Question

What are the oxidation numbers of the atoms in the compound $C{{F}_{4}}$?
A. +4, -1
B. +1, -4
C. +3, -2
D. +2, -3
E. +4, -4

Answer 1

Hint: Oxidation numbers are the values that an atom gets when present in a chemical compound. All the atoms, or elements present in a chemical compound contribute to each other’s oxidation states. Electronegativity of any atom also decides its oxidation number.

Complete answer:
Oxidation numbers or oxidation states of any atom are calculated in its molecular form. The individual oxidation number of all the atoms is determined on their capability to gain or lose electrons. The electronegative character of any element decides that the oxidation number of that element will be positive or negative.
We have been given the compound$C{{F}_{4}}$, we have to find the oxidation numbers of 4 fluorine, F and 1 carbon, C in this molecule.
As fluorine is electronegative than carbon, so it will have a negative sign, while the oxidation state of fluorine is 1, due to the absence of 1 electron in its outer shell, so its oxidation number is -1. Therefore, the oxidation number of 4 fluorine atoms will be, $4\times (-1)$= -4.
Now, we have carbon, which is less electronegative than fluorine, so its sign will be positive. Carbon has 4 electrons in its outer shell to lose, so, either it can lose 4 electrons or gain 4 electrons. So, the oxidation number of carbon will be +4.
Hence, the oxidation numbers of C and F in $C{{F}_{4}}$ are +4 and -4 respectively.

So, option E is correct.

Note:
The oxidation number of a free element, like helium, nitrogen gas, is always 0. While oxidation numbers of metals are positive integers of 1,2, and 3. While that of non metals in the negative integers.