Question

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

Explanation:

Step1: Analyze \(BH_3\)

Boron (\(B\)) in \(BH_3\) has 3 valence electrons. It forms 3 covalent bonds with hydrogen atoms. So the number of valence electrons around \(B\) is \(3 + 3=6\) (using the formula: valence electrons of central atom + number of bonding electrons from surrounding atoms). It does not have an octet (8 - electron) configuration.

Step2: Analyze \(NH_4^+\)

Nitrogen (\(N\)) in \(NH_4^+\) has 5 valence electrons. It forms 4 covalent bonds (one of which is a coordinate covalent bond) with hydrogen atoms. The number of valence electrons around \(N\) is \(5+4 - 1=8\) (the - 1 is because of the + 1 charge). It has an octet configuration.

Step3: Analyze \(H_3O^+\)

Oxygen (\(O\)) in \(H_3O^+\) has 6 valence electrons. It forms 3 covalent bonds with hydrogen atoms and has one lone - pair. The number of valence electrons around \(O\) is \(6+3 - 1=8\) (the - 1 is because of the + 1 charge). It has an octet configuration.

Step4: Analyze \(NH_3\)

Nitrogen (\(N\)) in \(NH_3\) has 5 valence electrons. It forms 3 covalent bonds with hydrogen atoms and has one lone - pair. The number of valence electrons around \(N\) is \(5 + 3=8\). It has an octet configuration.

Answer:

A. \(BH_3\)