The metal ion with total number of electrons | vedclass

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Question

(B) Total number of electrons in $S^{2-} = 16 + 2 = 18$.Total number of electrons present in the ions given in options are as follows:$(I)$ $Na^{+}$:

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$Ca^{2+}$

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(B) Total number of electrons in $S^{2-} = 16 + 2 = 18$.Total number of electrons present in the ions given in options are as follows:$(I)$ $Na^{+}$: Total electrons = $11 - 1 = 10$.$(II)$ $Ca^{2+}$: Total electrons = $20 - 2 = 18$.$(III)$ $Mg^{2+}$: Total electrons = $12 - 2 = 10$.$(IV)$ $Sr^{2+}$: Total electrons = $38 - 2 = 36$.Thus,$S^{2-}$ and $Ca^{2+}$ have the same number of electrons.

The metal ion with total number of electrons same as $S^{2-}$ is

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