1
MCQ (Single Correct Answer)

NEET 2017

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
MCQ (Single Correct Answer)

NEET 2017

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
MCQ (Single Correct Answer)

NEET 2016 Phase 1

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
MCQ (Single Correct Answer)

NEET 2016 Phase 1

When copper is heated with conc. HNO3 it produces
A
Cu(NO3)2, NO and NO2
B
Cu(NO3)2 and N2O
C
Cu(NO3)2 and NO2
D
Cu(NO3)2 and NO

Explanation

Cu +4HNO3 (conc.) $$ \to $$ Cu(NO3)2 + 2NO2 + 2H2O

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