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In lanthanide series, the element well known to exhibit +4 oxidation state is
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The correct answer is:
$\mathrm{Ce}$
$$
\begin{aligned}
\mathrm{Ce}(58) & =[\mathrm{Xe}] 4 f^1 5 d^1 6 s^2 \\
\mathrm{Ce}^{+4} & =[\mathrm{Xe}] 4 f^0 5 d^0 6 s^0
\end{aligned}
$$
Cerium exihibit +4 oxidation state.
\begin{aligned}
\mathrm{Ce}(58) & =[\mathrm{Xe}] 4 f^1 5 d^1 6 s^2 \\
\mathrm{Ce}^{+4} & =[\mathrm{Xe}] 4 f^0 5 d^0 6 s^0
\end{aligned}
$$
Cerium exihibit +4 oxidation state.
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