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 :

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

EXAM MAP

Joint Entrance Examination

JEE Advanced JEE Main

Medical

NEET

Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI