1

### 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
2

### 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

### 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

### 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 :

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