The ionization energy of \(\mathrm{Xe} ightarrow \mathrm{Xe}^{+}+\mathrm{e}^{-}\)is very close to that of
- \(\mathrm{N}_{2} ightarrow \mathrm{N}_{2}^{+}+\mathrm{e}^{-}\)
- \(\mathrm{O}_{2} ightarrow \mathrm{O}_{2}^{+}+\mathrm{e}^{-}\)
- \(\mathrm{C}_{2} ightarrow \mathrm{C}_{2}^{+}+\mathrm{e}^{-}\)
- \(\mathrm{B}_{2} ightarrow \mathrm{B}_{2}^{+}+\mathrm{e}^{-}\)
Solution
Asked in: JEE-TOPICTESTS-CHEMISTRY
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