Question

what is the closest bond angle around the central atom?

Explanation:

Step1: Determine the hybridization of central atom

The central atom is \(N\). In \(NH_4^+\), \(N\) forms 4 \(\sigma\) bonds and has no lone pairs. According to VSEPR theory, the hybridization is \(sp^3\).

Step2: Recall the bond - angle for \(sp^3\) hybridization

For \(sp^3\) hybridization with 4 bonding pairs (no lone pairs), the molecular geometry is tetrahedral. The bond angle in a tetrahedral geometry is \(109.5^{\circ}\).

Answer:

\(109.5^{\circ}\)