Question

What is the oxidation number of sulfur In $NaHSO_{4}^{{}}$?

Answer 1

Hint: The oxidation state is defined as charge on the atom.The sum of the oxidation state of an atom in a molecule is equal to charge on the molecule.The neutral molecules charge less species.It has zero charges.
Complete answer:
 The oxidation state of an atom is defined as the charge on an atom on forming an ionic bond. An atom which has higher electronegativity has a negative oxidation state and more electropositive elements have positive oxidation state
We are interested in determining the oxidation state of an atom in a molecule.
We know net charge on the neutral atom or molecule is always equal to zero.
Let us determine oxidation state of S in $NaHSO_{4}^{{}}$
The oxidation number of alkali metal that is sodium atom is equal to $+1$.
The oxidation state of hydrogen is $+1$.
Oxygen is an electronegative element.
It has $-2$ oxidation state.
The oxidation states are listed below.

Element	Oxidation state
$Na$	$+1$.
$H$	$+1$.
$O$	$-2$

We know that sum of oxidation state of an atom in a neutral molecule is equal to zero.Then we have,
$0$= O.S of Na + O.S of H + O.S of S + 4(O.S of O)
Substitute value of oxidation state.
$0$=$1$+$1$+O.S of S + 4($-2$)
O.S of S=$8$-$2$=+$6$
Hence, Oxidation state of sulphur in $NaHSO_{4}^{{}}$ is equal to +$6$.

Note:
Note that, along with +$6$oxidation state sulphur exists in $-2$,$0$,+$2$ and +$4$ oxidation state. The valence shell configuration of sulphur is $3{{s}^{2}}$ $3{{p}^{2}}$. Thus it exhibits a variable oxidation state. It loses six electrons in the outermost shell and exhibits $+6$ an oxidation state.