Among the isomeric alkanes of molecular formula $C_{5}H_{12}$,identify the one that on photochemical chlorination yields:
$(i)$ $A$ single monochloride.
$(ii)$ Three isomeric monochlorides.
$(iii)$ Four isomeric monochlorides.

(N/A) $(i)$ To have a single monochloride,there should be only one type of $H$ atom in the isomer of the alkane of the molecular formula $C_{5}H_{12}$. This is because replacement of any $H$ atom leads to the formation of the same product. The isomer is neopentane ($2,2-$dimethylpropane).
$CH_{3}-C(CH_{3})_{2}-CH_{3}$ (neopentane)
$(ii)$ To have three isomeric monochlorides,the isomer of the alkane of the molecular formula $C_{5}H_{12}$ should contain three different types of $H$ atoms. Therefore,the isomer is $n-$pentane. It can be observed that there are three types of $H$ atoms labelled as $a, b,$ and $c$ in $n-$pentane.
$CH_{3}^{c}-CH_{2}^{b}-CH_{2}^{a}-CH_{2}^{b}-CH_{3}^{c}$ ($n-$pentane)
$(iii)$ To have four isomeric monochlorides,the isomer of the alkane of the molecular formula $C_{5}H_{12}$ should contain four different types of $H$ atoms. Therefore,the isomer is $2-$methylbutane. It can be observed that there are four types of $H$ atoms labelled as $a, b, c,$ and $d$ in $2-$methylbutane.
$CH_{3}^{a}-CH^{b}(CH_{3}^{a})-CH_{2}^{c}-CH_{3}^{d}$ ($2-$methylbutane)