Consider the following changes : (1) $\mathrm{M}(\mathrm{s}) \longrightarrow \mathrm{M}(\mathrm{g})$ (2)…
(1) $\mathrm{M}(\mathrm{s}) \longrightarrow \mathrm{M}(\mathrm{g})$
(2) $\mathrm{M}(\mathrm{s}) \longrightarrow \mathrm{M}^{2+}(\mathrm{g})+2 \mathrm{e}^{-}$
(3) $\mathrm{M}(\mathrm{g}) \longrightarrow \mathrm{M}^{+}(\mathrm{g})+\mathrm{e}^{-}$
(4) $\mathrm{M}^{+}(\mathrm{g}) \longrightarrow \mathrm{M}^{2+}(\mathrm{g})+\mathrm{e}^{-}$
(5) $\mathrm{M}(\mathrm{g}) \longrightarrow \mathrm{M}^{2+}(\mathrm{g})+2 \mathrm{e}^{-}$
The second ionization energy of $\mathrm{M}$ could be calculated from the energy values associated with:
- $1+3+4$
- $2-1+3$
- $1+5$
- $5-3$
Solution
Asked in: JEE-TOPICTESTS-CHEMISTRY
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