1
MCQ (Single Correct Answer)

### AIPMT 2009

According to MO theory which of the lists ranks the nitrogen species in terms of increasing bond order ?
A
$N_2^{2 - } < N_2^ - < {N_2}$
B
${N_2} < N_2^{2 - } < N_2^ -$
C
$N_2^ - < N_2^{2 - } < {N_2}$
D
$N_2^ - < {N_2} < N_2^{2 - }$

## Explanation

According to MOT, the molecular orbital electronic configuration of

${N_2}:{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}$

$\therefore$ B.O = ${{10 - 4} \over 2} = 3$

${N_2}^ - :{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}{\left( {{\pi ^*}2{p_x}} \right)^1}$

$\therefore$ B.O = ${{10 - 5} \over 2} = 2.5$

${N_2}^{2 - }:{\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}{\left( {{\pi ^*}2{p_x}} \right)^1}{\left( {{\pi ^*}2{p_y}} \right)^1}$

$\therefore$ B.O = ${{10 - 6} \over 2} = 2$

Hence the order : $N_2^{2-}$ < $N_2^-$ < $N_2$
2
MCQ (Single Correct Answer)

### AIPMT 2009

What is the dominant intermolecular force or bond that must be overcome in converting liquid CH3OH to a gas?
A
Dipole-dipole interaction
B
Covalent bonds
C
London dispersion force
D
Hydrogen bonding

## Explanation

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

$- - - \mathop O\limits^{\mathop |\limits^{C{H_3}} } \hbox{---}H - - - \mathop O\limits^{\mathop |\limits^{C{H_3}} } \hbox{---}H - - - \mathop O\limits^{\mathop |\limits^{C{H_3}} } \hbox{---}H - - -$

As a result the order to convert liquid CH3OH to gaseous state, the strong hydrogen bonds must be broken.
3
MCQ (Single Correct Answer)

### AIPMT 2008

The correct order of increasing bond angles in the following triatomic species is
A
NO$_2^ +$ < NO2 < NO$_2^ -$
B
NO$_2^ +$ < NO$_2^ -$ < NO2
C
NO$_2^ -$ < NO$_2^ +$ < NO2
D
NO$_2^ -$ < NO2 < NO$_2^ +$

## Explanation

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$_2^ -$ < NO2 < NO$_2^ +$
4
MCQ (Single Correct Answer)

### AIPMT 2008

Four diatomic species are listed below. Identify the correct order in which the bond order is increasing in them
A
$NO < {O_2}^ - < {C_2}^{2 - } < He_2^ +$
B
${O_2}^ - < NO < {C_2}^{2 - } < He_2^ +$
C
${C_2}^{2 - } < He_2^ + < {O_2}^ - < NO$
D
$He_2^ + < {O_2}^ - < NO < {C_2}^{2 - }$

## Explanation

Molecular orbital configuration of NO (15 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^1}^ * \, = \,\pi _{2p_y^o}^ *$

$\therefore\,\,\,\,$ Nb = 10

Na = 5

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 5} \right]$ = 2.5

Molecular orbital configuration of $O_2^ -$ (17 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\sigma _{2p_z^2}}\,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,\pi _{2p_x^2}^ * \, = \,\pi _{2p_y^1}^ *$

$\therefore\,\,\,\,$ Nb = 10

Na = 7

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 7} \right]$ = 1.5

Molecular orbital configuration of $C_2^ {2-}$ (14 electrons) is

${\sigma _{1{s^2}}}\,\sigma _{1{s^2}}^ * \,{\sigma _{2{s^2}}}\,\sigma _{2{s^2}}^ * \,{\pi _{2p_x^2}}\, = \,{\pi _{2p_y^2}}\,{\sigma _{2p_z^2}}$

$\therefore\,\,\,\,$ Nb = 10

Na = 4

$\therefore\,\,\,\,$ BO = ${1 \over 2}\left[ {10 - 4} \right]$ = 3

$\,\,\,$ Configuration of $He_2^ +$ (3 electrons) is = ${\sigma _{1{s^2}}}$ $\sigma _{1{s^1}}^ *$

$\therefore\,\,\,$ Bond order = ${1 \over 2}$ (2 $-$1) = 0.5

$\therefore$ Correct order is :

$He_2^ + < {O_2}^ - < NO < {C_2}^{2 - }$

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