- ✓boron
- Bberyllium
- Cmagnesium
- Daluminium
The electronic configuration of
$B ^{+} \rightarrow \,1 s ^2 \,2 s ^2$
$Be ^{+} \rightarrow \,1 s ^2\, 2 s ^1$
$Mg ^{+} \rightarrow \,1 s ^2\, 2 s ^2\, 2 p ^6\, 3 s ^1$
$Si ^{+} \rightarrow \,1 s ^2 \,2 s ^2 \,2 p ^6 \,3 s ^2 \,3 p ^1$
So, in $B ^{+}$the electron is to be removed from completely filled $s$-subshell, which is difficult. Thus, the second ionization energy of boron is maximum.
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$(V=33\, volt)$ $(m_p = 1.672 \times 10^{-27} \,kg)$
$\frac{1}{2}C{l_2}(g)\xrightarrow{{\frac{1}{2}{\Delta _{diss}}{H^\Theta }}}Cl(g)\xrightarrow{{{\Delta _{eg}}{H^\Theta }}}$ $C{l^ - }(g)\xrightarrow{{{\Delta _{Hyd}}{H^\Theta }}}C{l^ - }(aq)$
(using the data,
${{\Delta _{diss}}H_{C{l_2}}^\Theta } = 240\,kJ\,mol^{-1}, {{\Delta _{eg}}H_{C{l}}^\Theta }= -349 \,kJ\,mol^{-1},$${{\Delta _{Hyd}}H_{C{l}}^\Theta }= -381 \,kJ\,mol^{-1}$ ) will be ............. $\mathrm{kJ\,mol}^{-1}$
