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The pair in which both species have same magnetic moment (spin onlyvalue) is:
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$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}_{6}\right)^{2+},\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{3+}\right.$
$\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right]^{2+} \mathrm{Cr}$ is in $\mathrm{Cr}^{2+}$ form
In $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+} \mathrm{Fe}^{2+}$ form. Both will have 4
unpaired electrons.
In $\left[\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)\right]^{2+} \mathrm{Fe}^{2+}$ form. Both will have 4
unpaired electrons.
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