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NEET chemistry

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Question
CBSEENCH11008377
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Which of the following molecules acts as a Lewis acid?

  • (CH)3B

  • (CH3)O

  • (CH3)P

  • (CH3)2)O

Solution

A.

(CH)3B

According to Lewis concept " Acids are electron acceptor and bases are electron donor" Since, electron deficient compounds also have an ability to accept electrons, these are regarded as acids.
Trimethylborane (CH)3B have incomplete octet thus, act as Lewis acid.

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Question
CBSEENCH11008378
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From the following bond energies:

H-H bond energy: 431.37 kJ mol-

C=C  bond energy: 606.10 kJ mol-

C- C bond energy: 336.49 kJ mol-

C-H bond energy: 410.50 kJ mol-

Enthalpy for the reaction,


will be


  • 1523.6 kJ  mol-

  • -243.6 kJ mol-

  • -1200. kJ mol-

  • 553.0 kJ mol-

Solution

C.

-1200. kJ mol-

For a reaction

increment straight H subscript straight r space equals space left square bracket space 4 space straight x space BE subscript left parenthesis straight C minus straight H right parenthesis end subscript space plus space 1 space straight x space BE subscript straight C equals straight C space end subscript space plus space 1 space straight x space BE subscript left parenthesis straight H minus straight H right parenthesis end subscript right square bracket minus 6 straight x space BE subscript left parenthesis straight C minus straight H right parenthesis end subscript space plus space 1 space straight X space BE subscript left parenthesis straight C minus straight C right parenthesis end subscript equals space left parenthesis space 4 space straight x space 410.50 space plus space 1 space straight x space 606.10 space plus space 1 space straight x space 431.37 right parenthesis minus left square bracket 6 space straight x space 410.50 right parenthesis space plus 1 space straight x space 336.49 right parenthesis right square bracket space equals negative space 120.0 space kJ space mol to the power of minus

Question
CBSEENCH11008379
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Which one of the element with the following outer orbital configuration may exhibit the largest number of oxidation states?

  • 3d5,4s2

  • 3d5,4s1

  • 3d5,4s2

  • 3d2,4s2

Solution

C.

3d5,4s2

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

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

Question
CBSEENCH11008380
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The ionisation constant of ammonium hydroxide is 1.77 x 10-5 at 298 K. Hydrolysis constant of ammonium chloride is 

  • 5.65 x 10-10

  • 6.50 x 10-12

  • 5.65 x 10-13

  • 5.65 x 10-12

Solution

A.

5.65 x 10-10

Given comma space straight K subscript straight a space left parenthesis NH subscript 4 OH right parenthesis space equals space 1.77 space straight x space 10 to the power of negative 5 end exponent NH subscript 4 OH space leftwards harpoon over rightwards harpoon space NH subscript 4 to the power of plus space plus space OH to the power of minus straight K subscript straight a space equals space fraction numerator left square bracket NH to the power of plus subscript 4 right square bracket left square bracket OH to the power of minus right square bracket over denominator left square bracket NH subscript 4 OH right square bracket end fraction space equals space 1.77 space straight x space 10 to the power of negative 5 end exponent space... space left parenthesis straight i right parenthesis
Hydrolysis of NH4Cl takes place as,
NH4Cl + H2O → NH4OH + HCl
or 
NH4+ +H2O → NH4OH + H+
Hydrolysis constant, Kh

straight K subscript straight h space equals space fraction numerator left square bracket NH subscript 4 OH right square bracket left square bracket straight H to the power of plus right square bracket over denominator left square bracket NH subscript 4 to the power of plus right square bracket end fraction space... space left parenthesis ii right parenthesis or space straight K subscript straight h space equals space fraction numerator left square bracket NH subscript 4 OH to the power of plus right square bracket left square bracket OH to the power of minus right square bracket over denominator left square bracket NH subscript 4 to the power of plus right square bracket left square bracket OH to the power of minus right square bracket end fraction space... space left parenthesis iii right parenthesis from space eqs. space left parenthesis straight i right parenthesis space and space left parenthesis iii right parenthesis straight K subscript straight h space equals space straight K subscript straight w over straight K subscript straight a space left square bracket space because space left square bracket OH to the power of minus right square bracket left square bracket straight H to the power of plus right square bracket space equals space straight K subscript straight w right square bracket equals space fraction numerator 10 to the power of negative 14 end exponent over denominator 1.77 space straight x space 10 to the power of negative 5 end exponent end fraction space equals space 5.65 space straight x space 10 to the power of minus to the power of 10

Question
CBSEENCH11008381
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Oxidation number of P in PO43-, of S in SO42- and that of Cr in Cr2O72- are respectively,

  • +5,+6 and +6

  • +3, + 6 and +5

  • +5,+3 and +6

  • -3,+6 and +6

Solution

A.

+5,+6 and +6

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

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