MCQ
The correct order of $2^{nd}$ ionisation potential of carbon, nitrogen, oxygen and fluorine is
- A$C > N > O > F$
- B$O > N > F > C$
- ✓$O > F > N > C$
- D$F > O > N > C$
✓
Answer
Correct option: C.
$O > F > N > C$
c
$\mathrm{C}(6) \quad 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}_{\mathrm{x}}^{1} 2 \mathrm{p}_{\mathrm{y}}^{1}$
$\mathrm{C}(6) \quad 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}_{\mathrm{x}}^{1} 2 \mathrm{p}_{\mathrm{y}}^{1}$
$\mathrm{N}(7) \quad 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}_{\mathrm{x}}^{1} 2 \mathrm{p}_{\mathrm{y}}^{\mathrm{i}} 2 \mathrm{p}_{\mathrm{z}}^{1}$
$\mathrm{O}(\mathrm{8}) \quad 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}_{\mathrm{x}}^{2} 2 \mathrm{p}_{\mathrm{y}}^{1} 2 \mathrm{p}_{\mathrm{z}}^{1}$
$\mathrm{F}(9) \quad 1 \mathrm{s}^{2} 2 \mathrm{s}^{2} 2 \mathrm{p}_{\mathrm{x}}^{2} 2 \mathrm{p}_{\mathrm{y}}^{2} 2 \mathrm{p}_{\mathrm{z}}^{1}$
$O$ has highest $2^{\text {nd }}$ ionisation potential i.e ionisation energy because after losing one electron, it acquires noble gas configuration.
$\mathrm{F}$ is smaller than $\mathrm{N}$, therefore, it has higher second ionisation potential. $C$ has least due to bigger size.
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