1
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
2
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.
3
MCQ (Single Correct Answer)

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
MCQ (Single Correct Answer)

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 :

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

EXAM MAP

Joint Entrance Examination

JEE Advanced JEE Main

Graduate Aptitude Test in Engineering

GATE CE GATE ECE GATE ME GATE IN GATE EE GATE CSE GATE PI

Medical

NEET

CBSE

Class 12