1
MCQ (Single Correct Answer)

AIPMT 2010 Prelims

In which of the following pairs of molecules/ ions, the central atoms have sp2 hybridisation ?
A
NO$$_2^{ - }$$ and NH3
B
BF3 and NO$$_2^{ - }$$
C
NH$$_2^{ - }$$ and H2O
D
BF3 and NH$$_2^{ - }$$

Explanation

The hybridisation of the central atom can be calculated as

H = $${1 \over 2}\left[ \matrix{ \left( \matrix{ No.\,of\,electrons \hfill \cr in\,valence\,shell \hfill \cr of\,atom \hfill \cr} \right) + \left( \matrix{ No.\,of\,monovalent \hfill \cr atoms\,around \hfill \cr central\,atom \hfill \cr} \right) \hfill \cr - \left( \matrix{ Ch{\mathop{\rm arge}\nolimits} \,on \hfill \cr cation \hfill \cr} \right) + \left( \matrix{ Ch{\mathop{\rm arge}\nolimits} \,on \hfill \cr anion \hfill \cr} \right) \hfill \cr} \right]$$

$$ \therefore $$ For BF3, H = $${1 \over 2}\left[ {(3) + (3) - (0) + (0)} \right] \Rightarrow 3 \Rightarrow $$ sp2 hybridisation

$$ \therefore $$ For NO2-, H = $${1 \over 2}\left[ {(5) + (0) - (0) + (1)} \right] \Rightarrow 3 \Rightarrow $$ sp2 hybridisation
2
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 : $$N_2^{2-}$$ < $$N_2^-$$ < $$N_2$$
3
MCQ (Single Correct Answer)

AIPMT 2009

In which of the following molecules/ions BF3, NO$$_2^ - $$, NH$$_2^ - $$ and H2O, the central atom is sp2 hybridised?
A
NH$$_2^ - $$ and H2O
B
NO$$_2^ - $$ and H2O
C
BF3 and NO$$_2^ - $$
D
NO$$_2^ - $$ and NH$$_2^ - $$

Explanation

BF3 $$ \to $$ sp2

NO2- $$ \to $$ sp2

NH2- $$ \to $$ sp3

H2O $$ \to $$ sp3
4
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.

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