Question

Which of the following actinoids show oxidation states upto $+7?$

• A. $Pu$
• B. $Np$
• C. $Am$
• D. $U$

Answer 1

Answer: (A), (B)

$Np$

Analyzing the given options

In the actinide series, the oxidation state first increases and then decreases on moving left to right.

The maximum oxidation state first increases up to the middle of the series and then decreases i.e. it increases from $+4$ for Th to $+5,+6$ and $+7$ for $Pa,U$ and $Np\left(\text{and}Pu\right)$ respectively and then decreases in the succeeding elements.

Option $\left(A\right)$

The electronic configuration of Am $\left(\text{Americium}\right)$ is $\left[Rn\right]5f^{7}7s^2$ and it can exhibit oxidation states such as $+3,+4,+5,$ and $+6.$

Option$\left(B\right)$

The electronic configuration of $Pu\left(\text{Plutonium}\right)$ is $\left[Rn\right]5f^{6}7s^2,$ thus it can exhibit oxidation states such as $+3,+4,+5,+6$ and $+7.$

Option $\left(C\right)$

The electronic configuration of $U\left(\text{Uranium}\right)$ can be written as $\left[Rn\right]5f^{3}6d^{1}7s^2$ and it can exhibit oxidation states such as $+3,+4,+5,$ and $+6.$

Option $\left(D\right)$

The electronic configuration of $Np\left(\text{Neptunium}\right)$ is $\left[Rn\right]5f^{4}6d^{1}7s^2,$ thus it can exhibit oxidation states such as $+3,+4,+5,+6$ and $+7.$

Thus, $Pu$ and $Np$ can show oxidation states upto $+7.$

Hence, the correct options are $\left(B\right)$ and $\left(D\right).$