Question

What is the oxidation state of cr the complex [CrCl2(H2O)4]^+?

Answer 1

The oxidation number of `"Cr"` is +3.

Explanation:

We can use the rules for oxidation numbers. The essential rules for this problem are:

1. The oxidation number of a halogen in a compound is -1 unless it is with a more electronegative element.
2. The sum of the oxidation numbers of all atoms in a neutral compound is 0.
3. The sum of the oxidation numbers of all atoms in an ion equals charge on the ion.

Per Rule 1, the oxidation number of `"Cl"` is -1.

Write the oxidation number above the `"Cl"` in the formula:

`"Cr"stackrelcolor(blue)("-1")("Cl")_2("H"_2"O")_4^"+"`

The two `"Cl"` atoms together have a total oxidation number of -2.

Write this number below the `"Cl"` atom:

`"Cr"stackrelcolor(blue)("-1")("Cl")_2("H"_2"O")_4^"+"`
`color(white)(mll)stackrelcolor(blue)("-2")()`

Per Rule 2, the sum of all the oxidation numbers in `"H"_2"O"` must be 0.

Put zeroes above and below the formula for water.

`"Cr"stackrelcolor(blue)("-1")("Cl")_2(stackrelcolor(blue)(0)("H"_2"O"))_4^"+"`
`color(white)("Cr"stackrelcolor(blue)("-2")("Cl")_2(stackrelcolor(blue)(0)("H"_2"O"))_4^"+")`

Per Rule 3, the sum of all the oxidation numbers equals +1.

Let `x =` the oxidation number of `"Cr"`. Then

`x-2+0 = 1`

`x = 1+2 = 3`

There is only one `"Cr"` atom, so its oxidation number must be +3.

Write the oxidation number above and below the `"Mn"`.

`color(white)(m)stackrelcolor(blue)("+3")("Cr")stackrelcolor(blue)("-1")("Cl")_2(stackrelcolor(blue)(0)("H"_2"O"))_4^"+"`
`color(white)((stackrelcolor(blue)("+3")("Cr")stackrelcolor(blue)("-2")("Cl")_2(stackrelcolor(blue)(0)("H"_2"O"))_4^"+")`

The oxidation number of `"Cr"` is +3.