Correct order of first IP among following elements $\mathrm{Be}, \mathrm{B}, \mathrm{C}, \mathrm{N},…
- $\quad \mathrm{B} < \mathrm{Be} < \mathrm{C} < \mathrm{O} < \mathrm{N}$
- $\quad \mathrm{B} < \mathrm{Be} < \mathrm{C} < \mathrm{N} < \mathrm{O}$
- $\quad \mathrm{Be} < \mathrm{B} < \mathrm{C} < \mathrm{N} < \mathrm{O}$
- $\quad \mathrm{Be} < \mathrm{B} < \mathrm{C} < \mathrm{O} < \mathrm{N}$
Solution
$\mathrm{N}-1 s^{2} 2 s^{2} 2 p^{3 ;} \mathrm{O}-1 s^{2} 2 s^{2} 2 p^{4}$. IP increases
along the period. But IP of $\mathrm{Be}>\mathrm{B}$. Further IP of $\mathrm{O} < \mathrm{N}$ because atoms with fully or partly filled orbitals are most stable and hence have high ionisation energy. .
Asked in: JEE-TOPICTESTS-CHEMISTRY
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