Electronic configuration of calcium is $\ce{1s^2 2s^2 2p^6 3s^2 3p^6 4s^2}$. The orbitals are fully filled with electrons so would its valency be two or zero?


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    $\begingroup$ The configuration $$1s^2 2s^2 2p^6 3s^2 3p^6 4s^2$$ has a valency of zero since there are as many electrons as protons. The configuration $$1s^2 2s^2 2p^6 3s^2 3p^6 $$ is two electrons short, so it has a valency of +2. $\endgroup$ – MaxW Dec 17 '15 at 15:34

Yes, the valency is 2. It's all orbitals are completely filled, but it's outer most orbit is incomplete. Ca can easy donate 2 electrons and can become stable cation.


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