Which is a stronger reducing agent $C r^{2 +}$ or $F e^{2 +}$ and why?

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Solution

$C r^{+ 2} = 3 d^{4}$

$F e^{+ 2} = 3 d^{6}$

reducing agent means it will reduce the other & itself gets oxidized.
Means, one which will have a greater tendency to lose $e^{⊖}$ , would be better reducing agent.

$∴ F e^{+ 2}$ having $3 d^{6}$

The configuration will be more reducing because after losing an electron it will have $d^{5}$ configuration which is stable as it is half-filled configuration.
Hence, $F e^{+ 2}$ is stronger reducing agent than $C r^{+ 2}$

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