Match List - I with List - II: List - I (Species) (a) $\mathrm{SF}_{4}$ (b) $\mathrm{IF}_{5}$ (c) $\mathrm{NO}_{2}^{+}$ (d) $\mathrm{NH}_{4}^{+}$ List - II (Hybrid Orbitals) (i) $\mathrm{sp}^{3}\mathrm{d}^{2}$ (ii) $\mathrm{d}^{2}\mathrm{sp}^{3}$ (iii) $\mathrm{sp}^{3}\mathrm{d}$ (iv) $\mathrm{sp}^{3}$ (v) sp Choose the correct answer from the options given below:

(A) (a)-(iii), (b)-(i), (c)-(v) and (d)-(iv)

(B) (a)-(ii), (b)-(i), (c)-(iv) and (d)-(v)

(C) (a)-(i), (b)-(ii), (c)-(v) and (d)-(iii)

(D) (a)-(iv), (b)-(iii), (c)-(ii) and (d)-(v)

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Chemical Bonding and Molecular Structure

Match List - I with List - II:

List - I (Species)
(a)

SF_{4}

(b)

IF_{5}

(c)

NO_{2}^{+}

(d)

NH_{4}^{+}

List - II (Hybrid Orbitals)
(i)

sp^{3} d^{2}

(ii)

d^{2} sp^{3}

(iii)

sp^{3} d

(iv)

sp^{3}

(v) sp

Choose the correct answer from the options given below:

Answer: (A)

Hybridisation follows from the number of sigma bonds and lone pairs around the central atom. Sulphur in

S F_{4}

has four bond pairs and one lone pair, five electron domains, and is

s p^{3} d

hybridised. Iodine in

I F_{5}

has five bond pairs and one lone pair, six domains, and is

s p^{3} d^{2}

hybridised. Nitrogen in

N O_{2}^{+}

has two sigma bonds and no lone pair, giving two domains and sp hybridisation. Nitrogen in

N H_{4}^{+}

has four sigma bonds and no lone pair, giving

s p^{3}

hybridisation. The pairing is

S F_{4}

with

s p^{3} d

I F_{5}

with

s p^{3} d^{2}

N O_{2}^{+}

with sp and

N H_{4}^{+}

with

s p^{3}

JEE Main · 2021 · 22 Jul 2021 · Shift 2 · Single correct option