Q22 / 120 · VITEEE 2015 · Chemistry
Consider the following changes
$\begin{array}{l}
M(s) \longrightarrow M(g) \quad \ldots(1) \\
M(g) \longrightarrow M^{2+}(g)+2 e^{-} \quad \ldots(2) \\
M(g) \longrightarrow M^{+}(g)+e^{-} \quad \ldots(3) \\
M^{+}(g) \longrightarrow M^{2+}(g)+e^{-} \ldots(4) \\
M(g) \longrightarrow M^{2+}(g)+2 e^{-} \ldots(5)
\end{array}$
The second ionisation energy of $M$ could be determined from the energy values associated with
- A$1+2+4$
- B$1+5-3$
- C$2+3-4$
- Correct answerD$5-3$
Solution
The amount of energy required to take out an electron from the monopositive cation is called second ionisation energy
$\begin{array}{l}
\mathrm{M}(\mathrm{g}) \longrightarrow \mathrm{M}^{2+}(\mathrm{g})+2 \mathrm{e}^{-} ...(v) \\
\mathrm{M}(\mathrm{g}) \longrightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-} ...(iii)
\end{array}$
On subtracting eq(iii) form eq. (v) we get,
$\mathrm{M} \longrightarrow \mathrm{M}^{2+}+\mathrm{e}^{-}$
Hence, the correct answer is option: 'D'.