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Which of the following cation has only two electrons in its last shell?

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a

{Sn}^{+ 4}

b

{Bi}^{+ 5}

c

{Pb}^{+ 2}

d

{Hg}^{+ 2}

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detailed solution

Correct option is C

Ground-state electronic configuration of given $Pb$ elements is $[Xe] 6 s^{2} 4 f^{14} 5 d^{10} 6 p^{2}$ . Now, by disposing $6 p$ electrons we get ${Pb}^{2 +}$ with Ground-state electronic configuration of $[Xe] 6 s^{2} 4 f^{14} 5 d^{10}$ . Hence, the correct option is C.

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