Sponsor Area
What is valency of an element?
What are magic numbers?
Which property is the basis of long form of periodic table?
How many elements are placed in each period of the p-block?
What is the number of groups in: (i) p-block (ii) d-block?
(i) p-block: Six groups
(ii) d-block: Ten groups.
How many elements are present in:
(i) second period
(ii) fourth period
(iii) sixth period?
Why are there 10 elements in each series of d-block?
As the d-subshell can have a maximum of ten electrons.
In terms of electronic configuration,
what is common in a given period and group?
Give the general electronic configuration of
(i) Transition elements (ii) Inner transition elements.
(i) Transition elements:
(n - 1)d1-10 ns1-2.
(ii) Inner transition elements:
(n - 2)f1-14 (n - 1)d0-1 ns2.
Tips: -
Name the group and period to which an element with Z = 15 belongs ?
What are the advantages of periodic classification of elements?
(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.
Sponsor Area
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.
Which of the following statements related to the modern periodic table is incorrect?
(a) The p-block has 6 columns, because a maximum of 6 electrons can occupy all the orbitals in a p-shell.
(b) The d-block has 8 columns, because a maximum of 8 electrons can occupy all the orbitals in a d-subshell.
(c) Each block contains a number of columns equal to the number of electrons that can occupy that subshell.
(d) The block indicates value of azimuthal quantum number (l) for the last subshell that received electrons in building up the electronic configuration.
Anything that influences the valence electrons will effect the chemistry of the element. Which one of the following factors does not affect the valence shell?
(a) Valence principal quantum number (n)
(b) Nuclear charge (Z)
(c) Nuclear mass
(d) Number of core electrons.
Write the atomic number of the element present in the third period and seventeenth group of the periodic table.
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.
An element A with atomic number 112 has recently been discovered. Predict its electronic configuration and suggest the group in which this element be placed.
How is the atomic radius of an element actually expressed?
The bond length of HCl bond is 136 pm. Covalent radius of hydrogen atom is 37 pm. What is the covalent radius of chlorine atom?
Covalent radius of chlorine atom
= 136 pm - 37 pm = 99 pm.
Which radius, vander Waal's or covalent, is larger for an atom?
What is the correlation between atomic size and ionisation enthalpy?
Arrange the following in increasing order of size: Cl-, S2-, Ca2+, Ar
Arrange, l, l-, l+ in increasing order of size.
Arrange the following in decreasing order of ionic size:
Mg2+, O2-, Na+, F-, N3-.
Which one out of O and Se is bigger in size and why?
Which one out of Na+ and Cl- is bigger in size and why?
Out of P3-, S2- and Cl- ions, which has the smallest value of ionic radius and why?
Out of the following which has the smallest ionic 'Size and why: Li+, Na+, K+?
What is the trend of ionisation enthalpy in a period?
It increases from left to right in a period.
What is the trend of ionisation enthalpy in a group?
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?
The element with highest IE1 value: Kr (Noble gas).
The element with lowest IE1 value: K (Potassium)
For each of the following pairs state which one has greater ionisation enthalpy:
(i) I, I- (ii) Br, K.
(i) I- (ii) Br has the greater ionisation enthaply.
From amongst the following, choose the element with the lowest IE: Be, B and C.
Boron (B) has lowest Ionization energy.
Sponsor Area
What is the relationship between the first ionisation enthaplies and metallic properties?
Higher the value of ionisation enthalpy of an atom, greater will be its electronegative and non-metallic character.
What is the relationship between the first ionisation enthalpy and reducing power and basic nature of element?
Lower the Ionisation enthalpy, greater will be the reducing power and basic nature of the element.
Arrange the following in order of increasing first ionisation enthalpy:
Mg, Al, Si, Na.
Out of Na and Na+, which has higher ionisation enthalpy?
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.
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
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
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
The second ionisation enthalpy of an atom is greater than its first ionisation enthalpy because __________.
More enthalpy is requried to remove an electron held more firmly by the unipositive ion.
As the atomic number increases, the ionisation enthalpy _________ (in a period).
Generally increases.
Which elements occupy the maxima of ionisation potential curve?
Noble gases elements occupy the maxima of ionisation potential curve.
Al atom loses electrons successively to form Al+, Al2+ and Al3+ ions. Which step will have highest ionsiation enthalpy?
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.
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–
From the electronic configuration of the elements A, B and C given below:
(i) Which one has least negative electron gain enthalpy and
(ii) Which one has highest ionisation enthalpy value?
(i) Element having lowest negative electron gain enthalpy is C.
(ii) Element having highest I.E1 value is C.
Are the oxidation state and covalency of Al in same?
Valency of chlorine = (-1)
Valency of water = 0
Let the oxidation state of Al be x.
x -1 0
[Al Cl (H2O)6]2+ 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.
What is periodic table? What do you mean by classification of elements?
What was the need for the classification of elements?
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.
Discuss the earlier attempts to classify elements.
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
Triad Chlorine Bromine Iodine Mean atomic mass
Atomic masses 35.5 80 127
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.
Which important property did Mendeleev use to classify the elements in his Periodic Table? and did he stick to that?
What is the basis of Mendeleev's Periodic table?
Give a brief discussion of the Mendeleev's Periodic table.
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).
What are the merits and applications of Mendeleev's Periodic table?
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.
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.
State and explain Modern Periodic Law.
What is periodicity? What is its cause?
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.
What is long form of the periodic table? Discuss in brief its salient structural features.
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.
Why atomic number is a better basis for classifying the elements than atomic mass?
Why is long form periodic table regarded better than the Mendeleev's periodic table?
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. |
What are the defects of long form of the periodic table?
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.
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.
What would be the IUPAC name and symbol for the element with atomic number 120?
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.
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.
What are the advantages of classifying the elements into s, p, d and f-block elements?
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.
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.
Sponsor Area
Which element do you think would have been named by:
(i) Lawrence Berkeley Laboratory
(ii) Seaborg's group?
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.
Why do elements in the same group have similar physical and chemical properties?
In terms of period and group were would you locate the element with Z = 114?
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.
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. |
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.
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.
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
Account for the fact that the 4th period has eighteen and not eight elements.
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
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.
Lanthanoides and actinodes are placed in separate rows at the bottom of the periodic table. Explain the reason for this arrangement.
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.
How will you predict the position of an element in the periodic table?
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.
Name the block to which the elements with following valence shell electronic configurations belong:
(i) 3s2 3p5
(ii) 3d104s2
(iii) 3p64s2
(iv) 6s2 4f0
(v) 4s1 3d5
(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
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.
(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.
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.
(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 2p6 3s2 3p6 4s2 3d10 4p6.
(iv) The first halogen: The atomic number is 9 and electronic configuration is, 1s2 2s22p5.
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.
(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
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.
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?
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
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.
(i) The electronic configuration of element with Z = 117 can be represented as
This element belongs to Group 17 (halogen family).
(ii) The electronic configuration of element with Z = 120 can be represented as
This element belongs to Group 2 (alkaline earth metals).
What are metals, non-metals and metalloids?
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.
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. |
Arrange the following elements in the increasing order of their metallic character: Si, Be, Mg, Na, P.
What are periodic properties of the elements? Name a few important periodic properties.
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.
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.
What do you mean by covalent radius?
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.
How do atomic radius vary in a period and in a group? How do you explain the variation?
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.
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?
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.
Atomic radii of noble gases are essentially the vander Waal's radii. Comment.
What is ionic radius?
Why is the radius of cation less than that of parent atom?
Why is the radius of anion more than that of the parent atom?
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)
What does atomic radius and ionic radius really mean to you?
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.
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+
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+
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.
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.
(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+
(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.
Give the formula of a species which will be isoelectronic with the following atoms or ions:
(i) Na2+
(ii) Cl-
(iii) Ca2+
(iv) Rb+.
(i) Na+: Mg2+ (10 electrons)
(ii) Cl-: Ar (18 electrons)
(iii) Ca2+: K+ (18 electrons)
(iv) Rb+: Kr (36 electrons)
Which of the following species will have the largest and the smallest size?
Mg, Mg2+, Al, Al3+.
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+ .
Arrange the following ions in order of increasing size:
Be2+, Cl-, S2-, Na+, Mg2+, Br-.
Sponsor Area
In each of the following pairs, which species has a larger size? Explain.
(i) K or K+
(ii) Br or Br-
(iii) O2- or F-
(iv) Li+ or Na+
(v) P or As
(vi) Na+ or Mg2+
(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, Na+ ion 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.
Define ionsiation enthalpy. State its units.
What are successive ionisation enthalpies? Explain why the second ionisation enthalpy is higher than the first ionisation enthalpy?
Discuss in brief the factors which influence the magnitude of ionsiation enthalpy.
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.
डायफ़्रामHow does ionisation enthalpy vary along a group?
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.
What are the various factors due to which the ionization enthalpy of the main group elements tends to decrease down a group?
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.
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?
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.
Why is ionisation enthalpy of Be more than that of B?
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.
Why ionsiation enthalpy of N is more than that of O even though oxygen has higher nuclear charge than nitrogen?
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.
Ionsiation enthalpy of fluorine is greater than that of oxygen. Explain.
Why do noble gases have very high value of ionisation enthalpy?
Nitrogen and phosphorus have high value of ionisation enthalpies. Explain.
Be and Mg have high value of ionisation enthalpies. Explain.
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.
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.
(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.
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 ?
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).
The first (IE1) and second (IE2) ionisation enthalpies (kJ mol-1) of a few elements by Roman numerals are as shown:
IE1 | IE2 | |
i | 2372 | IE2 |
ii | 520 | 7300 |
iii | 900 | 1760 |
iv | 1680 | 380 |
which of the above elements is likely to be:
(i) a reactive metal
(ii) a reactive non-metal
(iii) a noble gas
(iv) a metal that forms a stable binary halide of the formula AX3 (X = halogen).
(i) Reactive metal II.
(ii) Reactive non-metal IV.
(iii) A noble gas I.
(iv) Metal forming halide III.
The IE1, of carbon, is more than that of boron while its IE2 value is smaller. Explain.
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 C+ ion.
Arrange the following in order of increasing ionisation enthalpies and assign reason:
(i) K+, Ar, Cl-
(ii) Fe, Fe2+, Fe3+
(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.
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?
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 Mg+ is 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.
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?
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.
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.
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
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:
Second electron gain enthalpy:
Discuss in brief the factors which influence the magnitude of electron gain enthalpy.
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.
How does electron gain enthalpy vary along a group?
How does electron gain enthalpy vary along a period?
Why do halogens have very higher value of electron gain enthalpy?
Why do noble gases have large positive electron gain enthalpies?
The electron gain enthalpy of fluorine has less negative value than that of chlorine while it is expected to be more. Explain.
Explain which of the following will have the most negative electron gain enthalpy and which the least negative: P, S, CI, F.
Which of the following elements has more negative electron gain enthalpy?
(i) [Ne] 3s2sp3
(ii) [Ne] 3s23p4
(iii) [Ne] 3s2 3p5
Which of the following pairs of elements would have a more negative electron gain enthalpy?
(i) O or F (ii) F or Cl
(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.
What is the significance of the terms - 'isolated gaseous atom' and 'ground state' while defining the ionization enthalpy and electron gain enthalpy?
(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.
Explain, which of the following pairs of elements would have a more negative electron gain enthalpy:
(i) N or O (ii) O or S
(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.
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.
(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.
Differentiate between ionisation enthalpy and electron gain enthalpy ?
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. |
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?
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.
(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).
The increasing order of reactivity among group 1 elements is Li < Na < K < Rb is F > CI > Br > I. Explain.
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.
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.
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:
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.
Define electronegativity. Explain the factors influencing electronegativity. How does it vary along periods and groups? what are its units?
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).
What is the importance of electronegativity?
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).
Explain giving various reasons for the variation of electronegativity along a period and along a group.
(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.
What is the basic difference between the terms electronegativity and electron gain enthalpy?
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. |
How would you react to the statement that the electronegativity of N on Pauling scale is 3.0 in all the nitrogen compounds?
Give a brief account of valence ?
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.
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?
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 |
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.
Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:
silicon and oxygen
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.
Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:
aluminium and bromine
Predict the formulae of the stable binary compounds that would be formed by the following pairs of elements:
calcium and iodine
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.
(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.
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.
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).
What do you understand by diagonal relationship?
Show by a chemical reaction with water that Na2O is a basic oxide and Cl2O7 is an acidic oxide.
What is the basic theme of organisation in the periodic table?
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.
What is the basic difference in approach between the Mendeleev’s Periodic Law and the Modern Periodic Law?
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.
On the basis of quantum numbers, justify that the sixth period of the periodic table should have 32 elements.
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.
Explain why cation are smaller and anions larger in radii than their parent atoms?
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.
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 than B
(ii) O has lower than N and F?
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, of Be is Higher than that of
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, 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.
What is the basic difference between the terms electron gain enthalpy and electronegativity?
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.
The first and the second
ionization enthalpies (in kJ mol-1) and the
electron gain enthalpy (in kJ mol-1) of a few elements are given below:
Elements
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)?
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).
Which of the following atoms has the highest first ionisation energy?
Na
K
Sc
Rb
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.
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
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
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
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
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-
C.
Ca2+, K+, Cl-, S2-
For isoelectronic species, as the z/e decreases, ionic radius increases
such as
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+
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.
The charge/size ratio of a cation determines its polarizing power. Which one of the following sequences represents the increasing order of the polarizing power of the cationic species, K+, Ca2+, Mg2+, Be2+?
Mg2+ < Be2+ < K+ < Ca2+
Be2+ < K+ < Ca2+ < Mg2+
K+ < Ca2+ < Mg2+ < Be2+
Ca2+ < Mg2+ < Be2+ < K+
C.
K+ < Ca2+ < Mg2+ < Be2+
Following statements regarding the periodic trends of chemical reactivity of the alkali metals and the halogens are given. Which of these statements gives the correct picture?
The reactivity decreases in the alkali metals but increases in the halogens with increase in atomic number down the group
In both the alkali metals and the halogens the chemical reactivity decreases with increase in atomic number down the group
Chemical reactivity increases with increase in atomic number down the group in both the alkali metals and halogens
In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group
D.
In alkali metals the reactivity increases but in the halogens it decreases with increase in atomic number down the group
In which of the following arrangements the order is NOT according to the property indicated against it?
Al3+ < Mg2+ < Na+ < F- Increasing ionic size
B < C < N < O Increasing first ionization enthalpy
I < Br < F < Cl Increasing electron gain enthalpy (with negative sign)
Li < Na < K < Rb Increasing metallic radius
B.
B < C < N < O Increasing first ionization enthalpy
Which of the following statements in relation to the hydrogen atom is correct?
3s orbital is lower in energy than 3p orbital
3p orbital is lower in energy than 3d orbital
3s and 3p orbitals are of lower energy than 3d orbital
3s, 3p and 3d orbitals all have the same energy
D.
3s, 3p and 3d orbitals all have the same energy
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
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
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)
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.
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+
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.
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
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.
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 )
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+
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.
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.
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
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
Which of the following pairs has the same size?
Fe2+ , Ni2+
Zr4+, Ti4+
Zr4+, Hf4+
Zn2+, Hf4+
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.
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+
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
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
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
Among the following which one has the highest cation to anion size ratio?
CsI
CsF
LiF
NaF
B.
CsF
The order of size of given cations is
Li+ < Na+ < Cs+
and the order of the size of given anions is
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
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.
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
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
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
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
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
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.]
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+
B.
Cs+ > Rb+ > K+ >Na+
The smaller the size of ion, the greater the degree of hydration.
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
Which of the following oxidation states are the most characteristics for lead and tin respectively?
+4,+2
+2,+4
+4,+4
+2,+2
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.
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 |
ii, iii and iv
i, ii, iii and iv
ii, iv and v
i and iii
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' 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 |
With which of the following electronic configuration an atom has the lowest ionisation enthalpy?
1s2 2s22p5
1s2 2s22p3
1s2 2s22p5 3s1
1s2 2s22p6
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.
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-
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-
The element Z = 114 has been discovered recently. It will belong to which of the following family group and electronic configuration?
Halogen family, [Rn] 5f14 6d10 7s2 7p5
Carbon family, [Rn] 5f14 6d10 7s2 7p2
Oxygen family,[Rn] 5f14 6d10 7s2 7p4
Nitrogen family, [Rn] 5f14 6d10 7s2 7p3
B.
Carbon family, [Rn] 5f14 6d10 7s2 7p2
Z = 114 belong to Group 14, carbon family Electronic configuration , [Rn] 5f14 6d10 7s2 7p2
Sponsor Area
Sponsor Area