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The number of $\mathrm{d}$ - electrons in $\mathrm{Cr}^{3+}$ ion $(\mathrm{Z}=24)$ is
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$3$
$\mathrm{Cr}=[\mathrm{Ar}] 3 \mathrm{~d}^5 4 \mathrm{~s}^1$ so $\mathrm{Cr}^{3+}=[\mathrm{Ar}] 3 \mathrm{~d}^3$
$\Rightarrow \quad$ Number of d-electrons $=3$
$\Rightarrow \quad$ Number of d-electrons $=3$
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