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1

AIPMT 2010 Prelims

MCQ (Single Correct Answer)
Oxidation states of P in H4P2O5, H4P2O6, H4P2O7 are respectively
A
+3, +5, +4
B
+5, +3, +4
C
+5, +4, +3
D
+3, +4, +5

Explanation

The oxidation state can be calculated as :

H4P2O5 : +4 + 2x + 5(– 2) = 0 $$ \Rightarrow $$ 2x – 6 = 0 $$ \Rightarrow $$ x = +3

H4P2O6 : +4 + 2x + 6(– 2) = 0 $$ \Rightarrow $$ 2x – 8 = 0 $$ \Rightarrow $$ x = +4

H4P2O7 : +4 + 2x + 7(– 2) = 0 $$ \Rightarrow $$ 2x – 10 = 0 $$ \Rightarrow $$ 2x = 10 $$ \Rightarrow $$ x = +5
2

AIPMT 2010 Prelims

MCQ (Single Correct Answer)
The correct order of increasing bond angles in the following species is
A
Cl2O < ClO2 < ClO$$_2^ - $$
B
ClO2 < Cl2O < ClO$$_2^ - $$
C
Cl2O < ClO$$_2^ - $$ < ClO2
D
ClO$$_2^ - $$ < Cl2O < ClO2

Explanation

The correct order of increasing bond angles is

ClO$$_2^ - $$ < Cl2O < ClO2
3

AIPMT 2010 Prelims

MCQ (Single Correct Answer)
The tendency of BF3, BCl3 and BBr3 to behave as Lewis acid decreases in the sequence
A
BCl3 > BF3 > BBr3
B
BBr3 > BCl3 > BF3
C
BBr3 > BF3 > BCl3
D
BF3 > BCl3 > BBr3

Explanation

In BF3, p-p overlap between B and F is maximum due to identical size and energy of p-orbitals, so electron deficiency in boron of BF3 is neutralized partially to the maximum extent by back donation. Also, the tendency to back donate decreases from F to I. So the order will be:

BBr3 > BCl3 > BF3
4

AIPMT 2010 Prelims

MCQ (Single Correct Answer)
Which one of the following molecular hydrides acts as a Lewis acid?
A
NH3
B
H2O
C
B2H6
D
CH4

Explanation

Among the given molecules, only diborane(B2H6) is electron deficient i.e., it does not complete octet.

Thus, it acts as a Lewis acid.

NH3 and H2O being electron rich species behave as Lewis base.

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