Question

Which of the following shows non – zero multiple oxidation state?
A. S
B. O
C. Zn
D. H

Answer 1

Oxidation state of any element is their ability to gain or lose electrons. This is the property that an atom exhibits in a molecule due to the other bonded atoms. The metallic character of elements has an effect on oxidation states. Oxidation states are positive and negative depending on the electronegativities of bonded atoms.

Complete answer:
An oxidation state is the ability of atoms to gain or lose electrons. Oxidation state of any atom depends on the nature of bonded atoms, whether they are electronegative or electropositive, this gives the atom, the sign of positive and negative, depending on its electronegativity and the ability to gain or lose electrons.
Many elements have the ability to possess more than one oxidation state. These are known as multiple oxidation states of any atom. The multiple oxidation states are due to the ability of an element to donate or gain electrons to complete its octet or achieve a stable noble gas configuration (fully filled configuration).
We have been given various elements, through their electronic configuration, we can see how many electrons they can donate or gain to achieve a fully filled configuration, and this will decide their oxidation states.
S, sulphur has a configuration, $[Ne]3{{s}^{2}}3{{p}^{4}}$ , this means from the p orbital it has 2 electrons remaining so oxidation state of 2. While from the outer shell, sulphur can donate all of its 6 electrons due to its less electronegativity to achieve a noble gas configuration of neon gas. So the oxidation states can vary from, -2 to +6, which are,
S = -2, -1, 0, +1, +2, +3, +4, +5, +6
O, oxygen has the configuration, $[He]2{{s}^{2}}2{{p}^{4}}$ , this means it can posses oxidation states between -2 to +2 because it can gain or lose 2 electrons, it is not exceeded from 2 due to the more electronegative character of oxygen. So oxygen can have,
O = -2, -1, +1, +2
Zn, zinc is the transition element with configuration, $[Ar]4{{s}^{2}}3{{d}^{10}}$ , it has fully filled d orbital, so it loses electrons from s orbital, so it can posses an oxidation state,
Zn = -2, +1, +2
H, hydrogen has the configuration, $1{{s}^{1}}$ which means it can either gain 1 electron or lose this 1 electron. So it has oxidation state,
H = -1, +1

Hence, S, sulphur has multiple non – zero oxidation states, so option A is correct.

Note:
These oxidation states of elements vary in every molecule, different molecules have different oxidation states of their atoms. Though oxygen and sulphur are from the same group, sulphur has more oxidation states due to its less electronegativity and greater size. Zinc has only 1 and 2 oxidation states as it does not have the energy to get excited that can result in higher oxidation states.