Concept: Dipole moment is a measure of the polarity of a molecule. A molecule has a zero dipole moment if the individual bond dipoles cancel each other out due to a symmetrical molecular geometry. The hybridization of the central atom determines the molecular geometry.

Why (C) is correct:

For a molecule $M{X}_3$ to have a zero dipole moment, it must possess a symmetrical geometry where the bond dipoles cancel out. This typically occurs in a trigonal planar geometry. A trigonal planar geometry is achieved when the central atom M undergoes $s{p}^2$ hybridization and has no lone pairs of electrons.

In $s{p}^2$ hybridization, one s-orbital and two p-orbitals mix to form three equivalent $s{p}^2$ hybrid orbitals, which are oriented at 120° to each other in a trigonal planar arrangement. If the three X atoms are identical, the individual bond dipoles cancel out, resulting in a net zero dipole moment. Examples include $B{F}_3$ and $S{O}_3$.

Option Analysis:

• A) pure p: If pure p orbitals were used for bonding, the geometry would be angular (like $H_{2}S$ if only p orbitals were involved, though hybridization is more complex for such molecules), which would result in a non-zero dipole moment.
• B) sp hybrid: $sp$ hybridization leads to a linear geometry (e.g., $BeC{l}_2$). For $M{X}_3$, this hybridization is not possible as it requires three bonding orbitals.
• D) sp 3 hybrid: $s{p}^3$ hybridization leads to a tetrahedral geometry (e.g., $C{H}_4$) or pyramidal geometry (e.g., $N{H}_3$) if lone pairs are present. For $M{X}_3$, if it were $s{p}^3$ hybridized with one lone pair (like $N{H}_3$), it would have a pyramidal geometry and a non-zero dipole moment. If it were $s{p}^3$ hybridized with no lone pairs, it would form $M{X}_4$ (tetrahedral), not $M{X}_3$.

Therefore, for $M{X}_3$ to have a zero dipole moment, the central atom M must be $s{p}^2$ hybridized, leading to a trigonal planar geometry.

Correct Answer: (C)