$\mathrm{IE}=E_{\infty}-E_{1}=0-E_{1}=-E_{1}$
$\therefore \mathrm{E}_{1}$ of $\mathrm{He}^{+}=-19.6 \times 10^{-18} $$\mathrm{J}\; atom ^{-1}$
Energy of a species at $n$ state,
$\left(\mathrm{E}_{\mathrm{n}}\right)_{species}$$=\left(\mathrm{E}_{\mathrm{n}}\right)_{\mathrm{hydrogen }} \times \mathrm{z}^{2}$
$\therefore\left(\mathrm{E}_{1}\right)_{\text {hydrogen }}=\frac{-19.6 \times 10^{-18}}{4}$$[\text { For } \mathrm{He}, \mathrm{Z}=2]$
$\left(E_{1}\right)_{L^{*}}=\frac{-19.6 \times 10^{-18}}{4} \times 3^{2}$
$=-4.41 \times 10^{-17}\, \mathrm{J}$ $atom$ $^{-1}$
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| Reaction | Energy Change (in $kJ$ ) |
| $Li(s) \to Li(g)$ | $161$ |
| $Li(g) \to Li^+(g)$ | $520$ |
| $\frac {1}{2}F_2(g)\,\to F(g)$ | $77$ |
| $F(g) + e^- \to F^-(g)$ | (Electron gain enthalpy) |
| $Li^+ (g) + F^-(g) \to LiF(s)$ | $-1047$ |
| $Li (s) + \frac {1}{2}F_2(g)\to LiF(s)$ | $-617$ |
Based on data provided, the value of electron gain enthalpy of fluorine would be.....$kJ\,mol^{-1}$
$(i)$ $\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
$A$ in the above reaction is: