Question

In the form of dichromate, $Cr\left(VI\right)$ is a strong oxidizing agent in acidic medium but $Mo\left(VI\right)$ in $Mo{O}_3$ and $W\left(VI\right)$ in $W{O}_3$ are not because ___________.

• A. $Cr\left(VI\right)$ is more stable than $Mo\left(VI\right)$ and $W\left(VI\right)$
• B. " Higher oxidation states of heavier members of group-6 of transition series are more stable"
• C. $Mo\left(VI\right)$ and $W\left(VI\right)$ are more stable than $Cr\left(VI\right)$
• D. Lower oxidation states of heavier members of group$-6$ of transition series are more stable

Answer 1

Answer: (B), (C)

$Mo\left(VI\right)$ and $W\left(VI\right)$ are more stable than $Cr\left(VI\right)$

The $Cr\left(VI\right)$ is not the only stable oxidation state of Chromium. So, $Cr\left(VI\right)$ gains electrons to reduce itself to its other stable oxidation states like $Cr\left(III\right),$ thereby oxidizing other species.

Electronic configuration of $Cr$ is $\left[Ar\right]3d^{5}4s^1$

Electronic configuration of $C{r}^{3+}$ is $\left[Ar\right]3d^3$

This makes its half-filled (highly stable) at the $t_{2g}$​ level which means three $sub-d$ orbitals i.e. $d_{xy}{d}_{yz}{d}_{zx}$ are half filled.

Hence, it acts as an oxidizing agent.

Down the group, stability of heavier transition metals in their higher oxidation states increases.

This is due to the similarity of the energy of the $d-$orbitals with the outer $s$ orbitals. Hence the $d-$orbitals are more available for bond formation.

As $Mo$ and $W$ are heavier elements, these elements are more stable in their $+6$ oxidation state. Thus, these metals have little or no tendency to get reduced to their lower oxidation states, that’s why $Mo\left(VI\right)$ in $Mo{O}_3$ and $W\left(VI\right)$ in $W{O}_3$ are not strong oxidizing agents.

Hence, $Mo\left(VI\right)$ and $W\left(VI\right)$ are more stable than $Cr\left(VI\right)$ and higher oxidation states of heavier members of group$-6$ of transition series are more stable.

So correct answers are options $\left(B\right)$ and $\left(C\right)$