Among the following compounds, which one has the shortest C-Cl bond?

To determine which compound has the shortest $$C-Cl$$ bond, we examine the hybridization of the carbon atom bonded to chlorine and the effect of resonance on the bond.

The bond length of a $$C-Cl$$ bond depends primarily on two factors. First, as the $$s$$-character of the carbon hybrid orbital increases, the bond becomes shorter. Thus, the trend is

$$sp<sp^2<sp^3$$

in terms of bond length. Second, if resonance imparts partial double bond character to the $$C-Cl$$ bond, the bond becomes shorter and stronger than a normal single bond.

In tert-butyl chloride, the carbon attached to chlorine is $$sp^3$$ hybridized, so the $$C-Cl$$ bond is a normal single bond and relatively long.

In allyl chloride, although the molecule contains a double bond, the carbon directly bonded to chlorine is still $$sp^3$$ hybridized. Therefore, the $$C-Cl$$ bond remains essentially a single bond.

In methyl chloride, the carbon bonded to chlorine is also $$sp^3$$ hybridized, giving a normal single $$C-Cl$$ bond.

In vinyl chloride, the chlorine atom is directly attached to an $$sp^2$$ hybridized carbon of a double bond. Moreover, the lone pair on chlorine participates in conjugation with the adjacent $$\pi$$ bond, leading to resonance as shown below:

$$CH_2=CH-\ddot{Cl};\leftrightarrow;{}^{-}CH_2-CH=\overset{+}{Cl}.$$

This resonance imparts partial double bond character to the $$C-Cl$$ bond, making it significantly shorter than a normal single bond.

Therefore, vinyl chloride (Option D) has the shortest $$C-Cl$$ bond because the carbon is $$sp^2$$ hybridized and the bond possesses partial double bond character due to resonance.