Question

Determine the number of unpaired electrons in the octahedral coordination complex [fex6]3–, where x = any halide.

Answer 1

There will be four unpaired electrons
The metal complex is [FeX₆]³⁻
X being the halogen ligand 
X = F, CL, Br, and I
The oxidation of metal state is +3
The ground state configuration is
₂₆Fe =Is² 2s²2p⁶ 3s² 3p⁶ 3d⁶ 4s²
Metal, Fe(III) ion electron configures
₂₆Fe³⁺ = Is2 2s² 2p⁶ 3s² 3p⁶ 3d⁵

Answer 2

Unpaired electron is the electron, that occupies the a place in orbital without the pair of electron. The number of unpaired electron in 3d sub shell of octahedral coordination complex $[FeX_6]^{2-}$ is 3.

The given octahedral coordination complex in the problem is ${FeX_6]^{2-}$

Here, $X$ is the halogen.

What is unpaired electron?

Unpaired electron is the electron, that occupies the a place in orbital without the pair of electron.

In the given complex ion +3 oxidation state $F$ (iron) represents in,

$_{26}F^{3+}=1s^2, 2s^2 2p^6, 3s^23p^63d^6,4s^2$

$F^{3+}=1s^2, 2s^2 2p^6, 3s^23p^63d^6$

As the unpaired electron in 3d sub shell is 3.

Hence the number of unpaired electron in 3d sub shell of octahedral coordination complex $[FeX_6]^{2-}$ is 3.

Learn more about the unpaired electron here;https://brainly.com/question/2292932