Question

The correct order of spin-only magnetic moments among the following is : (Atomic number : Mn=25, Co=27, Ni=28, Zn=30)

• A. $\left[ ZnCl _{4}\right]^{2-}>\left[ NiCl _{4}\right]^{2-}>\left[ CoCl _{4}\right]^{2-}>\left[ MnCl _{4}\right]^{2}$
• B. $\left[ CoCl _{4}\right]^{2^{2}}>\left[ MnCl _{4}\right]^{2^{-}}>\left[ NiCl _{4}\right]^{2^{2}}>\left[ ZnCl _{4}\right]^{2-}$
• C. $\left[ NiCl _{4}\right]^{2}>\left[ CoCl _{4}\right]^{2-}>\left[ MnCl _{4}\right]^{2}>\left[ ZnCl _{4}\right]^{2-}$
• D. $\left[ MnCl _{4}\right]^{2^{2}}>\left[ CoCl _{4}\right]^{2-}>\left[ NiCl _{4}\right]^{2}>\left[ ZnCl _{4}\right]^{2-}$

Answer 1

The Correct Option is D

The question requires us to determine the order of spin-only magnetic moments of the given complex ions: \( \left[ ZnCl_4 \right]^{2-} \), \( \left[ NiCl_4 \right]^{2-} \), \( \left[ CoCl_4 \right]^{2-} \), and \( \left[ MnCl_4 \right]^{2-} \).

To calculate the spin-only magnetic moment (\(\mu_s\)), we use the formula:

\mu_s = \sqrt{n(n+2)} Bohr Magneton (BM)

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

1. Manganese (Mn) in \( \left[ MnCl_4 \right]^{2-} \):

   Mn has an atomic number of 25. In the \(+2\) oxidation state, Mn loses two electrons from its 4s orbital: 3d⁵. This means it has 5 unpaired electrons.

   So, for \( \left[ MnCl_4 \right]^{2-} \), n = 5.

   The magnetic moment is \mu_s = \sqrt{5(5+2)} = \sqrt{35} \approx 5.92 BM.

2. Cobalt (Co) in \( \left[ CoCl_4 \right]^{2-} \):

   Co has an atomic number of 27. In the \(+2\) oxidation state, Co loses two electrons, making 3d⁷. It has 3 unpaired electrons.

   So, for \( \left[ CoCl_4 \right]^{2-} \), n = 3.

   The magnetic moment is \mu_s = \sqrt{3(3+2)} = \sqrt{15} \approx 3.87 BM.

3. Nickel (Ni) in \( \left[ NiCl_4 \right]^{2-} \):

   Ni has an atomic number of 28. In the \(+2\) oxidation state, Ni loses two electrons, making 3d⁸. It has 2 unpaired electrons.

   So, for \( \left[ NiCl_4 \right]^{2-} \), n = 2.

   The magnetic moment is \mu_s = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 BM.

4. Zinc (Zn) in \( \left[ ZnCl_4 \right]^{2-} \):

   Zn has an atomic number of 30. In the \(+2\) oxidation state, Zn loses two electrons, making 3d¹⁰. It has 0 unpaired electrons.

   So, for \( \left[ ZnCl_4 \right]^{2-} \), n = 0.

   The magnetic moment is \mu_s = \sqrt{0(0+2)} = 0 BM.

The correct order of spin-only magnetic moments is, therefore: \( \left[ MnCl_4 \right]^{2-} \gt \left[ CoCl_4 \right]^{2-} \gt \left[ NiCl_4 \right]^{2-} \gt \left[ ZnCl_4 \right]^{2-} \).

Hence, the correct answer is: \( \left[ MnCl_4 \right]^{2-} \gt \left[ CoCl_4 \right]^{2-} \gt \left[ NiCl_4 \right]^{2-} \gt \left[ ZnCl_4 \right]^{2-} \).