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Chemical Bonding & Molecular Structure

NEET Chemistry · 95 questions · Page 7 of 10 · Click an option or "Show Solution" to reveal answer

Q61
Which one of the following species does not exist under normal conditions?
A Be 2 +
B Be 2
C B 2
D Li 2
Correct Answer
Option B
Solution

Be 2 does not exists Be 2 has an electronic configuration of :

σ1s2σ∗1s2σ2s2σ∗2s2\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}

∴\therefore Bond order =

4−42=0{{4 - 4} \over 2} = 0

Thus, Be 2 does not exists

Q62
In which one of the following species the central atom has the type of hybridization which is not the same as that present in the other three ?
A SF 4
B I 3 −-
C SbCl 5 2−-
D PCl 5
Correct Answer
Option C
Solution

The hybridisation of the central atom can be calculated as H =

12[(No. of electronsin valence shellof atom)+(No. of monovalentatoms aroundcentral atom)−(Charge oncation)+(Charge onanion)]{1 \over 2}\left[ \begin{array}{ll}\left( \begin{array}{ll}No.\,of\,electrons \\ in\,valence\,shell \\ of\,atom \end{array} \right) + \left( \begin{array}{ll}No.\,of\,monovalent \\ atoms\,around \\ central\,atom \end{array} \right) \\ - \left( \begin{array}{ll}Ch{\mathop{\rm arge}\nolimits} \,on \\ cation \end{array} \right) + \left( \begin{array}{ll}Ch{\mathop{\rm arge}\nolimits} \,on \\ anion \end{array} \right) \end{array} \right]

Applying that formula we find that all the given species except [SbCl 5 ] 2- have central atom with sp 3 d ( corresponding to H = 5) hybridisation.

In [SbCl 5 ] 2- Sb is sp 3 d 2 hybridized.

Q63
According to MO theory which of the lists ranks the nitrogen species in terms of increasing bond order ?
A N22−<N2−<N2N_2^{2 - } < N_2^ - < {N_2}
B N2<N22−<N2−{N_2} < N_2^{2 - } < N_2^ -
C N2−<N22−<N2N_2^ - < N_2^{2 - } < {N_2}
D N2−<N2<N22−N_2^ - < {N_2} < N_2^{2 - }
Correct Answer
Option A
Solution

According to MOT, the molecular orbital electronic configuration of

N2:(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(π2pz)2{N_2}:{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}

∴\therefore B.O =

10−42=3{{10 - 4} \over 2} = 3
N2−:(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(π2pz)2(π∗2px)1{N_2}^ - :{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}{\left( {{\pi ^*}2{p_x}} \right)^1}

∴\therefore B.O =

10−52=2.5{{10 - 5} \over 2} = 2.5
N22−:(σ1s)2(σ∗1s)2(σ2s)2(σ∗2s)2(π2px)2(π2py)2(π2pz)2(π∗2px)1(π∗2py)1{N_2}^{2 - }:{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}{\left( {{\pi ^*}2{p_x}} \right)^1}{\left( {{\pi ^*}2{p_y}} \right)^1}

∴\therefore B.O =

10−62=2{{10 - 6} \over 2} = 2

Hence the order :

N22−N_2^{2-}

<

N2−N_2^-

<

N2N_2
Q64
What is the dominant intermolecular force or bond that must be overcome in converting liquid CH 3 OH to a gas?
A Dipole-dipole interaction
B Covalent bonds
C London dispersion force
D Hydrogen bonding
Correct Answer
Option D
Solution

Methanol can undergo intermolecular association through H-bonding as the - OH group in alcoholos is highly polarised.

−−−O∣CH3—H−−−O∣CH3—H−−−O∣CH3—H−−−- - - \mathop O\limits^{\mathop |\limits^{C{H_3}} } \hbox{---}H - - - \mathop O\limits^{\mathop |\limits^{C{H_3}} } \hbox{---}H - - - \mathop O\limits^{\mathop |\limits^{C{H_3}} } \hbox{---}H - - -

As a result the order to convert liquid CH 3 OH to gaseous state, the strong hydrogen bonds must be broken.

Q65
In which of the following molecules/ions BF 3 , NO2−_2^ - , NH2−_2^ - and H 2 O, the central atom is sp 2 hybridised?
A NH2−_2^ - and H 2 O
B NO2−_2^ - and H 2 O
C BF 3 and NO2−_2^ -
D NO2−_2^ - and NH2−_2^ -
Correct Answer
Option C
Solution

BF 3 →\to sp 2 NO 2 - →\to sp 2 NH 2 - →\to sp 3 H 2 O →\to sp 3

Q66
The correct order of increasing bond angles in the following triatomic species is
A NO2+_2^ + < NO 2 < NO2−_2^ -
B NO2+_2^ + < NO2−_2^ - < NO 2
C NO2−_2^ - < NO2+_2^ + < NO 2
D NO2−_2^ - < NO 2 < NO2+_2^ +
Correct Answer
Option D
Solution

We know that higher the number of lone pair of electron on central atom, greater is the lp – lp repulsion between Nitrogen and oxygen atoms.

Thus smaller is bond angle.

The correct order of bond angle is NO

2−_2^ -

< NO 2 < NO

2+_2^ +
Q67
Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them
A NO<O2−<C22−<He2+NO < {O_2}^ - < {C_2}^{2 - } < He_2^ +
B O2−<NO<C22−<He2+{O_2}^ - < NO < {C_2}^{2 - } < He_2^ +
C C22−<He2+<O2−<NO{C_2}^{2 - } < He_2^ + < {O_2}^ - < NO
D He2+<O2−<NO<C22−He_2^ + < {O_2}^ - < NO < {C_2}^{2 - }
Correct Answer
Option D
Solution

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\,\,\,\,

N b = 10 N a = 5

∴    \therefore\,\,\,\,

BO =

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

= 2.5 Molecular orbital configuration of

O2−O_2^ -

(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}^ *
∴    \therefore\,\,\,\,

N b = 10 N a = 7

∴    \therefore\,\,\,\,

BO =

12[10−7]{1 \over 2}\left[ {10 - 7} \right]

= 1.5 Molecular orbital configuration of

C22−C_2^ {2-}

(14 electrons) is

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

N b = 10 N a = 4

∴    \therefore\,\,\,\,

BO =

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

= 3

   \,\,\,

Configuration of

He2+He_2^ +

(3 electrons) is =

σ1s2{\sigma _{1{s^2}}}
σ1s1∗\sigma _{1{s^1}}^ *
∴   \therefore\,\,\,

Bond order =

12{1 \over 2}

(2 −-1) = 0.5 ∴\therefore Correct order is :

He2+<O2−<NO<C22−He_2^ + < {O_2}^ - < NO < {C_2}^{2 - }
Q68
The correct order of C −- O bond length among CO, CO32−{_3^{2 - }}, CO 2 is
A CO < CO32−{_3^{2 - }} < CO 2
B CO32−{_3^{2 - }} < CO 2 < CO
C CO < CO 2 < CO32−{_3^{2 - }}
D CO 2 < CO32−{_3^{2 - }} < CO
Correct Answer
Option C
Solution

All these structures exhibits resonance and can be represented by the following resonating structures.

More single bond character in resonance hybrid, more is the bond length.

Hence the increasing bond length is CO << CO 2 << CO

32−{_3^{2 - }}
Q69
In which of the following pairs, the two species are isostructural?
A SO32−{_3^{2 - }} and NO3−{_3^{ - }}
B BF 3 and NF 3
C BrO3−{_3^{ - }} and XeO 3
D SF 4 and XeF 4
Correct Answer
Option C
Solution

Hybridisation of Br in BrO 3 – : =

12(7+0−0+1){1 \over 2}\left( {7 + 0 - 0 + 1} \right)

= 4 Four hybrid orbital means sp 3 hybridisation. Hybridisation of Xe in XeO 3 : =

12(8+0−0+0){1 \over 2}\left( {8 + 0 - 0 + 0} \right)

= 4 Four hybrid orbital means sp 3 hybridisation.

Thus, both BrO 3 – and XeO 3 are sp 3 hybridised with three bond pairs of electrons and one lone pair of electrons and results in trigonal pyramidal shape.

Q70
Which of the following is not isostructural with SiCl 4 ?
A NH 4 −-
B SCl 4
C SO 4 2−-
D PO 4 3−-
Correct Answer
Option B
Solution

SiCl 4 , NH 4 + , SO 4 2- and PO 4 3- ions are the examples of molecules/ions which are of AB 4 type and have tetrahedral structure.

SCl 4 is AB 4 (lone pair) types species.

Although the arrangement of five sp 3 d hybrid orbitals in space is trigonal bipyramidal, due to the presense of one lone pair of electron in the basal hybrid orbital, the shape of AB 4 (lone pair) species gets distorted and becomes distorted tetrahedral or see-saw.

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