Q88 E - Chemistry (OpenStax)

Question

Why is the \({\rm{H - N - H}}\) angle in \({\rm{N}}{{\rm{H}}_{\rm{3}}}\) smaller than the \({\rm{H - C - H}}\) bond angle in \({\rm{C}}{{\rm{H}}_{\rm{4}}}\)? Why is the \({\rm{H - N - H}}\) angle in \({\rm{NH}}_{\rm{4}}^{\rm{ + }}\) identical to the \({\rm{H - C - H}}\) bond angle in \({\rm{C}}{{\rm{H}}_{\rm{4}}}\)?

Step-by-Step Solution

Answer

Due to presence of lone pairs the \({\rm{H - N - H}}\) angle in \({\rm{N}}{{\rm{H}}_{\rm{3}}}\) is smaller than the \({\rm{H - C - H}}\) bond angle in \({\rm{C}}{{\rm{H}}_{\rm{4}}}\). The \({\rm{H - N - H}}\) angle in \({\rm{NH}}_{\rm{4}}^{\rm{ + }}\) is identical to the \({\rm{H - C - H}}\) bond angle in \({\rm{C}}{{\rm{H}}_{\rm{4}}}\)as both have \({\rm{4}}\) bonding pairs and no lone pairs.

Step 1: Concept Introduction

The angle created by three atoms across at least two bonds is known as a bond angle. The torsional angle is the angle produced between the first three atoms and the plane formed by the last three atoms for four atoms linked together in a chain.

Step 2: Comparison of bond angles

Lone pair takes up more room than bonding pair of electrons. Therefore, presence of lone pair makes the angle small.

Now, \({\rm{N}}{{\rm{H}}_{\rm{3}}}\) contains \({\rm{3}}\) bonding pairs of electrons and a lone pair of electrons whereas both \({\rm{CH}}\), contain \({\rm{4}}\) bonding pairs and no lone pairs.

So, because of the presence of lone pair, \({\rm{H - N - H}}\) angle is smaller than the \({\rm{H - C - H}}\) bond angle. Now, both \({\rm{C}}{{\rm{H}}_{\rm{4}}}\) and \({\rm{NH}}_{\rm{4}}^{\rm{ + }}\) contain \({\rm{4}}\) bonding pairs and no lone pairs.

 

Therefore \({\rm{H - C - H}}\) bond angle and \({\rm{H - N - H}}\) bond angle is similar.