1

### AIPMT 2002

Which of the following is isoelectronic?
A
CO2, NO2
B
NO2$-$, CO2
C
CN$-$, CO
D
SO2, CO2

## Explanation

In CO, the number of electrons

$\Rightarrow$ 6 + 8 = 14 [ Z of C = 6 and O = 8]

Electronic configuration of molecular orbital of CO :

${\left( {\sigma 1s} \right)^2}{\left( {{\sigma ^*}1s} \right)^2}{\left( {\sigma 2s} \right)^2}{\left( {{\sigma ^*}2s} \right)^2}{\left( {\pi 2{p_x}} \right)^2}{\left( {\pi 2{p_y}} \right)^2}{\left( {\pi 2{p_z}} \right)^2}$

CN- also get (6 + 7 + 1) 14 electrons and the configuration is similar to that CO.

So, CN and CO is isoelectronic.
2

### AIPMT 2002

In NO3$-$ ion number of bond pair and lone pair of electrons on nitrogen atom are
A
2, 2
B
3, 1
C
1, 3
D
4, 0

## Explanation

${O^ - }\hbox{---}\mathop N\limits^{\mathop \uparrow \limits^O } = O$

In NO3- ion, nitrogen has 4 bond pair of electrons and no lone pair of electrons.
3

### AIPMT 2002

Which of the following has p$\pi$ $-$ d$\pi$ bonding?
A
NO3$-$
B
SO32$-$
C
BO33$-$
D
CO32$-$

## Explanation

Electronic structure of S atom in excited state In 'S' unhybride d- orbital is present, which will involved in bond formation with oxygen atom.

In oxygen two unpaired p- orbital is present in these one is involved in $\sigma$ bond formation while other is used in $\pi$ bond formation.

So, in SO32$-$, p$\pi$ $-$ d$\pi$ bonding present.
4

### AIPMT 2001

In X - H --- Y, X and Y both are electronegative elements. Then
A
electron density on X will increase and on H will decrease
B
in both electron density will increase
C
in both electron density will decrease
D
on X electron density will decrease and on H increases.

## Explanation

As X and Y both are electronegative elements thus both attracts the electron density from H thus, electron density on H decreases and on X it increases.