Question

What are the oxidation states of phosphorus in the following compounds?
\[{{H}_{3}}P{{O}_{2}}\],\[{{H}_{3}}P{{O}_{4}}\] , $M{{g}_{2}}{{P}_{2}}{{O}_{7}}$, $P{{H}_{3}}$, \[HP{{O}_{3}}\]
(a) \[+1,\text{ }+3\text{ },+3\text{ },+3\text{ },+5\]
(b)\[+3,\text{ }+3\text{ },+5\text{ },+5\text{ },+5\]
(c) \[+1,\text{ }+2\text{ },+3\text{ },+5,+5\]
(d) \[+1,\text{ }+5\text{ },+5\text{ },-3\text{ },+5\]

Answer 1

Hint: To calculate the oxidation number of P in all of the above compounds, we should know the oxidation number of the other atoms in them i.e. the oxidation number of\[H=\text{ }+1\] , the oxidation number of \[O=\text{ }-2\] , the oxidation number of \[Mg=\text{ }+2\] etc. Now, with the help of these oxidation numbers we can easily find the oxidation number of P .

Complete step by step answer:
By the term oxidation number, we simply mean the total number electrons in the valence shell of an atom when it loses or gains the electrons while undergoing the chemical reaction.
Now, let’s find the oxidation states of the given compounds.
1.\[{{H}_{3}}P{{O}_{2}}\]
Let, the Oxidation state of \[P=x\]
Oxidation state of \[H=\text{ }+3\]
Oxidation state of \[O=\text{ }2\left( -2 \right)=-4\]
Oxidation state of P in \[{{H}_{3}}P{{O}_{2}}\] is
\[\begin{array}{*{35}{l}}
   \Rightarrow 3 + x - 4 = 0 \\
   ~\Rightarrow ~x - 1 = 0 \\
   ~\Rightarrow x=\text{ }+1 \\
\end{array}\]

2. \[{{H}_{3}}P{{O}_{4}}\]
Let, the Oxidation state of \[P=x\]
Oxidation state of \[H=\text{ }+3\]
Oxidation state of \[O=\text{ 4}\left( -2 \right)=-8\]
Oxidation state of P in \[{{H}_{3}}P{{O}_{4}}\] is
\[\begin{array}{*{35}{l}}
   \Rightarrow 3+x-8=0 \\
   ~\Rightarrow x-5=0 \\
   ~\Rightarrow x=\text{ }+5 \\
\end{array}\]

3.$M{{g}_{2}}{{P}_{2}}{{O}_{7}}$
Let, the Oxidation state of \[P=x\]
Oxidation state of \[Mg = 2(+2) = +4\]
Oxidation state of \[O=\text{ 7}\left( -2 \right)=-14\]
Oxidation state of P in $M{{g}_{2}}{{P}_{2}}{{O}_{7}}$ is
\[\begin{align}
 & \Rightarrow 4 + 2x - 14 = 0\\
 & \Rightarrow 2x - 10 = 0\\
 & \Rightarrow 2x = \text{ }10\\
 & \Rightarrow x = ~\dfrac{10}{2}\\
 & \Rightarrow x = \text{ }+5 \\
\end{align}\]

4. $P{{H}_{3}}$
Let, the Oxidation state of \[P=x\]
Oxidation state of \[H=\text{ }+3\]
Oxidation state of P in $P{{H}_{3}}$ is
\[\begin{array}{*{35}{l}}
   \Rightarrow x+3=0 \\
   \Rightarrow x=\text{ }-3 \\
\end{array}\]

5. \[HP{{O}_{3}}\]
Let, the Oxidation state of \[P=x\]
Oxidation state of \[H=\text{ }+1\]
Oxidation state of \[O=\text{ 3}\left( -2 \right)=-6\]
Oxidation state of P in \[HP{{O}_{3}}\] is
\[\begin{array}{*{35}{l}}
   \Rightarrow 1 + x - 6 = 0 \\
   ~\Rightarrow x - 5 = 0 \\
   ~\Rightarrow x=\text{ }+5 \\
\end{array}\]
So, the oxidation states of P in \[{{H}_{3}}P{{O}_{2}}\],\[{{H}_{3}}P{{O}_{4}}\] , $M{{g}_{2}}{{P}_{2}}{{O}_{7}}$, $P{{H}_{3}}$, \[HP{{O}_{3}}\] are \[+1,\text{ }+5\text{ },+5\text{ },-3\text{ }and\text{ }+5\] respectively.
The correct answer is option “D” .

Note: The oxidation state of any free element is always zero, for monatomic ions it is the same as the charge on them. In peroxides , it has an oxidation number as 1 and for polyatomic ions it is equal to the net charge of the ion.