# Question #41ebf

##### 1 Answer

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, *calcium carbide*,

Now, notice that **one formula unit** of calcium carbide contains

,one atomof calcium#1 xx "Ca"# ,two atomsof carbon#color(red)(2) xx "C"#

You can thus say that you must add the oxidation number of **three atoms** to get zero. More specifically, if you take

#overbrace(1 xx x)^(color(darkgreen)("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 **ions**, the oxidation number is * equal* to the overall charge of the ion.

This tells you that the oxidation number of calcium will be

#1 xx color(blue)((+2)) + color(red)(2) xx y = 0#

Rearrange to find

#2y = 0 - 2#

#2y = -2 implies y = ((-2))/2 = -1#

Therefore, carbon has a

#stackrel(color(blue)(+2))("Ca") stackrel(color(blue)(-1))("C") _2#