Question

The d-electronic configuration of an octahedral Co(II) complex having a magnetic moment of 3.95 BM is:

• A. \( t_{2g}^6 e_g^1 \)
• B. \( t_{2g}^3 e_g^0 \)
• C. \( t_{2g}^5 e_g^2 \)
• D. \( e_g^4 t_{2g}^3 \)

Hint:

For d-block elements, determining the electronic configuration in an octahedral complex can be done by considering the splitting of the d-orbitals into \( t_{2g} \) and \( e_g \) orbitals. The magnetic moment is also a useful indicator to determine the number of unpaired electrons in the complex.

Answer 1

The Correct Option is C

To solve this question, we need to determine the d-electronic configuration of a cobalt(II) complex in an octahedral environment with a given magnetic moment. Let's break it down step by step.

1. First, note that the formula for calculating the magnetic moment (\(\mu\)) in Bohr Magnetons (BM) is: \(\mu = \sqrt{n(n+2)}\) where \(n\) is the number of unpaired electrons. 
2. Given, \(\mu = 3.95\) BM. We can estimate the number of unpaired electrons by solving: \(\sqrt{n(n+2)} = 3.95\)

Solving for \(n\):

1. \(\mu^2 = n(n+2)\)
   \((3.95)^2 \approx 15.6 = n(n+2)\)\)
   \(n(n+2) = 16\)\) (approximate to the nearest whole number)
   Solving \(n^2 + 2n - 16 = 0\)\) gives \(n = 3\)\).
2. Cobalt(II) ion, \(Co^{2+}\), originates from \(Co\) with a ground state electronic configuration of\)
3. Considering that an octahedral complex is formed:
   • In octahedral complexes, the d-orbitals split into \(t_{2g}\) and \(e_g\) levels.
   • We need a configuration where there are 3 unpaired electrons, which fits the description \(3d^7\)\).
   • Thus, the most likely distribution with three unpaired electrons in an octahedral complex is: \(t_{2g}^5 e_g^2\)\) This involves placing 5 electrons in the lower energy \(t_{2g}\) orbitals and 2 unpaired electrons in the higher energy \(e_g\) orbitals.
4. Therefore, the electron configuration of the given Co(II) complex in an octahedral field with the given magnetic moment is indeed \(t_{2g}^5 e_g^2\).

Thus, the correct answer is: \(t_{2g}^5 e_g^2\).

Answer 2

To solve the problem of determining the d-electronic configuration of an octahedral Co(II) complex with a magnetic moment of 3.95 BM, we need to consider the following:

1. Cobalt (II) Ion: Cobalt in its +2 oxidation state (Co²⁺) means it has lost 2 electrons. The electronic configuration of neutral cobalt (atomic number 27) is [Ar] 3d⁷4s². Thus, Co²⁺ becomes [Ar] 3d⁷.
2. Magnetic Moment: The given magnetic moment is 3.95 BM. The magnetic moment (\(\mu\)) in Bohr Magneton (BM) is calculated using the formula: \(\mu = \sqrt{n(n+2)}\), where n is the number of unpaired electrons.
3. Calculate Unpaired Electrons: From the given magnetic moment:\[\mu = \sqrt{n(n+2)} \approx 3.95\]Squaring both sides to simplify:\[15.6 \approx n(n+2)\]Solving \(n^2 + 2n - 15.6 = 0\) suggests \(n\) is approximately 3. Consequently, Co²⁺ has 3 unpaired electrons.
4. d-Electronic Configuration: In an octahedral field, high-spin complexes generally distribute electrons in the \(t_{2g}\) and \(e_g\) orbitals such that maximally unpaired electrons are favored. Therefore, with 7 electrons:\(t_{2g}^5 e_g^2\). This distribution gives 3 unpaired electrons, consistent with a magnetic moment of 3.95 BM.

In conclusion, the d-electronic configuration of the Co(II) complex is: \(t_{2g}^5 e_g^2\).