The major product formed in the Friedel-Craft acylation of chlorobenzene is

During Friedel-Crafts acylation, an acylium ion (such as $$\text{CH}_3\text{C}^+=\text{O}$$) attacks the benzene ring.

1. Attack at the ortho position experiences significant steric hindrance due to the relatively large size of both the chlorine atom and the incoming acyl group.
2. Attack at the para position faces minimal steric repulsion.

Consequently, the para-substituted product is formed as the major product, while the ortho-isomer is the minor product.

$$\text{Chlorobenzene} + \text{CH}_3\text{COCl} \xrightarrow{\text{Anhydrous AlCl}_3} \underbrace{\text{4-chloroacetophenone}}_{\textbf{Major (Para)}} + \underbrace{\text{2-chloroacetophenone}}_{\text{Minor (Ortho)}}$$

Hence, correct option is