MCQ
The electronic configuration of calcium ion $(C{a^{2 + }})$ is
- A$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6},4{s^2}$
- B$1{s^2},2{s^2}s{p^6},3{s^2}3{p^6},4{s^1}$
- C$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6}3{d^2}$
- ✓$1{s^2},2{s^2}2{p^6},3{s^2}3{p^6},4{s^0}$
The calcium atom loses two electrons for the formation of $Ca ^{2+}$.
So, two electrons are removed from the $4 s$ orbital for the formation of $Ca ^{2+}$
Hence, the electronic configuration of $Ca ^{2+}$ is $1 s ^2\, 2 s ^2 \,2 p ^6 \,3 s ^2 \,3 p ^6$.
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