Match List-I with List-II

(a) SF$$_4$$	(i) sp$$^3$$d$$^2$$
(b) IF$$_5$$	(ii) d$$^2$$sp$$^3$$
(c) NO$$_2^+$$	(iii) sp$$^3$$d
(d) NH$$_4^+$$	(iv) sp$$^3$$
	(v) sp

Choose the correct answer from the options given below:

We determine the hybridisation of the central atom in each species.

SF$$_4$$: Sulphur has 6 valence electrons; 4 are used for bonding with F, leaving 1 lone pair. Total electron pairs around S = 4 bonding pairs + 1 lone pair = 5, giving $$sp^3d$$ hybridisation (iii).

IF$$_5$$: Iodine has 7 valence electrons; 5 are used for bonding with F, leaving 1 lone pair. Total electron pairs = 5 bonding + 1 lone pair = 6, giving $$sp^3d^2$$ hybridisation (i).

NO$$_2^+$$: This cation has no lone pairs on nitrogen (N loses one electron, giving a linear structure). With 2 double bonds and no lone pairs, the steric number is 2, giving $$sp$$ hybridisation (v).

NH$$_4^+$$: Nitrogen forms 4 equivalent N-H bonds with no lone pairs (the lone pair donated into the extra bond). Total electron pairs = 4, giving $$sp^3$$ hybridisation (iv).

The correct matching is (a)-(iii), (b)-(i), (c)-(v) and (d)-(iv), which corresponds to option (3).