The bond dissociation energy (E) and bond length (R) of O 2, N 2 and F 2 follow the order as: (E) (R) (a) O 2 N 2 F 2 F 2 O 2 N 2 (b) N 2 F 2 O 2 N 2 O 2 F 2 (c) N 2 O 2 F 2 F 2 O 2 N 2 (d) N 2 O 2 F 2 N 2 O 2 F 2

Question

The bond dissociation energy (E) and bond length (R) of O 2, N 2 and F 2 follow the order as: (E) (R) (a) O 2 > N 2 > F 2 F 2> O 2> N 2 (b) N 2> F 2> O 2 N 2> O 2> F 2 (c) N 2> O 2> F 2 F 2> O 2> N 2 (d) N 2> O 2> F 2 N 2> O 2> F 2 (A) a (B) b (C) c (D) d

Tardigrade Admin · Accepted Answer

Greater the bond order, shorter is the bond length and thus, stronger is the bond.

Since, for the given molecules, bond order follows the trend

(B . O = 3) N_{2} > (B . O = 2) O_{2} > (B . O = 1) F_{2}

Thus, the bond length decreases as

F_{2} > O_{2} > N_{2}

and bond dissociation energy follows the trend

N_{2} > O_{2} > F_{2}

	(E)	(R)
(a)	$O_{2} > N_{2} > F_{2}$	$F_{2} > O_{2} > N_{2}$
(b)	$N_{2} > F_{2} > O_{2}$	$N_{2} > O_{2} > F_{2}$
(c)	$N_{2} > O_{2} > F_{2}$	$F_{2} > O_{2} > N_{2}$
(d)	$N_{2} > O_{2} > F_{2}$	$N_{2} > O_{2} > F_{2}$

Q. The bond dissociation energy (E) and bond length (R) of O2​,N2​ and F2​ follow the order as: (E) (R) (a) O2​>N2​>F2​ F2​>O2​>N2​ (b) N2​>F2​>O2​ N2​>O2​>F2​ (c) N2​>O2​>F2​ F2​>O2​>N2​ (d) N2​>O2​>F2​ N2​>O2​>F2​

a

b

c

d