The difference in the number of unpaired electrons of a metal ion in its high spin and low-spin octahedral complexes is two. The metal ion is :
Show Hint
- \(\mathrm{Mn^{2+}}\)
- \(\mathrm{Ni^{2+}}\)
- \(\mathrm{Co^{2+}}\)
- \(\mathrm{Fe^{2+}}\)
The Correct Option is C
Solution and Explanation
\(\mathrm{Ni^{2+}}\): d8. Only one arrangement, \(t_{2g}^6 e_g^2\), 2 unpaired in both. Difference = 0.
\(\mathrm{Co^{2+}}\): d7. High spin \(t_{2g}^5 e_g^2\) has 3 unpaired; low spin \(t_{2g}^6 e_g^1\) has 1. Difference = 2.
\(\mathrm{Fe^{2+}}\): d6. High spin \(t_{2g}^4 e_g^2\) has 4 unpaired; low spin \(t_{2g}^6\) has 0. Difference = 4.
Difference of two is obtained only for \(\mathrm{Co^{2+}}\).
Answer: (3) \(\mathrm{Co^{2+}}\)
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