Question

The correct order of the spin-only magnetic moment of metal ions in the following low spin complexes, ${[V(CN)_6]^{4-}, [Fe(CN)6]^{4-}, [Ru (NH3)6]^{3+},}$ and ${[Cr(NH3)6]^{2+}}$, is :

• A. ${V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}}$
• B. ${V^{2+} > Ru^{3+} > Cr^{2+}> Fe^{2+}}$
• C. ${Cr^{2+} > V^{2+} > Ru^{3+} > Fe^{2+}}$
• D. ${ Cr^{2+} > Ru^{3+} > Fe^{2+} > V^{2+} }$

Answer 1

The Correct Option is A

 To determine the correct order of the spin-only magnetic moment for the given low spin complexes, we need to calculate or determine the number of unpaired electrons in each complex. The spin-only magnetic moment \((\mu)\) is given by the formula:

\(\mu = \sqrt{n(n+2)}\)

where \(n\) is the number of unpaired electrons.

1. Complex: \([V(CN)_6]^{4-}\)
2. Complex: \([Fe(CN)_6]^{4-}\)
3. Complex: \([Ru(NH_3)_6]^{3+}\)
4. Complex: \([Cr(NH_3)_6]^{2+}\)

This analysis shows that the complexes should be ranked based on the number of paired electrons and their ability to pair-off completely in the \(t_{2g}\) orbitals under low spin condition. However, considering paramagnetism intensity, \(Cr^{2+}\) ion often just has more tendency in paramagnetic as compared to \(V^{2+}\) ions but both could end at zero unpaired electrons for low spin complex. Hence, considering options and subtle knowledge of how spin crossover can be in practice, at the exam, the efficient order is \(V^{2+} > Cr^{2+} > Ru^{3+} > Fe^{2+}\) identifying properly paramagnetic stability.