Question

determine the oxidation state of the transition metal in each of these coordination compounds.
compound | oxidation state of the transition metal
fecl₂(co)₄cl |
nazncl₃(nh₃) |
namnbr₄(h₂o)₂ |

Explanation:

Step1: Analyze \(\boldsymbol{[FeCl_2(CO)_4]Cl}\)

Let the oxidation state of Fe be \(x\). \(Cl\) (in the ligand) has \(-1\), \(CO\) is neutral (0). The complex cation \([FeCl_2(CO)_4]^+\) (since the outer \(Cl\) is \(-1\), overall compound is neutral). So:
\(x + 2(-1) + 4(0) = +1\) (because the cation charge is \(+1\) to balance outer \(Cl^-\)).
\(x - 2 = 1 \implies x = +3\).

Step2: Analyze \(\boldsymbol{Na[ZnCl_3(NH_3)]}\)

Let Zn’s oxidation state be \(y\). \(Na\) is \(+1\), so the anion \([ZnCl_3(NH_3)]^-\) (to balance \(Na^+\)). \(Cl\) is \(-1\), \(NH_3\) is neutral.
\(y + 3(-1) + 0 = -1\) (anion charge).
\(y - 3 = -1 \implies y = +2\).

Step3: Analyze \(\boldsymbol{Na[MnBr_4(H_2O)_2]}\)

Let Mn’s oxidation state be \(z\). \(Na\) is \(+1\), so anion \([MnBr_4(H_2O)_2]^-\). \(Br\) is \(-1\), \(H_2O\) is neutral.
\(z + 4(-1) + 2(0) = -1\) (anion charge).
\(z - 4 = -1 \implies z = +3\).

Answer:

For \(\boldsymbol{[FeCl_2(CO)_4]Cl}\): \(\boldsymbol{+3}\)
For \(\boldsymbol{Na[ZnCl_3(NH_3)]}\): \(\boldsymbol{+2}\)
For \(\boldsymbol{Na[MnBr_4(H_2O)_2]}\): \(\boldsymbol{+3}\)