Octahedral d6 Complexes and Spin States

An octahedral complex with a d⁶ electron configuration can exist in two main spin states depending on the crystal field strength: high spin and low spin.

• High Spin (Weak field): The electrons occupy the \(t_{2g}\) and \(e_g\) orbitals according to Hund's rule, maximizing the number of unpaired electrons. The configuration is \(t_{2g}^4 e_g^2\).
• Low Spin (Strong field): Electrons preferentially pair up in the lower energy \(t_{2g}\) orbitals before occupying the \(e_g\) orbitals. The configuration is \(t_{2g}^6 e_g^0\).

Let's determine the number of unpaired electrons for each case.

Spin State	Electron Configuration	Number of Unpaired Electrons (n)
High Spin	\(t_{2g}^4 e_g^2\)	4
Low Spin	\(t_{2g}^6 e_g^0\)	0

Spin-Only Magnetic Moment Calculation

The spin-only magnetic moment (\(\mu_s\)) is calculated using the formula: \[ \mu_s = \sqrt{n(n+2)} \text{ B.M.} \] where \(n\) is the number of unpaired electrons and B.M. stands for Bohr Magnetons. * For the high spin d⁶ complex (n=4): \[ \mu_s = \sqrt{4(4+2)} = \sqrt{4 \times 6} = \sqrt{24} \approx 4.90 \text{ B.M.} \] * For the low spin d⁶ complex (n=0): \[ \mu_s = \sqrt{0(0+2)} = \sqrt{0} = 0 \text{ B.M.} \] Based on the magnetic moment values in the options, the complex is likely high spin, corresponding to a magnetic moment of approximately 4.9 B.M.

Electronic Transitions in Octahedral d6 Complexes

Electronic transitions involve the movement of an electron from a ground state energy level to an excited state energy level. For a transition to be spin-allowed, the spin multiplicity (\(2S+1\)) must remain the same in both the ground and excited states (\(\Delta S = 0\)). * High Spin d⁶: * Ground state configuration: \(t_{2g}^4 e_g^2\) (\(S=2\), \(2S+1=5\)). The ground state term symbol is \(^5T_{2g}\). * Lowest excited state configuration: \(t_{2g}^3 e_g^3\) (\(S=2\), \(2S+1=5\)). The term symbol is \(^5E_g\). * The spin-allowed transition is \(^5T_{2g} \rightarrow ^5E_g\). This transition is written as Excited State \(\leftarrow\) Ground State. So it is \(^5E_g \leftarrow ^5T_{2g}\). This gives one spin-allowed absorption band. * Low Spin d⁶: * Ground state configuration: \(t_{2g}^6 e_g^0\) (\(S=0\), \(2S+1=1\)). The ground state term symbol is \(^1A_{1g}\). * Lowest excited state configurations: \(t_{2g}^5 e_g^1\). This gives rise to excited states with term symbols \(^1T_{1g}\) and \(^1T_{2g}\) (among others). * The spin-allowed transitions from the ground state are \(^1A_{1g} \rightarrow ^1T_{1g}\) and \(^1A_{1g} \rightarrow ^1T_{2g}\). This typically results in two spin-allowed absorption bands (though they might overlap or one might be weak).

Determining the Correct Case

The question states that the complex has a single spin-allowed absorption band.
Comparing the two spin states:

• High spin d⁶ gives one spin-allowed transition (\(^5T_{2g} \rightarrow ^5E_g\)).
• Low spin d⁶ gives typically two spin-allowed transitions (\(^1A_{1g} \rightarrow ^1T_{1g}\) and \(^1A_{1g} \rightarrow ^1T_{2g}\)).

Therefore, the complex must be in the high spin configuration.

Final Properties

For the high spin d⁶ octahedral complex:

• Spin-only magnetic moment \(\approx\) 4.9 B.M.
• Electronic transition: \(^5E_g \leftarrow ^5T_{2g}\).

Let's examine the options to find the one that matches these properties.

Option	Magnetic Moment (B.M.)	Electronic Transition	Match?
1	0	\(^1T_{1g} \leftarrow ^1A_{1g}\)	No (Low spin, two transitions expected)
2	4.9	\(^5T_{2g} \leftarrow ^5E_g\)	No (Correct magnetic moment, but transition direction is reversed)
3	4.9	\(^5E_g \leftarrow ^5T_{2g}\)	Yes (Correct magnetic moment and transition)
4	0	\(^1T_{2g} \leftarrow ^1A_{1g}\)	No (Low spin, two transitions expected)

Option 3 matches the calculated spin-only magnetic moment of approximately 4.9 B.M. and the single spin-allowed electronic transition \(^5E_g \leftarrow ^5T_{2g}\) expected for a high spin d⁶ octahedral complex.