Solution:

• The magnetic moment of a complex is given by the formula:

Magnetic moment = √n(n+2)BM

Where n is the total number of unpaired electrons in the complex and BM is the Bohr magneton (9.27 × 10^-24 joule/tesla).

• Given the magnetic moment of the Co(II) complex is 4.0 BM, we can calculate the value of n as follows:

4.0 BM = √n(n+2)BM

Squaring both sides:

16.0 = n(n+2)

n^2 + 2n - 16 = 0

(n+4)(n-2) = 0

n = -4 or n = 2

Since the number of unpaired electrons cannot be negative, we take n = 2.

• The electronic configuration of Co(II) is [Ar]3d^7.

• The complex is octahedral, which means that the d-orbitals split into two sets - t2g (dxy, dyz, dxz) and eg (dx^2-y^2, dz^2).

• The number of electrons in the t2g set is given by (n - Δo)/2, where Δo is the crystal field splitting energy.

• The number of electrons in the eg set is given by Δo/2.

• We can use the fact that the complex has 2 unpaired electrons to determine the value of Δo.

• If the complex had a high-spin configuration (t2g4 eg2), then Δo would be small and the magnetic moment would be less than 4.0 BM.

• Therefore, the complex must have a low-spin configuration.

• The only possible low-spin configuration with 2 unpaired electrons is t2g4 eg1.

• Therefore, the correct answer is option B) t2g6 eg1.