Question

which of the following molecules would you expect to have sp hybrid orbitals around the central atom?
co₂
h₂o
ch₄
alcl₃

Explanation:

• For \(CO_2\): The central \(C\) atom has 2 electron - group (two double bonds with \(O\) atoms). According to the VSEPR (Valence - Shell Electron - Pair Repulsion) theory and hybridization rules, when the number of electron - groups (\(n\)) around the central atom is 2, the hybridization is \(sp\). The Lewis structure of \(CO_2\) is \(O = C=O\), and the electron - group geometry is linear.
• For \(H_2O\): The central \(O\) atom has 4 electron - groups (2 bonding pairs and 2 lone pairs). Using the formula \(n=\text{number of bonding pairs}+\text{number of lone pairs}\), \(n = 2 + 2=4\). When \(n = 4\), the hybridization is \(sp^{3}\).
• For \(CH_4\): The central \(C\) atom has 4 bonding pairs (no lone pairs). So \(n=4\) (using \(n=\text{number of bonding pairs}+\text{number of lone pairs}\), here number of lone pairs \(=0\)), and the hybridization is \(sp^{3}\).
• For \(AlCl_3\): The central \(Al\) atom has 3 bonding pairs (no lone pairs). Using \(n=\text{number of bonding pairs}+\text{number of lone pairs}\), \(n = 3+0 = 3\). When \(n = 3\), the hybridization is \(sp^{2}\).

Answer:

\(CO_2\)