The correct order of bond dissociation energy among N2, O2, O2- is shown in which of the following arrangements ?

Question

The correct order of bond dissociation energy among N2, O2, O2- is shown in which of the following arrangements ? (A) N2> O2->O2 (B) O-2> O2>N2 (C) N2> O2>O2- (D) O2> O2->N2

Tardigrade Admin · Accepted Answer

Bond energy propto Bond order bondorder :- N2 =Nb=10, Na=4 B.O.=(N2)=(10-4/2)=3 O2=Nb=10, Na=6 B.O(o2)=(10-6/2)=2 O2-=Nb=10, Na=7 B.O.(o2)=(10-7/2)=(3/2)=1.5 Hence the order of B.O. N2> O2 > O2-

Q. The correct order of bond dissociation energy among $N_{2}, O_{2}, O_{2}^{-}$ is shown in which of the following arrangements ?

$N_{2} > O_{2}^{-} > O_{2}$

$O_{2}^{-} > O_{2} > N_{2}$

$N_{2} > O_{2} > O_{2}^{-}$

$O_{2} > O_{2}^{-} > N_{2}$

Q. The correct order of bond dissociation energy among N2​,O2​,O2−​ is shown in which of the following arrangements ?

N2​>O2−​>O2​

O2−​>O2​>N2​

N2​>O2​>O2−​

O2​>O2−​>N2​

Bond energy ∝ Bond order bondorder :- N2​=Nb=10,Na=4 B.O.=(N2​)=210−4​=3 O2​=Nb=10,Na=6 B.O(o2​)​=210−6​=2 O2−​=Nb=10,Na=7 B.O.(o2​)​=210−7​=23​=1.5 Hence the order of B.O. N2​>O2​>O2−​

Q. The correct order of bond dissociation energy among $N_{2}, O_{2}, O_{2}^{-}$ is shown in which of the following arrangements ?

$N_{2} > O_{2}^{-} > O_{2}$

$O_{2}^{-} > O_{2} > N_{2}$

$N_{2} > O_{2} > O_{2}^{-}$

$O_{2} > O_{2}^{-} > N_{2}$