Question

question 1 (2 points)
which compound is likely to have an incomplete octet?
n₂o
so₃
bh₃
nh₃

Explanation:

• For \(N_2O\): Nitrogen and oxygen are in the second period. In \(N_2O\), the central atom (either \(N\)) forms double - bonds. All atoms ( \(N\) and \(O\)) follow the octet rule. For example, in the resonance structures of \(N_2O\), each atom has 8 valence electrons around it.
• For \(SO_3\): Sulfur is in the third period. In \(SO_3\), sulfur forms double - bonds with oxygen atoms. Although sulfur can have an expanded octet (more than 8 electrons due to the availability of \(d\) - orbitals), the oxygen atoms follow the octet rule.
• For \(BH_3\): Boron is in the second period. Boron has 3 valence electrons. In \(BH_3\), it forms three single bonds with hydrogen atoms. So, the number of valence electrons around boron is \(3 + 3=6\) (using the formula \(V = N_{valence\ of\ central\ atom}+N_{bonding\ electrons}\)), which is less than 8.
• For \(NH_3\): Nitrogen is in the second period. Nitrogen has 5 valence electrons. In \(NH_3\), it forms three single bonds with hydrogen atoms and has one lone pair. The total number of valence electrons around nitrogen is \(5 + 3=8\) (using the formula \(V = N_{valence\ of\ central\ atom}+N_{bonding\ electrons}\)).

Answer:

\(BH_3\)