The O-N-O bond angle is maximum in: | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(D) The bond angles for the given species are as follows:$NO_2^+$: The nitrogen atom is $sp$ hybridized,resulting in a linear geometry with a bond ang

Vedclass · Accepted Answer

$NO_2^+$

Vedclass · Answer

(D) The bond angles for the given species are as follows:$NO_2^+$: The nitrogen atom is $sp$ hybridized,resulting in a linear geometry with a bond angle of $180^\circ$.$NO_2$: The nitrogen atom is $sp^2$ hybridized with one unpaired electron,resulting in a bent geometry with a bond angle of approximately $134^\circ$.$NO_2^-$: The nitrogen atom is $sp^2$ hybridized with one lone pair,resulting in a bent geometry with a bond angle of approximately $115^\circ$.$NO_3^-$: The nitrogen atom is $sp^2$ hybridized,resulting in a trigonal planar geometry with a bond angle of $120^\circ$.Comparing these values,the $O-N-O$ bond angle is maximum in $NO_2^+$.Therefore,the correct option is $(D)$.

The $O-N-O$ bond angle is maximum in:

Similar Questions

The shape of a molecule which has $EB_5L_2$ molecular formula
{$E =$ Central atom,$B =$ Terminal atom,$L =$ Lone pair}

$SbF_5$ reacts with $XeF_4$ and $XeF_6$ to form ionic compounds $[XeF_3]^+ [SbF_6]^-$ and $[XeF_5]^+ [SbF_6]^-$ respectively. What are the molecular shapes of the $[XeF_3]^+$ and $[XeF_5]^+$ ions,respectively?

Provide the different spatial arrangements of $SF_4$ including its electron pairs.

The correct order of $H-N-H$ bond angles of ammonia,ammonium ion,and amide ion is:

Which of the following molecules has unequal bond lengths?

Vedclass Test Series

Exam Paper Generator

Online Exam Module