1
MCQ (Single Correct Answer)

### NEET 2016 Phase 2

The hybridizations of atomic orbitals of nitrogen in NO$_2^ +$, NO$_3^ -$  and  NH$_4^ +$  respectively are
A
sp, sp3 and sp2
B
sp2, sp3 and sp3
C
sp, sp2 and sp3
D
sp2, sp and sp3

## Explanation

X = $1 \over 2$ (VE + MA - c + a)

For $NO^+_2$, X = $1 \over 2$ (5 + 0 - 1) = 2, so sp hybridisation

For $NO^-_3$, X = $1 \over 2$ (5 + 0 + 1) = 3, so sp2 hybridisation

For $NH^+_4$, X = $1 \over 2$ (5 + 4 - 1) = 4, so sp3 hybridisation

2
MCQ (Single Correct Answer)

### NEET 2016 Phase 2

Which one of the following compounds shows the presence of intramolecular hydrogen bond?
A
H2O2
B
HCN
C
Cellulose
D
Concentrated acetic acid

## Explanation

H2O2, HCN and conc. CH3COOH form intermolecular hydrogen bonding while cellulose has intramolecular hydrogen bonding.
3
MCQ (Single Correct Answer)

### AIPMT 2015 Cancelled Paper

Which of the following species contains equal number of $\sigma$- and $\pi$-bonds ?
A
(CN)2
B
CH2(CN)2
C
HCO3$-$
D
XeO4

## Explanation

Number of $\sigma$ bonds = 4

Number of $\pi$ bonds = 4
4
MCQ (Single Correct Answer)

### AIPMT 2015

Decreasing order of stability of O2, O2$-$, O2+ and O22$-$ is
A
O22$-$  >  O2$-$  >  O2  >  O2+
B
O2  >  O2+  >  O22$-$  >  O2$-$
C
O2$-$  >  O22$-$  >  O2+  >  O2
D
O2+  >  O2  >  O2$-$  >  O22$-$

## Explanation

According to molecular orbital theory as bond order decreases stability of the molecule decreases.

Bond order = $1 \over 2$(Nb - Na)

Bond order of $O_2^+$ = $1 \over 2$(10 - 5) = 2.5

Bond order of $O_2$ = $1 \over 2$(10 - 6) = 2

Bond order of $O_2^-$ = $1 \over 2$(10 - 7) = 1.5

Bond order of $O_2^{2-}$ = $1 \over 2$(10 - 8) = 1.0

Hence the correct order is $O_2^+$ > $O_2$ > $O_2^-$ > $O_2^{2-}$

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