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NEET chemistry

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Question
CBSEENCH11008348
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In which of the following pairs of molecules/ions, the central atoms have sp2 hybridization?

  • NO2- and NH3

  • BF3 and NO2-

  • NH2- and H2O

  • BF3 and NH2-

Solution

B.

BF3 and NO2-

(i) NO2- ⇒ 2σ + 1 lp =3, ie, sp2 hybridisation
(ii) NH3 ⇒ 3σ + 1 lp = 4, ie, sp3 hybridisation
(iii) BF3 ⇒ 3σ + 0 lp = 3, i.e sp2 hybridisation
(iv) NH2- ⇒ 2σ + 2 lp = 4 i.e sp3 hybridisation
(v) H2O ⇒ 2σ + 2 lp = 4 i.e sp3 hybridisation
Thus, among the given pairs, only BF3 and NO2- have sp2 hybridizations.

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Question
CBSEENCH11008349
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If pH ofa saturated solution of Ba(OH)2 is 12, the value of its Ksp is 

  • 4.00 x 10-6 M3

  • 4.00 x 10-7 M3

  • 5.00 x 10-6 M3

  • 5.00 x 10-7 M3

Solution

C.

5.00 x 10-6 M3

Given pH of Ba(OH)2 = 12
therefore, [H+] = [1 x 10-12]
and [OH-] = 1 x 10-14 / 1 x 10-12   {[H+][OH-] = 1 x 10-14}
Ba left parenthesis OH right parenthesis subscript 2 space equals space stack Ba to the power of 2 plus end exponent space with straight s below plus space stack 2 OH to the power of minus with 2 straight s below Ksp space equals space left square bracket Ba to the power of 2 plus end exponent right square bracket left square bracket OH to the power of minus right square bracket squared space equals space left square bracket straight s right square bracket left square bracket 2 straight s squared right square bracket equals space open square brackets fraction numerator 1 space straight x space 10 to the power of negative 2 end exponent over denominator 2 end fraction close square brackets left parenthesis 1 space straight x space 10 to the power of negative 2 end exponent right parenthesis squared equals space 0.5 space straight x space 10 to the power of negative 6 end exponent space equals space 5.0 space straight x space 10 to the power of negative 7 end exponent space straight M cubed

Question
CBSEENCH11008350
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Which one of the following species does not exist under normal conditions?

  • Be2+

  • Be2

  • B2

  • Li2

Solution

B.

Be2

Molecules with zero bond order do not exist.
a) Be2+ (4 + 4 -1 = 7) = σ1s2 σ*1s2, σ2s2, σ2s1
BO = 4 - 3 / 2  = 0.5

b) Be2 (4 + 4 = 8 )  = σ1s2, σ*1s2, σ2s2, σ*2s2

BO = 4 - 4 / 2 = 0
c) B2 = (5 + 5) = 10
= σ1s2 , σ*1s2, 2σs2, σ*2s2, π2px1 = π2py1
BO = 6 - 4 /2 = 1
d) Li2 (3 + 3) = 6
= σ1s2 , σ*1s2,σ2s2

BO = 4 - 2 / 2 = 1
Thus, bond order of Be2 does not exist under normal conditions.

Question
CBSEENCH11008353
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The number of atoms in 0.1 moles of a triatomic gas is (NA = 6.02 x 1023 mol-1)

  • 6.026 x 1022

  • 1.806 x 1023

  • 3.600 x 1023

  • 1.800 x 1022

Solution

B.

1.806 x 1023

Number of atoms = number of moles x NA x atomicity
= 0.1 x 6.02 x 1023 x 3
= 1.806 x 1023 atoms

Question
CBSEENCH11008354
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Which of the following represents the correct order of increasing electron gain enthalpy with negative sign for the elements O, S, F and Cl?

  • Cl < F < O < S

  • O < S < F< Cl

  • F < S < O < Cl

  • S < O < Cl < F

Solution

B.

O < S < F< Cl

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

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