Question

Determine the oxidation state of the metal in each of

the following coordination complexes:$\left[{\mathbf{Mn}}_{\mathbf{2}}{\left(\mathbf{CO}\right)}_{\mathbf{10}}\right]$,${\left[{\mathbf{Re}}_{\mathbf{3}}{\mathbf{Br}}_{\mathbf{12}}\right]}^{\mathbf{3}\mathbf{-}}$, ${\left[\mathbf{Fe}{\left({\mathbf{OH}}_{\mathbf{2}}\right)}_{\mathbf{4}}{\left(\mathbf{OH}\right)}_{\mathbf{2}}\right]}^{\mathbf{+}}$,${\left[\mathbf{CO}{\left({\mathbf{NH}}_{\mathbf{3}}\right)}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}\right]}^{\mathbf{+}}$

Short answer

The oxidation state of metal manganese is 0; rhenium +3, iron is +3, and cobalt is +3.

Step-by-step solution

Oxidation state

The oxidation state is thenumber assigned to the compound/complex element, indicating the amount of electron lost/gained by that element, or it can say the “charge present on that element.”

Calculation of oxidation of metal

The oxidation state of manganese metal (Mn) in complex$\left[{\text{Mn}}_{\text{2}}{\left(\text{CO}\right)}_{\text{10}}\right]$ determined as follows:

$\begin{array}{rcl}& & \text{2x + 10}\times \text{0}\text{=} \text{0}\\ & & \text{2x =}\text{0}\\ & & \text{x}\text{=}0\end{array}$

The oxidation state of rhenium metal (Re) in complex${\left[{\text{Re}}_{\text{3}}{\text{Br}}_{\text{12}}\right]}^{\text{3 -}}$determined as follows:

$\begin{array}{rcl}& & \text{3x + 12}\times \left(\text{- 1}\right)\text{=}\text{- 3}\\ & & \text{3x - 12}\text{=}\text{- 3}\\ & & \text{3x}\text{=}+9\\ & & \text{x}\text{=}+3\end{array}$

The oxidation state of iron metal (Fe) in complex${\left[\text{Fe}{\left({\text{OH}}_{\text{2}}\right)}_{\text{4}}{\left(\text{OH}\right)}_{\text{2}}\right]}^{\text{+}}$determined as follows:

$\begin{array}{rcl}& & \text{x + 4}\times \text{0 + 2}\times \left(\text{- 1}\right)\text{=}+1\\ & & \text{x - 2}\text{=}+\text{1}\\ & & \text{x}\text{=}\text{+ 3}\end{array}$

The oxidation state of cobalt metal (CO) in complex${\left[\text{CO}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}\right]}^{\text{+}}$determined as follows:

$\begin{array}{rcl}& & \text{x + 4}\times \text{0 + 2}\times \left(\text{- 1}\right)\text{=}\text{+ 1}\\ & & \text{X}\text{- 2 =}\text{+ 1}\\ & & \text{x} \text{= + 3}\end{array}$