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$(i)$ $\begin{gathered}
HCN\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons {H_3}{O^ + }\left( {aq} \right) + C{N^ - }\left( {aq} \right) \hfill \\
{K_a} = 6.2 \times {10^{ - 10}} \hfill \\
\end{gathered} $
$(ii)$ $\begin{gathered}
C{N^ - }\left( {aq} \right) + {H_2}O\left( l \right) \rightleftharpoons HCN\left( {aq} \right) + O{H^ - }\left( {aq} \right) \hfill \\
{K_b} = 1.6 \times {10^{ - 5}} \hfill \\
\end{gathered} $
These equilibria show the following order of the relative base strength
| List $-I$ (Atomic Number) | List $-II$ (Block of periodic table) |
| $A$ $37$ | $I$ $p-$block |
| $B$ $78$ | $II$ $d-$block |
| $C$ $52$ | $III$ $f-$block |
| $D$ $65$ | $IV$ $s-$block |
નીચે આપેલા વિકલ્પોમાંથી યોગ્ય ઉત્તરની પસંદગી કરો.
$\Psi_{2 s}=\frac{1}{2 \sqrt{2 \pi}}\left(\frac{1}{a_0}\right)^{1 / 2}\left(2-\frac{ r }{ a _0}\right) e ^{- r / 2 a _0}$
At $r=r_0$, radial node is formed. Thus, $r_0$ in terms of $a_0$