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1

NEET 2017

MCQ (Single Correct Answer)
Match the interhalogen compounds of column-I with the geometry in column-II and assign the correct code.

Column $${\rm I}$$ Column $${\rm I}$$$${\rm I}$$
(A) XX' (i) T-shape
(B) XX'3 (ii) Pentagonal bipyramidal
(C) XX'5 (iii) Linear
(D) XX'7 (iv) Square pyramidal
(v) Tetrahedral
A
(A) $$ \to $$ (iii);  (B) $$ \to $$ (i);  (C) $$ \to $$ (iv);  (D) $$ \to $$ (ii)
B
(A) $$ \to $$ (v);  (B) $$ \to $$ (iv);  (C) $$ \to $$ (iii);  (D) $$ \to $$ (ii)
C
(A) $$ \to $$ (iv);  (B) $$ \to $$ (iii);  (C) $$ \to $$ (ii);  (D) $$ \to $$ (i)
D
(A) $$ \to $$ (iii);  (B) $$ \to $$ (iv);  (C) $$ \to $$ (i);  (D) $$ \to $$ (ii)

Explanation

XX' $$ \to $$ Linear (e.g. ClF, BrF)

XX3' $$ \to $$ T-Shape (e.g. ClF3, BrF3)

XX5' $$ \to $$ Square pyramidal (e.g. BrF5, IF5)

XX7' $$ \to $$ Pentagonal bipyramidal (e.g. IF7)
2

NEET 2017

MCQ (Single Correct Answer)
It is because of inability of ns2 electrons of the valence shell to participate in bonding that
A
Sn2+ is oxidising while Pb4+ is reducing
B
Sn2+ and Pb2+ are both oxidising and reducing
C
Sn4+ is reducing while Pb4+ is oxidising
D
Sn2+ is reducing while Pb4+ is oxidising

Explanation

The inertness of s-subshell electrons towards bond formation is known as inert pair effect. This effect increases down the group thus, for Sn, +4 oxidation state is more stable, whereas, for Pb, +2 oxidation state is more stable, i.e., Sn2+ is reducing while Pb4+ is oxidising
3

NEET 2016 Phase 1

MCQ (Single Correct Answer)
Among the following the correct order of acidity is
A
HClO2 < HClO < HClO3 < HClO4
B
HClO4 < HClO2 < HClO < HClO3
C
HClO3 < HClO4 < HClO2 < HClO
D
HClO < HClO2 < HClO3 < HClO4

Explanation

The acidic character of the oxoacids increases with increase in oxidation number of the halogen atom i.e.,

HClO < HClO2 < HClO3 < HClO4

Since the stability of the anion decreases in the order.

ClO4- > ClO3- > ClO2- > ClO

Acid strength also decreases in the same order.
4

NEET 2016 Phase 1

MCQ (Single Correct Answer)
Which one of the following orders is correct for the bond dissociation enthalpy of halogen molecules?
A
Br2 > I2 > F2 > Cl2
B
F2 > Cl2 > Br2 > I2
C
I2 > Br2 > Cl2 > F2
D
Cl2 > Br2 > F2 > I2

Explanation

Bond dissociation enthalpy decreases as the bond distance increases from F2 to I2. This is due to increase in the size of the atom, on moving from F to I.

F – F bond dissociation enthalpy is smaller then Cl – Cl and even smaller than Br – Br. This is because F atom is very small and hence the three lone pairs of electrons on each F atom repel the bond pair holding the F-atoms in F2 molecules.

The order of bond dissociation enthalpy is :

Cl2 > Br2 > F2 > I2

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