Question

which of the following contains an atom (other than hydrogen) which lacks an octet of valence electrons?
a bh₃
b nh₃
c nh₄⁺
d h₃o⁺

Explanation:

• For \(BH_3\): Boron (\(B\)) has 3 valence electrons. It forms 3 single bonds with hydrogen atoms. Each bond contributes 1 electron to boron's valence shell. So, the number of valence electrons around boron is \(3 + 3=6\) (since each bond is 2 - electron sharing, but we count per atom's contribution).
• For \(NH_3\): Nitrogen (\(N\)) has 5 valence electrons. It forms 3 single bonds with hydrogen atoms and has 1 lone pair. The number of valence electrons around nitrogen is \(5+3 = 8\) (3 from bonds and 2 from the lone pair, total 8).
• For \(NH_4^+\): Nitrogen (\(N\)) has 5 valence electrons. It forms 4 single bonds with hydrogen atoms (due to the positive charge, one less electron from the original nitrogen's count). The number of valence electrons around nitrogen is \(5 + 4-1=8\) (4 from bonds, subtract 1 for the positive charge).
• For \(H_3O^+\): Oxygen (\(O\)) has 6 valence electrons. It forms 3 single bonds with hydrogen atoms and has 1 lone pair. The number of valence electrons around oxygen is \(6+3 - 1=8\) (3 from bonds, subtract 1 for the positive charge).

Answer:

A. \(BH_3\)