Question

What are the oxidation numbers of each element in `Ca(OH)_2`?

Answer 1

Well the sum of the oxidation numbers equals the charge on the salt or molecule....i.e. the sum of the oxidation numbers equals ZERO.

Explanation:

We gots calcium, a Group 2 alkaline earth, i.e. `Ca^(2+)`...and `2xxHO^-`. Now hydrogen generally takes a `+I` oxidation number, and it does so here. And finally, we give oxygen a `-II` oxidation number....

And `2_"Ca"+2xx(-2_"O")+2xx(+1_"H")=0` as required....

And so we gots `Ca(+II)`, `O(-II)`, and `H(+I)` as the oxidation numbers....

Answer 2

Here's how to do it.

Explanation:

You must follow several rules to determine oxidation numbers.

The important rules for this question are:

1. The oxidation number of Group 2 metals in a compound is +2.
2. The oxidation number of oxygen in a compound is usually -2.
3. The oxidation number of hydrogen in a compound is usually +1.
4. The sum of all oxidation numbers in an ion must equal the charge on the ion.
5. The sum of all oxidation numbers in a neutral compound must equal zero.

Per Rule 1, the oxidation number of `"Ca"` is +2.

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

`stackrelcolor(blue)("+2")("Ca")"(OH)"_2`.

We will write the sum of the oxidation numbers below each atom.

There is only one `"Ca"` atom, so we write +2 below the `"Ca"` atom:

`stackrelcolor(blue)("+2")("Ca")"(OH)"_2`
`color(white)(stackrelcolor(blue)("+2")("Ca")"(OH)"_2)`

The oxidation number of the calcium ion is +2.

Per Rule 2, the oxidation number of `"O"` is -2.

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

`stackrelcolor(blue)("+2")("Ca")(stackrelcolor(blue)("-2")("O")"H")_2`

There is only one `"O"` atom inside the parentheses. so we write -2 below the atom.

`stackrelcolor(blue)("+2")("Ca")(stackrelcolor(blue)("-2")("O")"H")_2`
`color(white)(l)stackrelcolor(blue)("+2")("")color(white)(m)stackrelcolor(blue)("-2")("")`

Per Rule 3, the oxidation number of `"H"` is +1.

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

`stackrelcolor(blue)("+2")("Ca")(stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H"))_2`
`color(white)(l)stackrelcolor(blue)("+2")("")color(white)(m)stackrelcolor(blue)("-2")("")`

There is only one `"H"` atom inside the parentheses. so we write +1 below the atom.

`stackrelcolor(blue)("+2")("Ca")(stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H"))_2`
`color(white)(l)stackrelcolor(blue)("+2")("")color(white)(m)stackrelcolor(blue)("-2")("")stackrelcolor(blue)("+1")("")`

Per Rule 4, the sum of all the oxidation numbers in an ion must equal the charge on the ion.

`stackrelcolor(blue)("+2")("Ca")(stackrelcolor(blue)("-2")("O")stackrelcolor(blue)("+1")("H"))_2`
`color(white)(l)stackrelcolor(blue)("+2")("")color(white)(m)color(blue)(underbrace(stackrelcolor(blue)("-2")("")stackrelcolor(blue)("+1")(""))_color(blue)("-1")`

Thus, the total oxidation number of the hydroxide ion is `"-2 + 1 = -1"`

We could also have argued that since the oxidation number of `"Ca"` is +2, the total oxidation numbers of the `"OH"` ions must be -2, the charge on each `"OH"` ion must
be -1.

Per Rule 5, the sum of all the oxidation numbers in a neutral compound must equal zero.

Thus, for `"Ca(OH)"_2, "+2 + 2"("-1") = 2 - 2 = 0"`