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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,

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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
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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

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Hydrolysis of NH4Cl takes place as,
NH4Cl + H2O → NH4OH + HCl
or 
NH4+ +H2O → NH4OH + H+
Hydrolysis constant, Kh

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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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