Question

The oxidation number of C in $C{H_2}O$ is:
A.-2
B.+2
C.0
D.+4

Answer 1

Oxidation number or oxidation state of an element can be defined as the degree of oxidation of an element in a given compound. In simpler terms, it can be understood as the number of the electrons gained or lost by an atom while forming a compound. This results in forming a net charge over this element, which is referred to as the oxidation state.

Complete step by step answer:
Before we move towards the solution of this question, let us understand some basic concepts.
Depending on the number of electrons present in the valence shell of an atom, an element may exhibit a single or in some cases multiple oxidation states, depending on the atoms they are combining with.
Now, forming the equation for calculating the oxidation states of iodine and chlorine in $C{H_2}O$ :
Net charge on $C{H_2}O$ = (O.S. of carbon) (no. of atoms of carbon) + (O.S. of hydrogen) (no. of atoms of hydrogen) + (O.S. of oxygen) (no. of atoms of oxygen)

$0 = \left( x \right)\left( 1 \right) + \left( { + 1} \right)\left( 2 \right) + \left( { - 2} \right)\left( { - 1} \right)$
$0 = x + 2 - 2$
$0 = x$
hence, the oxidation state of carbon in $C{H_2}O$ is zero.

Hence, Option C is the correct option.

Note:
Hydrogen tends to show two oxidation states: (-1) and (+1). To determine which oxidation state to use, you must look for the presence of a metal in the compound. Because in metal hydrides, the oxidation of hydrogen is (-1), otherwise it is always (+1)