1
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.
2
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$$_2^ + $$
3
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 :

$$He_2^ + < {O_2}^ - < NO < {C_2}^{2 - }$$
4
MCQ (Single Correct Answer)

AIPMT 2007

The correct order of C $$-$$ O bond length among CO, CO$${_3^{2 - }}$$, CO2 is
A
CO < CO$${_3^{2 - }}$$ < CO2
B
CO$${_3^{2 - }}$$ < CO2 < CO
C
CO < CO2 < CO$${_3^{2 - }}$$
D
CO2 < CO$${_3^{2 - }}$$ < CO

Explanation

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 $$<$$ CO2 $$<$$ CO$${_3^{2 - }}$$

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