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JEE Main 2024 · Chemistry · Coordination Compounds · medium

The total number of unpaired electrons in the complexes [Fe(H2O)6]^3+, [Fe(CN)6]^3-, [Co(NH3)6]^3+, [CoF6]^3-, [Ni(CN)4]^2- and [NiCl4]^2- taken…

Numerical answer

The total number of unpaired electrons in the complexes [Fe(H2O)6]3+[\mathrm{Fe(H_2O)_6}]^{3+}, [Fe(CN)6]3−[\mathrm{Fe(CN)_6}]^{3-}, [Co(NH3)6]3+[\mathrm{Co(NH_3)_6}]^{3+}, [CoF6]3−[\mathrm{CoF_6}]^{3-}, [Ni(CN)4]2−[\mathrm{Ni(CN)_4}]^{2-} and [NiCl4]2−[\mathrm{NiCl_4}]^{2-} taken together is
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Answer: 12

Use ligand strength and geometry. [Fe(H2O)6]3+[\mathrm{Fe(H_2O)_6}]^{3+}: d5d^5 high spin, 55. [Fe(CN)6]3−[\mathrm{Fe(CN)_6}]^{3-}: d5d^5 low spin, 11. [Co(NH3)6]3+[\mathrm{Co(NH_3)_6}]^{3+}: d6d^6 low spin, 00. [CoF6]3−[\mathrm{CoF_6}]^{3-}: d6d^6 high spin, 44. [Ni(CN)4]2−[\mathrm{Ni(CN)_4}]^{2-}: d8d^8 square planar, 00. [NiCl4]2−[\mathrm{NiCl_4}]^{2-}: d8d^8 tetrahedral, 22. Sum =5+1+0+4+0+2=12=5+1+0+4+0+2=12.
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