Which major product is formed when excess of methane is treated with chlorine in presence of UV light?

Methane reacts with chlorine under UV light through free radical substitution. The stepwise mechanism involves:

$\text{CH}_4 + \text{Cl}_2 \xrightarrow{\text{UV light}} \text{CH}_3\text{Cl} + \text{HCl}$

$\text{CH}_3\text{Cl} + \text{Cl}_2 \xrightarrow{\text{UV light}} \text{CH}_2\text{Cl}_2 + \text{HCl}$

$\text{CH}_2\text{Cl}_2 + \text{Cl}_2 \xrightarrow{\text{UV light}} \text{CHCl}_3 + \text{HCl}$

$\text{CHCl}_3 + \text{Cl}_2 \xrightarrow{\text{UV light}} \text{CCl}_4 + \text{HCl}$

With excess methane, chlorine radicals predominantly encounter unreacted methane rather than chlorinated products. This kinetic preference favors monochlorination, making chloromethane the major product.

The final answer is $\boxed{A}$.