Which of the following compounds will show retention in configuration on nucleophilic substitution by $OH^{-}$ ion?

The product $(C)$ in the following reaction sequence is:

Chloroethane with silver acetate forms $X$ and with $LiAlH_4$ forms $Y$. What are $X$ and $Y$?

For the following compounds,the correct statement$(s)$ with respect to nucleophilic substitution reactions is(are):
$I$: Benzyl bromide $(C_6H_5CH_2Br)$
$II$: Cyclohexylmethyl bromide $(C_6H_{11}CH_2Br)$
$III$: tert-Butyl bromide $((CH_3)_3CBr)$
$IV$: $1-$Phenylethyl bromide $(C_6H_5CH(CH_3)Br)$
$A$. $I$ and $III$ follow $S_{N}1$ mechanism
$B$. $I$ and $II$ follow $S_{N}2$ mechanism
$C$. Compound $IV$ undergoes inversion of configuration
$D$. The order of reactivity for $I$,$III$ and $IV$ is: $IV > I > III$

Consider the following anions:
$(I)$ $CF_3SO_3^-$
$(II)$ $C_6H_5SO_3^-$
$(III)$ $C_6H_5O^-$
$(IV)$ $CH_3COO^-$
When attached to $sp^3$ hybridized carbon,their leaving group ability in nucleophilic substitution reaction decreases in the order:

Arrange the leaving group ability of the following groups in decreasing order:
$(I)$ $CH_3-C_6H_4-SO_3^-$,$(II)$ $C_6H_5-SO_3^-$,$(III)$ $N_3^-$,$(IV)$ $Br^-$