How would you account for the following:
Of the $d^{4}$ species, ${C r}^{2 +}$ is strongly reducing while manganese (III) is strongly oxidising.

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Solution

${C r}^{2 +}$ is strongly reducing in nature. It has a $d^{4}$ configuration. While acting as a reducing agent, it gets oxidised to ${C r}^{3 +}$ (electronic configuration $d^{3}$ ). This $d^{3}$ configuration can be written as $t_{2}^{3}$ configuration, which is a more stable configuration.

In the case of ${M n}^{3 +}$ $(d^{4}),$ it gains one electron to become ${M n}^{2 +}$ $(d^{5})$ . This has an exactly half filled $d -$ orbital and is highly stable.

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C r^{2 +}

M n^{3 +}

C r^{2 +}

M n^{3 +}

d^{4}

C r^{2 +}

d -

(d^{4}) M n^{+ 3}

d^{4}

C r^{2 +}

(I I I)

(I I)

d^{1}

Explain why Cr2+ is reducing but Mn3+ is oxidising while both ions have d4 configuration?

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