Chemistry - PERIODIC TABLE Question with Solution | TestHub

$\mathrm{X}\to {\mathrm{X}}^{+}+\mathrm{e};\Delta \mathrm{H}={\mathrm{IE}}_1=\mathrm{aeV}$

$\mathrm{X}+\mathrm{e}\to {\mathrm{X}}^{-};\Delta \mathrm{H}={\mathrm{E}}_{{\mathrm{ga}}_1}=-\mathrm{beV}$

If $\mathrm{N}/2$ atoms of X lose electrons which are taken up by remaining $\mathrm{N}/2$ of X to give ${\mathrm{X}}^{-}$, then $\mathrm{a}\times \frac{\mathrm{N}}{2}-\mathrm{b}\times \frac{\mathrm{N}}{2}=\frac{409\times 10^3}{1.602\times 10^{-19}}\mathrm{eV}$

or $\mathrm{a}-\mathrm{b}=\frac{409\times 10^3}{1.602\times 10^{-19}\times 6.023\times 10^{23}}$

$\therefore \mathrm{a}-\mathrm{b}=8.477$

Now $\mathrm{N}/2$ of ${\mathrm{X}}^{-}$; lose two electrons to give ${\mathrm{X}}^{+}$

${\mathrm{X}}^{-}\to \mathrm{X}+\mathrm{e};\Delta \mathrm{H}=+\mathrm{b}$

$$\mathrm{X}\to {\mathrm{X}}^{+}+\mathrm{e};\Delta \mathrm{H}=+{\mathrm{IE}}_2=+\mathrm{a}$$

$\mathrm{a}\times \frac{\mathrm{N}}{2}+\mathrm{b}\times \frac{\mathrm{N}}{2}=\frac{733\times 10^3}{1.602\times 10^{-19}}$

$\mathrm{a}+\mathrm{b}=\frac{733\times 10^3\times 2}{1.602\times 10^{-19}\times 6.023\times 10^{23}}$

or $\mathrm{a}+\mathrm{b}=15.194$