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1

AIPMT 2002

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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

MCQ (Single Correct Answer)
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.

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