Question

Assign oxidation number to Zn in ZnS, P in ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ , and C in HCN.

Answer 1

An oxidation number is a number assigned to an element in a chemical combination that represents the number of electrons lost or gained by that particular atom of that element in the given compound. The oxidation number of an atom varies according to the chemical environment surrounding it.

Complete step by step answer:
Following are the rules to be followed while calculating the oxidation number of an element in a given compound:
Any free element has an oxidation number of zero.
For monatomic ions, the overall charge on the species is equal to the oxidation number of that atom.
In the case of a neutral compound, the sum of the oxidation numbers of all the atoms is zero.
In the case of polyatomic ions, the sum of the oxidation numbers of all the atoms is equal to the charge on that ion.
The hydrogen atom has an oxidation state of +1. But if it is bonded to a highly electropositive atom has an oxidation state of -1
The oxygen atom has an oxidation state of -2 and in the case of peroxides, it is -1.
All alkali metals exhibit an oxidation state of +1. And all alkaline earth metals have an oxidation state of +2.
Keeping all these points in mind we will calculate the oxidation number:
Zn in ZnS:
S has an oxidation number of -2 and lets us assume the oxidation number of Zn is x, therefore we have:
${\text{x - 2 = 0}}$, thus we have x= +2.
Therefore the oxidation number of Zn in ZnS is +2.
P in ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$:
H has an oxidation number of +1 and oxygen has -2, let us assume that the oxidation number of P be y, therefore:
${\text{3(1) + 4( - 2) + y = 0}}$
$3 - 8 + y = 0$
${\text{y = 5}}$
Therefore, the oxidation number of P in ${{\text{H}}_{\text{3}}}{\text{P}}{{\text{O}}_{\text{4}}}$ is +5.
C in HCN:
H has the oxidation state of +1, N has an oxidation state of -3, and let us assume the oxidation state of C is z, which gives us:
${\text{1 + z - 3 = 0}}$, which gives us, z = +2.
Therefore the oxidation state of C in HCN is +2

Note:
Transition metals (d-block elements) of the periodic table exhibit variable oxidation. This means that they do not have only one oxidation state but their oxidation state varies according to their chemical environment and this is because the last electron in these metals enters the penultimate shell. Osmium is the element known for having the highest number of oxidation states.