 60 elements in periodic table.

     An English physicist Moseley, in 1913 gave modern periodic law which states that the physical & chemical properties of the elements are the periodic function of their atomic number.This law becomes the base of modern periodic law.In the modern periodic table, there are 18 groups & 7 periods.

    Question 178
    CBSEENCH11005549

    On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.

    Solution

    A principal quantum number is a number which determines the main energy level or shell in which the electron is present.It gives the average distance of the electron from the nucleus & specifies the value of the energy of the electron. The azimuthal quantum number is the number which determines the subshell in a principal energy shell to which an electron belongs.

    In the periodic table of the elements, a period indicates the value of the principal quantum number (n) for the outermost shells. Each period begins with the filing of principal quantum number (n). The value of n for the sixth period is 6. For n = 6, azimuthal quantum number (l) can have values of 0, 1, 2, 3, 4.

    According to Aufbau’s principle, electrons are added to different orbitals in order of their increasing energies. The energy of the 6d subshell is even higher than that of the 7s subshell.

    In the 6th period, electrons can be filled in only 6s, 4f, 5d, and 6 p subshells. Now, 6s has one orbital, 4f has seven orbitals, 5d has five orbitals, and 6p has three orbitals. Therefore, there are a total of sixteen (1 + 7 + 5 + 3 = 16) orbitals available. According to Pauli’s exclusion principle, each orbital can accommodate a maximum of 2 electrons. Thus, 16 orbitals can accommodate a maximum of 32 electrons.

    Hence, the sixth period of the periodic table should have 32 elements.

    Question 179
    CBSEENCH11005550

    Explain why cation are smaller and anions larger in radii than their parent atoms?

    Solution

    The ionic radius of a cation is always smaller than the parent atom because a cation is formed by loss of one or more electrons by the neutral atom. The loss of one or more electrons increases the effective nuclear charge. As a result, the force of attraction of nucleus for the electrons increases and hence their ionic radii decrease. In contrast, the ionic radius of anion is formed by the addition of one or more electrons to the anion is formed by the addition of one or more electrons to the neutral atom. The addition of one  or more electrons decreases the effective nuclear charge. As a result, the force of attraction of the nucleus for the electrons decreases and hence the ionic radii increases.

    Question 180
    CBSEENCH11005551

    Among the second period elements the actual ionisation enthalpies are in the order 
    Li<B<Be<C<O<N<F<Ne.
    Explain why
    (i) Be has higher increment subscript straight i straight H than B
    (ii) O has lower increment subscript straight i straight H than N and F?

    Solution

    i) In case of Be ( 1s2 2s2) the outermost electron is present in 2s- orbital  while in B(1s22s22p1). IT is present in 29-orbital. Since 2s- electrons, due to greater penetration, are more strongly attracted by the nucleus than 2p-electrons, therefore, more amount of energy is required to knowck out a 2s-electron than a 2p-electron. Consequently, increment subscript straight i straight H of Be is Higher than that of increment subscript straight i straight H of B.

    ii) The electronic configuration of N(1s2 2s2 2p1x 2p1y 21z) in which 2p-orbital are exactly half filled is more stable than the electronic configuration of O( 1s2 2s2 2p2x2p1y 21z) in which the 12p-orbital are neither half filled nor completely filled. Therefore, it is  difficult to remove an electron from N than from O. As a result, increment subscript straight i straight H of O is less than that of N. Because of higher nuclear charge(+9) and smaller atomic size, the first ionization enthalpy of F is higher than that of O. Therefore, ionization enthalpy of O is less than that N as well as of F.

    Question 181
    CBSEENCH11005552

    What is the basic difference between the terms electron gain enthalpy and electronegativity?

    Solution

    Both electron gain enthalpy and electronegativity are measures of the tendency of the atom of an element to attract electrons. Whereas electron gain enthalpy refers to the tendency of an isolated gaseous atom to accept an electron to form  a negative ion, electronegativity refers to the tendency of the atom to attract the shared pair of electron  towards itself in a molecule.

    Question 182
    CBSEENCH11005553

    The first left parenthesis increment subscript straight i straight H subscript 1 right parenthesis and the second left parenthesis increment subscript straight i straight H subscript 2 right parenthesis ionization enthalpies (in  kJ mol-1) and the left parenthesis increment subscript eg straight H right parenthesis electron gain enthalpy (in kJ mol-1) of a few elements are given below:
    Elements            increment straight H subscript 1               increment straight H subscript 2                 increment subscript eg straight H
    I                        520                    7300                    -60
    II                       419                     3051                   -48
    III                      1681                    3374                  -328
    IV                       1008                    1846                 -295
    V                         2372                    5251                 +48
    VI                        738                      1451                -40  
    Which of the above elements is likely to be:
    (a) the least reactive element.
    (b) the most reactive metal.
    (c) the most reactive non-metal.
    (d) the least reactive non-metal.
    (e) the metal which can form a stable binary halide of the formula MX2(X =halogen).
    (f) the metal which can form a predominantly stable covalent halide of the formula MX(X= halogen)?

    Solution

    Element V is likely to be the least reactive element. This is because it has the highest first ionisation enthalpy (ΔiH1) and a positive electron gain enthalpy (ΔegH).It is a noble gas

     (b) Element II is likely to be the most reactive metal as it has the lowest first ionisation enthalpy (ΔiH1) and a low negative electron gain enthalpy (ΔegH).The with lowest first ionisation enthalpy is likely to be a reactive metal

    (c) Element III is likely to be the most reactive non–metal as it has a high first ionisation enthalpy (ΔiH1) but less than the noble gas elements and the highest negative electron gain enthalpy (ΔegH).

    (d) Element V is likely to be the least reactive non–metal since it has a very high first ionisation enthalpy (ΔiH2) and a positive electron gain enthalpy (ΔegH).

     (e) Element VI has a low negative electron gain enthalpy (ΔegH). Thus, it is a metal. Further, it has the lowest second ionisation enthalpy (ΔiH2). Hence, it can form a stable binary halide of the formula MX2 (X=halogen).

     (f) The element I have low first ionisation energy and high second ionisation energy. Therefore, it can form a predominantly stable covalent halide of the formula MX (X=halogen).

    Question 183
    CBSEENCH11008048

    Which of the following atoms has the highest first ionisation energy?

    • Na

    • K

    • Sc

    • Rb

    Solution

    C.

    Sc

    Order of first ionisation energy is Sc> Na > K > Rb. Due to poor shielding effect, removal of one electron from 4s orbital is difficult as compared to 3s-orbital.

    Question 184
    CBSEENCH11008072

    Which of the following represents the correct order of increasing first ionisation enthalpy for Ca, Ba, S, Se and Ar?

    • Ca <S<Ba<Se<Ar

    • S< Se<Ca<Ba< Ar

    • Ba< Ca<Se< S< Ar

    • Ca<Ba<S<Se<Ar

    Solution

    C.

    Ba< Ca<Se< S< Ar

    Ionisation energy increases along period from left to right and decreases down a group. The position of given elements in the periodic table is as;
    Increasing order of first ionization enthalpy is Ba < Ca < Se < S < Ar

    Question 185
    CBSEENCH11008074

    The first ionisation potential of Na is 5.1 eV. The value of electron gain enthalpy of Na+ will be

    • -2.55 eV

    • -5.1 eV

    • -10.2 eV

    • +2.55 eV

    Solution

    B.

    -5.1 eV

    Na → Na+ + e- (first IE)
    Na+ + e- →Na
    Since reaction is reverse, so electron gain enthalpy of Na+
    ΔH (eg)=-5.1 eV

    Question 186
    CBSEENCH11008088

    The increasing order of the ionic radii of the given isoelectronic species is 

    • Cl-, Ca2+, K+, S2-

    • S2-, Cl-, Ca2+,K+

    • Ca2+, K+, Cl-, S2-

    • K+, S2-,Ca2+,Cl-

    Solution

    C.

    Ca2+, K+, Cl-, S2-

    For isoelectronic species, as the z/e decreases, ionic radius increases
    such as Ionic space radii space proportional to space 1 over straight z

    Question 188
    CBSEENCH11008104

    The correct sequence which shows decreasing order of the ionic radii of the elements is

    • Al3+ > Mg2+ > Na+ > F- > O2-

    • Na+ > Mg2+ > Al3+ > O2- > F-

    • Na+ > F- > Mg2+ > O2- > Al3+

    • O2- >F- > Na+ > Mg2+ >Al3+

    Solution

    D.

    O2- >F- > Na+ > Mg2+ >Al3+

    O2-, F-, Na+, Mg2+,Al3+ are isoelectronic species (10 electrons each)
    Cation having a positive charge is smaller in size than neutral atom whereas anion having a negative charge is larger in size than neutral atom.

    Question 197
    CBSEENCH11008228

    The electronic configuration of Eu (Atomic no.63), Gd (Atomic no. 64) and Tb (Atomic no. 65) are

    • [Xe]4f6 5d1 6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f9  6s2 

    • [Xe] 4f6 5d1 6s2 , [Xe] 4f7 5d1 6s2 and [Xe] 4f8 5d1 6s2 

    • [Xe] 4f7  6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f9 6s2 

    • [Xe] 4f7  6s2,[Xe] 4f8  6s2 and [Xe] 4f8 5d1 6s2

    Solution

    C.

    [Xe] 4f7  6s2, [Xe] 4f7 5d1 6s2 and [Xe] 4f9 6s2 

    Electronic configuration of 63Eu [europium ]
    = [Xe] 4f7  6s2
    Electronic configuration of 64Gd [gadolinium ]
    = [Xe] 4f7 5d1 6s2 
    Electronic configuration 65Tb [terbium]
    = [Xe] 4f9 6s2 

    Question 198
    CBSEENCH11008231

    In which of the following options the order of arrangement does not agree with the variation of the property indicated against it?

    • B<C<N<O (increasing first ionisation enthalpy)

    • I<Br<Cl<F(increasing electron gain enthalpy)

    • Li<Na<K<Rb (increasing metallic radius)

    • Al3+ < Mg2+<Na+<F-(increasing ionic size)

    Solution

    B.

    I<Br<Cl<F(increasing electron gain enthalpy)

    The electron gain enthalpy is defined as the energy released when an electron is added to a neutral gaseous atom to convert it into a negative ion. Large negative values of electron gain enthalpy indicate that it is easier to add an electron to an atom to convert it into an anion, and hence large amount of energy is released. The electron gain enthalpies in halogens decreases in the order
    Cl>F>Br>I

    because as we proceed down a group because the size of the atom increases. Therefore the distance of the nucleus from the valence orbital (which accepts the additional electron) increases. Hence the force by which the additional electron is attracted by the nucleus also decreases and the electron gain enthalpies become less negative as we proceed down a group.

    Question 199
    CBSEENCH11008236

    The species Ar, K+ and Ca2+ contain the same number of electrons. In which order do their radii increase?

    • Ar < K+< Ca2+

    • Ca2+ < Ar < K+

    • Ca2+ < K+ < Ar

    • K+< Ar < Ca2+

    Solution

    B.

    Ca2+ < Ar < K+

    Ca2+ < K+ < Ar
     Ar, K+ and Ca2+ are isoelectronic i.e with the same number of electrons, 18. For isoelectronic species, ionic radii decrease with increases in effective (relative) positive charge. Also Ar, K and Ca belong to the same period.

    Question 200
    CBSEENCH11008237

    Solubility of the alkaline earth's metal sulphates in water decreases in the sequence

    • Mg>Ca>Sr>Ba

    • Ca>Sr>Ba>Mg>

    • Sr>Ca>Mg>Ba

    • Ba>Mg>Sr>Ca

    Solution

    A.

    Mg>Ca>Sr>Ba

    The solubility of the sulphates. The sulphates becomes less soluble as  you fo down the group i.e
    Mg> Ca>Sr> Ba
    The magnitude of the lattice energy remains almost constant as the size of the sulphate ion is so big that small increase in the size of the cation from Be to Ba does not make any difference. However, the hydration energy decreases from Be2+ to Ba2+ appreciably as the size of the cation increases down the group. The significantly high solubility of MgSO4 is due to high enthalpy of solvation on the smaller Mg2+ ions.

    Question 201
    CBSEENCH11008239

    The number of d- electrons in Fe2+ (Z = 26) is not equal to the number of electrons in which one of the following?

    • S- electronic in Mg (Z=12)

    • p-electrons in Cl( Z=17)

    • d- electrons in Fe (Z=26)

    • p-electrons in Ne (Z=10 )

    Solution

    B.

    p-electrons in Cl( Z=17)

    Electronic configuration of Fe2+ is [Ar]3d6 4s0 
    therefore, Number of electrons  = 6
    Mg- 1s2 2s2 2p6  (6s electrons)
    It matches with the 6d electrons of Fe2+ 
    Cl- 1s2 2s2 2p6 3s2 3p5 (11 p electrons)
    It does not match with the 6d  electrons of Fe2+
    Fe- [Ar] 3d6 4s2 (6d electrons)
    It does not match with the 6d electrons of Fe2+
    Ne- 1s2 2s2 2p6 3s2 3p 6 (6p electrons)
    It matches with the 6d electrons of Fe2+.
    Hence, Cl has 11 p electrons which do not match in number with 6d electrons of Fe2+

    Question 202
    CBSEENCH11008282

    Identify the wrong statement in the following.

    • Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius

    • Amongst isoelectronic species, greater the negative charge on the anion, larger is the ionic radius

    • Atomic radius of the elements increases as one moves down the first group of the periodic table

    • Atomic radius of the elements decreases as one moves across from left to right in the 2nd period of the periodic Table.

    Solution

    A.

    Amongst isoelectronic species, smaller the positive charge on the cation, smaller is the ionic radius

    The Atomic radius of the elements decreases across a period from left to right due to increasing in effective nuclear charge. On moving down a group, since, a number of shells increases, so atomic radius increases. Amongst isoelectronic species, ionic radius increases with increases in negative charge or decrease in positive charge. 

    Question 203
    CBSEENCH11008354

    Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements O, S, F and Cl?

    • Cl < F < O < S

    • O < S < F< Cl

    • F < S < O < Cl

    • S < O < Cl < F

    Solution

    B.

    O < S < F< Cl

    Electron gain enthalpy, generally, increases in a period from left to right and decreases in a group on moving downwards. However, members of III periods have somewhat higher electron gain enthalpy as compared to the corresponding members of the second period because of their small size.
    O and S belong to VI A (16) group and Cl and F belong to VII-A (17) group. Thus, the electron gain enthalpy of Cl and F is higher as compared to O and S.
    Cl and F > O and S
    Between Cl and F, Cl has higher electron gain enthalpy as in F, the incoming electron experiences a greater force of repulsion because of the small size of F atom. Similarly is true is a case of O and S ie, the electron gain enthalpy of S is higher as compared to O due to its small size. Thus, the correct order of electron gain enthalpy of given elements is 
    O < S < F < Cl

    Question 204
    CBSEENCH11008359

    Which of the following pairs has the same size? 

    • Fe2+ , Ni2+

    • Zr4+, Ti4+

    • Zr4+, Hf4+

    • Zn2+, Hf4+

    Solution

    C.

    Zr4+, Hf4+

    In general, the atomic and ionic radii increases on moving down a group. But the elements of second transition series (eg, Zr, Nb, Mo etc) have the almost same radii as the elements of third transition series (eg Hf, Ta, W etc.) This is because of lanthanide contraction i.e imperfect shielding of one 4f- electron by another.

    Question 205
    CBSEENCH11008364

    Which one of the following ion has electronic configuration [Ar] 3d6?
    (At. no:  Mn = 25, Fe = 26, Co = 27, Ni = 28)

    • Ni3+

    • Mn

    • Fe3+

    • Co3+

    Solution

    D.

    Co3+

    The electronic configurations of given ions are:
    Ni3+ (28) = [Ar] 3d7
    Mn3+ (25) = [Ar] 3d4
    Fe3+ (26) = [Ar] 3d5
    Co3+ (27) = [Ar] 3d6

    Question 206
    CBSEENCH11008366

    Among the elements Ca, Mg, P and Cl, the order of increasing atomic radii is 

    • Mg< Ca < Cl < P

    • Cl < P < Mg < Ca

    • P < Cl < Ca < Mg

    • Ca < Mg < P < Cl

    Solution

    B.

    Cl < P < Mg < Ca

    With the increase in the number of electron in the same shell, the atomic radii decrease due to increase in effective nuclear charge. However, atomic radii increases, as the number of shells increases. Thus, on moving down a group atomic radii increases.
    The electronic configuration of the given elements is
    Mg12 = [Ne] 3s2
    Ca20 = [Ar] 4s2
    P15 = [Ne]3s2 3p3
    Cl17 = [Ne] 3s2 3p5
    In Mg, P and Cl, the number of electrons is increasing in the same shell thus, the order of their atomic radii is
    Cl <P < Mg
    In Ca, the electron is entering in the higher shell, thus, it has the highest atomic radii among the given. Thus, the order of radii is
    Cl < P < Mg < Ca

    Question 207
    CBSEENCH11008369

    Among the following which one has the highest cation to anion size ratio?

    • CsI

    • CsF

    • LiF

    • NaF

    Solution

    B.

    CsF

    The order of size of given cations is 
    Li+ < Na+ < Cs+
    and the order of the size of given anions is 


    I- > F-
    Thus, when the cation is largest and anion is smallest the cation to anion size ratio is maximum. Hence cation to anion size ratio is maximum
    Question 208
    CBSEENCH11008379

    Which one of the element with the following outer orbital configuration may exhibit the largest number of oxidation states?

    • 3d5,4s2

    • 3d5,4s1

    • 3d5,4s2

    • 3d2,4s2

    Solution

    C.

    3d5,4s2

    A number of oxidation states exhibited by d-block elements is the sum of the number of electrons (unpaired) in d-orbitals and number of electrons in s - orbital.

    a) 3d3, 4s2 ⇒ Oxidation state = 3+2 = 5
    b) 3d5, 4s1 ⇒ Oxidation state = 5+1 = 6
    c) 3d5, 4s2 ⇒ Oxidation state = 5+2 = 7
    d) 3d4, 4s2 ⇒ Oxidation state = 2+ 2 = 4 
    Hence, an element with 3d5, 4s2 configuration exhibits the largest number of oxidation states. 

    Question 209
    CBSEENCH11008381

    Oxidation number of P in PO43-, of S in SO42- and that of Cr in Cr2O72- are respectively,

    • +5,+6 and +6

    • +3, + 6 and +5

    • +5,+3 and +6

    • -3,+6 and +6

    Solution

    A.

    +5,+6 and +6

    (i) Sum of oxidation states of all atoms = charge of ion.
    (ii) oxidation number of oxygen = -2
    Let the oxidation state of P in PO43- is x.
     PO43- 
    x + 4 (-2) = - 3
    x-8 = - 3
    x = +5
    Let the oxidation state of S in SO42- is y
    y + 4(-2) = -2
    y-8 = - 2
    y = +6
    Let the oxidation state of Cr in Cr2O72- is z.
    2 x z+7(-2) = -2
    2z-14 = - 2
    z=+6
    Hence, oxidation state of P, S and Cr are +5, +6 and +6

    Question 210
    CBSEENCH11008384

    Amongst the elements with following electronic configuration, which one of them may have the highest ionisation energy?

    • [Ne] 3s2 3p3

    • [Ne] 3s2 3p2

    • [Ar] 3d10, 4s2 4p3

    • [Ne] 3s2 3p1

    Solution

    A.

    [Ne] 3s2 3p3

    Ionisation energy usually increases from left to right in a period with a decrease in atomic size and decrease from up to down in a group with an increase in atomic size.
    [Ne] 3s2 3p3     = Group V
    [Ne] 3s2 3p2     = Group IV

    [Ar] 3d10, 4s2 4p3 = Group V
    [Ne] 3s2 3p1 = Group III

    Question 211
    CBSEENCH11008401

    The correct order of decreasing second ionisation enthalpy of Ti (22), V (23), Cr(24) and Mn (25) is 

    • Cr > Mn> V > Ti

    • V > Mn > Cr >Ti

    • Mn > Cr> Ti >V

    • Ti >V >Cr> Mn

    Solution

    A.

    Cr > Mn> V > Ti

    The amount of energy required to remove an electron from unipositive ions is called the second IP which generally increases in a period from left to right.
    The second ionisation energies of the first transition series increase almost regularly with an increase in atomic number. However, the value for Cr is sufficiently higher than those of its neighbour ie, (Mn). This is due to the stable configuration of Cr3+ (3d5 exactly half filled.)

    [Note: The half-filled and completely filled configuration are more stable.]

    Question 212
    CBSEENCH11008406

    The sequence of ionic mobility in aqueous solution is 

    • K+ > Na+ > Rb+ > Cs+

    • Cs+ > Rb+ > K+ >Na+

    • Rb+ > K+> Cs+ > Na+

    • Na+ > K+ >Rb+ >Cs+

    Solution

    B.

    Cs+ > Rb+ > K+ >Na+

    The smaller the size of ion, the greater the degree of hydration.
    Mobility space of space ion space in space aqueous space solution
space space space space proportional to space fraction numerator 1 over denominator size space of space ion space in space aqueous space solution end fraction
    Generally in a group size or radii of ions increases, So Li has the smallest size thus bind a maximum number of water molecules with itself and becomes largest in size (in aqueous solution). Hence, the order of ionic radii (size) of alkali metal ions in aqueous solution is 
    Li+ > Na+ > K+> Rb+ >Cs+
    Thus, the order of mobility 


    Li+ > Na+ > K+> Rb+ >Cs+
    Question 213
    CBSEENCH11008425

    Which of the following oxidation states are the most characteristics for lead and tin respectively? 

    • +4,+2

    • +2,+4

    • +4,+4

    • +2,+2

    Solution

    B.

    +2,+4

    The tendency to form +2 ionic state increase on moving down the group due to the inert pair effect.
    Most characteristic oxidation state for lead and tin are +2,+4 respectively.

    Question 214
    CBSEENCH11008428

    Consider the following sets of quantum numbers:


      n l m s
    i) 3 0 0 +1/2
    ii) 2 2 1 +1/2
    iii) 4 3 -2 -1/2
    iv) 1 0 -1 -1/2
    v) 3 2 3 +1/2

    Which of the following sets of quantum number is not possible? 
    • ii, iii and iv

    • i, ii, iii and iv

    • ii, iv and v

    • i and iii

    Solution

    C.

    ii, iv and v

    The value of l varies from 0 to n-i and the value of m varies from -l to +l through zero.
    the value of 's' negative-or-plus 1 half Which signifies the spin of the electron. The correct sets of quantum number are following:

      n l m s
    ii) 2 1 1 +1/2
    iv 1 0 0 -1/2
    v) 3 2 2 +1/2
    Question 215
    CBSEENCH11008432

    With which of the following electronic configuration an atom has the lowest ionisation enthalpy? 

    • 1s2 2s22p5

    • 1s2 2s22p3

    • 1s2 2s22p5 3s1

    • 1s2 2s22p6

    Solution

    C.

    1s2 2s22p5 3s1

    The electronic configuration 1s2 2s22p5 3s1 shows lowest ionisation energy because this configuration is unstable due to the presence of one electron is s- orbital. Hence, less energy is required to remove the electron.

    Question 216
    CBSEENCH11008434

    Identify the correct order of the size of the following:

    • Ca2+ < K+<Ar < S2- < Cl-

    • Ca2+ < K+ < Ar < Cl-< S2-

    • Ar < Ca2+ <K+ <Cl- <S2-

    • Ca2+ < Ar < K+ <Cl- <S2-

    Solution

    B.

    Ca2+ < K+ < Ar < Cl-< S2-

    A cation has always the lesser ionic size than a metal atom due to loss of an electron and an anion has always the greater size than metal atom due to a gain of electrons. So, the correct order of size is as 
    Ca2+ < K+ < Ar < Cl-< S2-

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