The decreasing order of bond angle is

Question

The decreasing order of bond angle is (A) NO 2> NO 2+> NO 2- (B) NO 2-> NO 2> NO 2+ (C) NO 2+> NO 2> NKO 2- (D) NO 2+> NO 2-> NO 2

Tardigrade Admin · Accepted Answer

As the number of unpaired electrons (lone pair of electrons) increases, bond angle decreases. Thus, the decreasing order of bond angle is <img class="img-fluid question-image" alt="image" src="https://cdn.tardigrade.in/img/question/chemistry/fbb061ce55a34b47a81b3826516847bd-.png" />

Q. The decreasing order of bond angle is

$N O_{2} > N O_{2}^{+} > N O_{2}^{-}$

$N O_{2}^{-} > N O_{2} > N O_{2}^{+}$

$N O_{2}^{+} > N O_{2} > N K O_{2}^{-}$

$N O_{2}^{+} > N O_{2}^{-} > N O_{2}$

As the number of unpaired electrons (lone pair of electrons) increases, bond angle decreases. Thus, the decreasing order of bond angle is

Q. The decreasing order of bond angle is

NO2​>NO2+​>NO2−​

NO2−​>NO2​>NO2+​

NO2+​>NO2​>NKO2−​

NO2+​>NO2−​>NO2​

As the number of unpaired electrons (lone pair of electrons) increases, bond angle decreases. Thus, the decreasing order of bond angle is

$N O_{2} > N O_{2}^{+} > N O_{2}^{-}$

$N O_{2}^{-} > N O_{2} > N O_{2}^{+}$

$N O_{2}^{+} > N O_{2} > N K O_{2}^{-}$

$N O_{2}^{+} > N O_{2}^{-} > N O_{2}$