Question

In which species does phosphorus have an oxidation state of zero? ${{P}_{4}}$ (Elemental phosphorus), $HP{{O}_{2}}$ , $L{{i}_{3}}P{{O}_{3}}$, ${{H}_{3}}P{{O}_{4}}$ , $P{{H}_{3}}$ .

Answer 1

To solve this question first of all, we are going to realize the oxidation state of Phosphorus in each molecule. When an atom forms a chemical bond with another atom, then the total range of electrons that either gains or losses by that atom is understood as oxidation state which is additionally known as oxidation state.

Complete answer:
When atom form a chemical bond with another atom, then the total range of electrons that an atom gains or losses is understood as oxidation state which additionally known as oxidation state, essentially it facilitates in simplifying the method of decisive what’s being oxidized or reduced within the reduction-oxidation reactions. Example, the oxidation state of ${{H}_{2}}S$ is $-2$ .
So to answer this question we will check the oxidation state of phosphorus atoms in each molecule one by one.
In ${{P}_{4}}$ :
In ${{P}_{4}}$ the electron density is shared equally among all of the atoms of phosphorus that means the phosphorus atoms in ${{P}_{4}}$ molecule have neither gained or lost any electron while forming the chemical bond. Thus, the oxidation state of each phosphorus atom in ${{P}_{4}}$ is zero.
In $HP{{O}_{2}}$ :
The general oxidation state of hydrogen is $+1$ and for oxygen it is $-2$ . So, the oxidation state of phosphorus atoms in $HP{{O}_{2}}$ is:
$\Rightarrow H=1 \times (+1)$
$\Rightarrow H=+1$
${{O}_{2}}=2\times O$
$\Rightarrow {{O}_{2}}=2\times \left( -2 \right)$
$\Rightarrow {{O}_{2}}=-4$
As we know the overall oxidation state of a neutral molecule is always zero. So:
$+1+P+\left( -4 \right)=0$
$\Rightarrow P+1-4=0$
$\Rightarrow P-3=0$
$\Rightarrow P=+3$
Thus, in the given molecule, the phosphorus atom exists in its $+3$ oxidation state.
In $L{{i}_{3}}P{{O}_{3}}$:
The oxidation state of $Li$ is $+1$ and oxygen is $-2$ . $-2$ . So, the oxidation state of phosphorus atoms in $L{{i}_{3}}P{{O}_{3}}$ is:
$L{{i}_{3}}=3\times Li$
$\Rightarrow L{{i}_{3}}=3\times \left( +1 \right)$
$\Rightarrow L{{i}_{3}}=+3$
${{O}_{3}}=3\times O$
$\Rightarrow {{O}_{3}}=3\times \left( -2 \right)$
$\Rightarrow {{O}_{3}}=-6$
As we know the overall oxidation state of a neutral molecule is always zero. So:
$+3+P+\left( -6 \right)=0$
$\Rightarrow P+3-6=0$
$\Rightarrow P-3=0$
$\Rightarrow P=+3$
Thus, in the given molecule, the phosphorus atom exists in its $+3$ oxidation state.
In ${{H}_{3}}P{{O}_{4}}$ :
The oxidation state of $H$ is $+1$ and of oxygen is $-2$ . So, the oxidation state of the phosphorus atom in ${{H}_{3}}P{{O}_{4}}$ is:
${{H}_{3}}=3\times H$
$\Rightarrow {{H}_{3}}=3\times \left( +1 \right)$
$\Rightarrow {{H}_{3}}=+3$
${{O}_{4}}=4\times O$
$\Rightarrow {{O}_{4}}=4\times \left( -2 \right)$
$\Rightarrow {{O}_{4}}=-8$
As we know the overall oxidation state of a neutral molecule is always zero. So:
$+3+P+\left( -8 \right)=0$
$\Rightarrow P+3-8=0$
$\Rightarrow P-5=0$
$\Rightarrow P=+5$
Thus, in the given molecule, the phosphorus atom exists in its $+5$ oxidation state.
In $P{{H}_{3}}$ :
The oxidation state of $H$ is $+1$ . So, the oxidation state of phosphorus atoms in $P{{H}_{3}}$ is:
${{H}_{3}}=3\times H$
$\Rightarrow {{H}_{3}}=3\times \left( +1 \right)$
$\Rightarrow {{H}_{3}}=+3$
As we know the overall oxidation state of a neutral molecule is always zero. So:
$P+3=0$
$\Rightarrow P=-3$
Thus, in the given molecule, the phosphorus atom exists in its $-3$ oxidation state.
Hence, we can conclude that the oxidation number (state) of the phosphorus atom is zero in ${{P}_{4}}$ (elemental phosphorus).

Note:
Always remember that the oxidation number of an atom in a molecule plays a very important role in determining that within a chemical reaction, which part of the molecule is reduced or oxidized. Moreover, the concept of oxidation state is very helpful in balancing the redox reactions.