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Chemical Bonding & Molecular Structure

JEE Chemistry · 150 questions · Page 6 of 15 · Click an option or "Show Solution" to reveal answer

Q51
The correct statement among the following is :
A (SiH3)3N is planar and more basic than (CH3)3N
B (SiH3)3N is pyramidal and more basic than (CH3)3N
C (SiH3)3N is pyramidal and less basic than (CH3)3N
D (SiH3)3N is planar and less basic than (CH3)3N
Correct Answer
Option D
Solution

In (CH3)3N, one lone pair and 3 sigma bond present.

So it's shape is pyramidal.

But in (SiH3)3N, due to backbonding lone pair of N shift to vacant 3d orbital of Si.

So now N have only 3 sigma bond and hybridization is sp2, that is why it has triangular planar geometry.

As in (SiH3)3N. nitrogen(N) don't have any lone pair to donate but in (CH3)3N, lone pair of nitrogen(N) can easily be donated.

That is why (CH3)3N is more basic.

Q52
During the change of O2 to O2- , the incoming electron goes to the orbital :
A π2py\pi 2{p_y}
B σ∗2pz{\sigma ^*}2{p_z}
C π∗2px{\pi ^*}2{p_x}
D π2px\pi 2{p_x}
Correct Answer
Option C
Solution

Molecular orbital configuration of O2 (16 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2py1∗{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^1}^ *

Molecular orbital configuration of O2- (17 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px2∗ = π2py1∗{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ *

So, the incoming electron goes to

π2px∗\pi _{2p_x}^ *

or

π2py∗\pi _{2p_y}^ *
Q53
HF has highest boiling point among hydrogen halides, because it has :
A lowest dissociation enthalpy
B strongest hydrogen bonding
C lowest ionic character
D strongest van der Waals' interactions
Correct Answer
Option B
Solution

Due to strong H-bonding between HF molecules, HF has highest boiling point among the hydrogen halides.

Q54
Among the following, the molecule expected to be stabilized by anion formation is : C2, O2, NO, F2
A C2
B NO
C O2
D F2
Correct Answer
Option A
Solution

C2 has 12 electrons. Moleculer orbital configuration of C2 is

σ1s2 σ1s2∗  σ2s2  σ2s2∗  π2px2= π2py2{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}

Bond order of C2 =

8−42{{8 - 4} \over 2}

= 2 C2- has 13 electrons. Moleculer orbital configuration of C2- is

σ1s2 σ1s2∗  σ2s2  σ2s2∗  π2px2= π2py2{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,\,{\sigma _{2{s^2}}}\,\,\sigma _{2{s^2}}^ * \,\,{\pi _{2p_x^2}} =\,{\pi _{2p_y^2}}
σ2pz1{\sigma _{2p_z^1}}

Bond order of C2- =

9−42{{9 - 4} \over 2}

= 2.5 As we know, higher is the bond order more will be the stability of the molecule.

So we can say, stability of C2 increases when it forms C2- ion.

Q55
The oxoacid of sulphur that does not contain bond between sulphur atoms is
A H2S2O7
B H2S2O3
C H2S4O6
D H2S2O4
Correct Answer
Option A
Solution

Here you can see, H2S2O7 does not have S – S linkage.

Q56
Of the species, NO, NO+, NO2+ and NO- , the one with minimum bond strength is :
A NO–
B NO
C NO+
D NO2+
Correct Answer
Option A
Solution

Note : (1)

    \,\,\,\,

Bond strength ∝\propto Bond order (2)

    \,\,\,\,

Bond length ∝\propto

1Bond  order{1 \over {Bond\,\,order}}

(3)

 \,

Bond order

=12= {1 \over 2}

[Nb −- Na] Nb = No of electrons in bonding molecular orbital Na == No of electrons in anti bonding molecular orbital (4)

    \,\,\,\,

upto 14 electrons, molecular orbital configuration is Here Na = Anti bonding electron == 4 and Nb = 10 (5)

    \,\,\,\,

After 14 electrons to 20 electrons molecular orbital configuration is - - - Molecular orbital configuration of NO (15 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2pyo∗{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ *
∴    \therefore\,\,\,\,

Nb = 10 Na = 5

∴    \therefore\,\,\,\,

BO =

12[10−5]{1 \over 2}\left[ {10 - 5} \right]

= 2.5 Similarly Molecular orbital configuration of NO+ (14 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 {\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,
∴    \therefore\,\,\,\,

Nb = 10 Na = 4

∴    \therefore\,\,\,\,

BO =

12[10−4]{1 \over 2}\left[ {10 - 4} \right]

= 3 Similarly Molecular orbital configuration of NO2+ (13 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py1 {\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^1}}\,
∴    \therefore\,\,\,\,

Nb = 9 Na = 4

∴    \therefore\,\,\,\,

BO =

12[9−4]{1 \over 2}\left[ {9 - 4} \right]

= 2.5 Molecular orbital configuration of NO- (16 electrons) is

σ1s2 σ1s2∗ σ2s2 σ2s2∗ σ2pz2 π2px2 = π2py2 π2px1∗ = π2py1∗{\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^1}^ *
∴    \therefore\,\,\,\,

Nb = 10 Na = 6

∴    \therefore\,\,\,\,

BO =

12[10−6]{1 \over 2}\left[ {10 - 6} \right]

= 2 As Bond strength ∝\propto Bond order ∴\therefore NO– will have minimum bond strength.

Q57
The hybridisations of the atomic orbitals of nitrogen in NO2−_2^ - , NO2+_2^ + and NH4+_4^ + respectively are.
A sp3, sp2 and sp
B sp, sp2 and sp3
C sp3, sp and sp2
D sp2, sp and sp3
Correct Answer
Option D
Solution

Hybridisation = Number of sigma bonds + number of lone pair of electrons + number of coordinate bond. For

NO2−NO_2^ -

hybridisation = 2 sigma NO bonds + 1 lone pair on N = 3; so

NO2−NO_2^ -

is sp2-hybridised. For

NO2+NO_2^ +

hybridisation = 2 sigma NO bonds = 2; so

NO2+NO_2^ +

sp-hybridised. For

NO4+NO_4^ +

Hybridisation = 4 sigma NH bonds = 4; so

NO4+NO_4^ +

is sp3-hybridised. Therefore, hybridisation of N in

NO2−NO_2^ -

,

NO2+NO_2^ +

,

NO4+NO_4^ +

are sp2, sp, sp3 respectively.

Q58
The shape / structure of [XeF5]– and XeO3F2, respectively, are
A Pentagonal planar and trigonal bipyramidal
B Trigonal bipyramidal and pentagonal planar
C Octahedral and square pyramidal
D Trigonal bipyramidal and trigonal bipyramidal
Correct Answer
Option A
Solution

[XeF5]– →\to 5BP + 2LP →\to sp3d3 hybridisation →\to Pentagonal planar XeO3F2 →\to 5BP + 0LP →\to sp3d hybridisation →\to Trigonal bipyramidal

Q59
The molecular geometry of SF6 is octahedral. What is the geometry of SF4 (including lone pair(s) of electrons, if any)?
A Tetrahedral
B Trigonal bipyramidal
C Square planar
D Pyramidal
Correct Answer
Option B
Solution

Here 4σ\sigma bonds +1 lone pair Electronic geometry is Trigonal bipyramidal.

Moleculer Geometry or Moleculer Shape is see-saw.

Note : The difference between Electronic Geometry and Moleculer Geometry(Moleculer Shape) is that in Electronic Geometry we include lone pair but in Moleculer Geometry(Moleculer Shape) we do not include lone pair.

Q60
The dipole of CCl4, CHCl3 and CH4 are in the order :
A CCl4 < CH4 < CHCl3
B CH4 = CCl4 < CHCl3
C CH4 < CCl4 < CHCl3
D CHCl3 < CH4= CCl4
Correct Answer
Option B
Solution

CHCl3 is polar while CH4 and CCl4 are nonpolar. So, dipole moment order is : CH4 = CCl4 < CHCl3

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