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In which of the following change both the electrons are removed from same orbital?

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\begin{aligned} Cu \to {Cu}^{+ 2} + 2 e^{-} \end{aligned}

\begin{aligned} Cr \to {Cr}^{+ 2} + 2 e^{-} \end{aligned}

Ge \to {Ge}^{+ 2} + 2 e^{-}

Zn \to {Zn}^{+ 2} + 2 e^{-}

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detailed solution

Correct option is D

The following change both the electrons are removed from same orbital are,

$\begin{array}{l} _{29} {Cu}_{} \to [Ar] 3 d^{10} 4 s^{1} \\ _{24} Cr \to [Ar] 3 d^{5} 4 s^{1} \\ _{32} Ge \to [Ar] 3 d^{10} 4 s^{2} \\ _{30} Zn \to [Ar] 3 d^{10} 4 s^{2} \end{array}$

The electrons are removed from same orbital in the $_{30} Zn$ .

Hence option D is correct.

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