The organic chloro compound, which shows complete stereochemical inversion during a $$S_N 2$$ reaction, is

The stereochemical outcome of a nucleophilic substitution depends on the mechanism followed by the substrate.

Key fact for $$S_N2$$ reactions:
• A single transition-state is formed when the nucleophile attacks from the side opposite to the leaving group (back-side attack).
• The carbon undergoing substitution must therefore invert its configuration (Walden inversion).
• $$S_N2$$ is strongly disfavoured by steric hindrance; hence the rate order is
$$\text{methyl} \gt 1^{\circ} \gt 2^{\circ} \gg 3^{\circ}$$

Now examine each option.

Option A : $$(C_2H_5)_2CHCl$$
This is a secondary chloride. Steric hindrance slows the $$S_N2$$ path, and $$S_N1$$ can also compete. The product would not show complete inversion.

Option B : $$(CH_3)_3CCl$$
A tertiary chloride cannot undergo $$S_N2$$ because the back-side is completely blocked. It reacts through the $$S_N1$$ mechanism, giving racemisation rather than inversion.

Option C : $$(CH_3)_2CHCl$$
Isopropyl chloride is again secondary. Although it can, in principle, react by $$S_N2$$, steric crowding makes the reaction slow and incomplete; $$S_N1$$ contribution leads to loss of pure inversion.

Option D : $$CH_3Cl$$
Methyl chloride is the least hindered substrate possible. $$S_N2$$ is the only viable pathway; there is no competing $$S_N1$$ route. Therefore every substitution occurs by a single back-side attack, giving 100 % inversion. (Because the carbon in $$CH_3Cl$$ is achiral, the inverted product looks identical, but mechanistically the inversion is complete.)

Thus, the chloro compound that shows complete stereochemical inversion in an $$S_N2$$ reaction is

Option D which is: $$CH_3Cl$$