# Question #41ebf

Jul 22, 2016

Here's what's going on here.

#### Explanation:

The thing to remember about oxidation numbers is that for a neutral compound, the sum of the oxidation numbers of all the atoms that make up the compound must be equal to zero.

In this case, you must find the oxidation number of carbon, $\text{C}$, and of calcium, $\text{Ca}$, in the ionic compound calcium carbide, ${\text{CaC}}_{\textcolor{red}{2}}$.

Now, notice that one formula unit of calcium carbide contains

• one atom of calcium, $1 \times \text{Ca}$
• two atoms of carbon, $\textcolor{red}{2} \times \text{C}$

You can thus say that you must add the oxidation number of three atoms to get zero. More specifically, if you take $x$ to be the oxidation number of calcium and $y$ to be the oxidation number of carbon, you have

${\overbrace{1 \times x}}^{\textcolor{\mathrm{da} r k g r e e n}{\text{one atom of Ca")) + overbrace(color(red)(2) xx y)^(color(purple)("two atoms of C}}} = 0$

Now, calcium is located in group 2 of the periodic table, so right from the start you know that it forms $2 +$ cations in ionic compounds. For ions, the oxidation number is equal to the overall charge of the ion.

This tells you that the oxidation number of calcium will be $\textcolor{b l u e}{+ 2}$. You thus have $x = \textcolor{b l u e}{+ 2}$, which means that

$1 \times \textcolor{b l u e}{\left(+ 2\right)} + \textcolor{red}{2} \times y = 0$

Rearrange to find $y$, the oxidation number of carbon

$2 y = 0 - 2$

$2 y = - 2 \implies y = \frac{\left(- 2\right)}{2} = - 1$

Therefore, carbon has a $\textcolor{b l u e}{- 1}$ oxidation number in calcium carbide.

${\stackrel{\textcolor{b l u e}{+ 2}}{\text{Ca") stackrel(color(blue)(-1))("C}}}_{2}$