1

AIPMT 2010 Prelims

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
SF4
B
I3$-$
C
SbCl52$-$
D
PCl5

Explanation

The hybridisation of the central atom can be calculated as

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

Applying that formula we find that all the given species except [SbCl5]2- have central atom with sp3d ( corresponding to H = 5) hybridisation. In [SbCl5]2- Sb is sp3d2 hybridized.
2

AIPMT 2010 Prelims

Which one of the following species does not exist under normal conditions?
A
Be2+
B
Be2
C
B2
D
Li2

Explanation

Be2 does not exists

Be2 has an electronic configuration of :

$\sigma 1{s^2}{\sigma ^*}1{s^2}\sigma 2{s^2}{\sigma ^*}2{s^2}$

$\therefore$ Bond order = ${{4 - 4} \over 2} = 0$

Thus, Be2 does not exists
3

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
4

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 : $N_2^{2-}$ < $N_2^-$ < $N_2$