Question

Determine the oxidation state of the metal in each of

the following coordination complexes: $\left[\mathbf{V}{\left({\mathbf{NH}}_{\mathbf{3}}\right)}_{\mathbf{4}}{\mathbf{Cl}}_{\mathbf{2}}\right]$, ${\left[\mathbf{CO}{\left({\mathbf{OH}}_{\mathbf{2}}\right)}_{\mathbf{2}}\left({\mathbf{NH}}_{\mathbf{3}}\right){\mathbf{Cl}}_{\mathbf{3}}\right]}^{\mathbf{-}}$, [Ni(CO)4].

Short answer

The oxidation state of metal vanadium is +2; molybdenum is +2, cobalt is +1, and nickel is 0.

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 vanadium metal (V) in complex$\left[\text{V}{\left({\text{NH}}_{\text{3}}\right)}_{\text{4}}{\text{Cl}}_{\text{2}}\right]$determined as follows:

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

The oxidation state of molybdenum metal (MO) in complex${\left[{\text{Mo}}_{\text{2}}{\text{Cl}}_{\text{8}}\right]}^{\text{4 -}}$ determined as follows:

$\begin{array}{rcl}& & \text{2x + 8}\times \left(\text{- 1}\right)\text{=}\text{- 4}\\ & & \text{2x - 8}\text{=}\text{- 4}\\ & & \text{2x}\text{=}+\text{4}\\ & & \text{x}\text{=}+\text{2}\end{array}$

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

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

The oxidation state of nickel metal (Ni) in complex$\left[\text{Ni}{\left(\text{CO}\right)}_{\text{4}}\right]$ determined as follows:

$\begin{array}{rcl}& & \text{x + 4}\times \text{0}\text{=}\text{0}\\ & & x\text{=}\text{0}\end{array}$