The difference in the number of unpaired electrons between the low-spin complex and the high-spin co — Coordination Compounds Chemistry Question
Question
The difference in the number of unpaired electrons between the low-spin $d^6$ complex $[Co(NH_3)_6]^{3+}$ and the high-spin $d^6$ complex $[Fe(H_2O)_6]^{2+}$ is:
Answer: C
💡 Solution & Explanation
$[Co(NH_3)_6]^{3+}$ pairs fully (0 unpaired). $[Fe(H_2O)_6]^{2+}$ is high spin $t_{2g}^4 e_g^2$ (4 unpaired). Difference is $4 - 0 = 4$ [65, 66].
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