The correct order of increasing bond angles in the following triatomic species is

Question

The correct order of increasing bond angles in the following triatomic species is (A) NO2- < NO2+ < NO2 (B) NO2- < NO2 < NO2+ (C) NO2+ < NO2 < NO2- (D) NO2+ < NO2- < NO2

Tardigrade Admin · Accepted Answer

According to VSEPR theory,
The order of repulsion of electrons is lp-lp repulsion > lp-bp repulsion > bp-bp repulsion
Higher the number of lone pair of electrons (greater is repulsion) lower is the bond angle and vice-versa.
In

N O_{2}^{-}

one lone pair of electron is present while in

N O_{2}^{+}

no lone xav of electron Is present. Hence, the correct order of bend angles is

N O_{2}^{-} < N O_{2} < N O_{2}^{+}

Q. The correct order of increasing bond angles in the following triatomic species is

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

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

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

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

Q. The correct order of increasing bond angles in the following triatomic species is

NO2−​<NO2+​<NO2​

NO2−​<NO2​<NO2+​

NO2+​<NO2​<NO2−​

NO2+​<NO2−​<NO2​

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

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

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

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