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What are the spin-only magnetic moments (in Bohr Magneton) for the complexes [Cr(NH₃)₆]³⁺ and [CuF₆]³⁻, given that the atomic numbers of Cr and Cu are 24 and 29, respectively?
- 3.87 and 2.84
- 4.90 and 1.73
- 3.87 and 1.73
- 4.90 and 2.84
Correct answer: 3.87 and 2.84
Solution
In [Cr(NH3)6]3+, Cr3+ is d3 with 3 unpaired electrons giving mu = sqrt(3*5) = 3.87 BM. In [CuF6]3-, Cu3+ is d8 and F- is a weak field, leaving 2 unpaired electrons giving mu = sqrt(2*4) = 2.83 ~ 2.84 BM. The correct pair is 3.87 and 2.84 (option 0); the stored answer is wrong.
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- Match the complexes listed in Group-I with their respective characteristics in Group-II based on the given information.
Group-I:
(I) [Cr(CN)₆]³⁻
(II) [RuCl₆]²⁻
(III) [Cr(H₂O)₆]²⁺
(IV) [Fe(H₂O)₆]²⁺
Group-II:
(P) The t₂g orbitals are occupied by 4 electrons.
(Q) The spin-only magnetic moment is 4.9 BM.
(R) Forms a low-spin complex.
(S) The metal ion has an oxidation state of +4.
(T) The complex is a d⁴ species.
Given: Chromium (Cr) has an atomic number of 24, Ruthenium (Ru) has an atomic number of 44, and Iron (Fe) has an atomic number of 26.
Choose the correct pairing of Group-I with Group-II.
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