Question

question 26 (1 point)
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what is the electronic geometry of the central atom of nh3?
octahedral
trigonal pyramidal
tetrahedral
trigonal bipyramidal
trigonal planar

Explanation:

To determine the electronic geometry of the central atom (\(N\)) in \(NH_3\):

• First, find the number of electron - pairs around the central atom. The formula for the number of electron - pairs (\(n\)) is \(n=\frac{V + M - C + A}{2}\), where \(V\) is the valence electrons of the central atom (\(N: V = 5\)), \(M\) is the number of monovalent atoms (\(H: M=3\)), \(C\) is the charge for cations (0 here as \(NH_3\) is neutral), and \(A\) is the charge for anions (0 here). So \(n=\frac{5 + 3}{2}=4\).
• According to VSEPR (Valence - Shell Electron - Pair Repulsion) theory, when the number of electron - pairs (\(n = 4\)), the electronic geometry (which considers all electron - pairs, bonding and non - bonding) is tetrahedral. The molecular geometry (which considers only bonding pairs) of \(NH_3\) is trigonal pyramidal (because it has 3 bonding pairs and 1 non - bonding pair), but the question asks for electronic geometry.

Answer:

tetrahedral