(i) $HCl \longrightarrow \underset{0.1}{ H }^{+}+\underset{0.1 M }{ Cl ^{-}}$
$\therefore \left[ H ^{+}\right]=0 \cdot 1 M$
$pH =-\log \left[ H ^{+}\right]=-\log 0 \cdot 1=1$
(ii) $NaCl$ is a salt of strong acid and strong base so it is not hydrolysed and hence its $pH$ is $7 .$
(iii) $NH _{4} Cl + H _{2} O \rightleftharpoons NH _{4} OH + HCl$
$\therefore $ The solution is acidic and its $pH$ is less than that of $0.1 \,M\, HCl$.
(iv) $NaCN + H _{2} O \rightleftharpoons NaOH + HCN$
$\therefore $ The solution is basic and its $pH$ is more than that of $0.1 \,M \,HCl$.
$\therefore $ Correct order for increase in $pH$ is
$HCl < NH _{4} Cl < NaCl < NaCN$.