Question

Match List-I with List-II.
$$\begin{array}{cccc}& \text{List-I}& & \text{List-II}\\ \left(P\right)& Fe\left(N{O}_3{)}_3\right(aq)& \left(1\right)& \text{Only cationic hydrolysis}\\ \left(Q\right)& KCl{O}_4\left(aq\right)& \left(2\right)& \text{Only anionic hydrolysis}\\ \left(R\right)& RCOONa\left(aq\right)& \left(3\right)& \text{Both cationic as anionic hydrolysis}\\ \left(S\right)& N{H}_{4}CN\left(aq\right)& \left(4\right)& \text{No hydrolysis}\end{array}$$

• A. $$\begin{array}{cccc}P& Q& R& S\\ 1& 2& 3& 4\end{array}$$
• B. $$\begin{array}{cccc}P& Q& R& S\\ 1& 4& 2& 3\end{array}$$
• C. $$\begin{array}{cccc}P& Q& R& S\\ 2& 1& 4& 3\end{array}$$
• D. $$\begin{array}{cccc}P& Q& R& S\\ 4& 3& 2& 1\end{array}$$

Answer 1

The correct option is B $$\begin{array}{cccc}P& Q& R& S\\ 1& 4& 2& 3\end{array}$$

$Fe(N{O}_3{)}_3$ is a salt of weak base $Fe(OH{)}_3$ and strong acid $HN{O}_3$.
So, the hydrolysis of this will make the solution slightly acidic and known as cationic hydrolysis.

$KCl{O}_4$ is a salt of strong acid $KOH$ and strong base $HCl{O}_4$.
So, there is no net hydrolysis of the salt.

$RCOONa$ is a salt of weak acid $RCOOH$ and strong base $NaOH$.
So, the hydrolysis of this will make the solution slightly basic and known as anionic hydrolysis.

$N{H}_{4}CN$ is a salt of weak acid $HCN$ and weak base $N{H}_{4}OH$.
So, the hydrolysis of this salt is both cationic and anionic hydrolysis.