Question

What is the oxidation state of sulphur in the following:
A. $S{O}_2$
B. $PbS$

Answer 1

Hint: The oxidation state of an atom present in a molecule can be calculated by summing up the oxidation states of the rest of the atoms and equating it with the net charge possessed by the molecule.
In case of neutral molecules the net charge possessed by them is zero.

Complete step by step answer:
In order to calculate the oxidation state of sulphur in sulphur dioxide and lead sulphide, we must first understand the concept of oxidation state or oxidation number of a molecule or atom.
Oxidation state is a number or state that indicates the extent of oxidation of an atom or a molecule in a given chemical compound in other words it is the hypothetical charge which represents the charge which is carried by an atom, in its ionic form. The values of Oxidation states are usually represented by the use of integers, which could be negative, positive, or even zero. The highest oxidation state which is known to us is $+8$ which could be observed in case of tetroxides of iridium xenon, ruthenium, osmium, and some other complexes which may involve plutonium whereas the lowest oxidation state known to us is $-4$ which is observed in cases of some elements which are present in the carbon group.
A.Now if we consider the question, we can see in the first part we are supposed to find the oxidation state of sulphur in sulphur dioxide, so we will suppose that the value of oxidation state of sulphur is $x$. Since two atoms of oxygen are attached to the sulphur, we will multiply the oxidation state of oxygen by a factor of $2$, so the general formula becomes,
$O.S_{(S)}+(2\times O.S_{(O)})=0$
We equated the equation with zero because the molecule has a net charge of zero. Now we know that the oxygen has six electrons in its outermost orbital and it takes two electrons to form a stable octet. So, gaining of two electrons, results in formation of $2^{-}$ charge on the oxygen. So, the oxidation state of oxygen would be $-2$. Now when we substitute this value in the above equation we will get,
$x+(2\times -2)=0$
After solving the above equation we get,
$x=+4$
Hence, the oxidation state of sulphur came out to be positive four.
B.Now if we consider the next molecule which is lead sulphide, we will proceed in a similar manner, so the general formula would be,
$$O.S_{(S)}+O.S_{(Pb)}=0$$
Again, the right side of the equation is zero, because the lead sulphide is a neutral molecule with zero net charge. Now, we know that the lead loses two electrons from its outermost orbital in order to gain a stable octet configuration. And losing electrons gives the lead a charge of $+2$. Now we will substitute this value in the equation and we get,
$x+2=0$
Now, after solving this equation, we will get,
$x=-2$
So the oxidation state of sulphur in lead sulphide came out to be $-2$.

Note: The oxidation state of sulphur in sulphur dioxide is $+4$, as the oxygen attached to the sulphur which is central atom, has an oxidation state $-2$ and there are two atoms of oxygen present in sulphur dioxide, hence the multiplication factor.
While in case of lead sulphide, the oxidation state is $-2$, because lead has an oxidation state of $+2$.