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1

### 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

### 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

### 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

### 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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NEET

Class 12