Question

The oxidation states of P atom in $POC{{l}_{3}},{{H}_{2}}P{{O}_{3}}$ and ${{H}_{4}}{{P}_{2}}{{O}_{6}}$ respectively are:

Answer 1

Hint: Draw the structures of $POC{{l}_{3}},{{H}_{2}}P{{O}_{3}}$ and ${{H}_{4}}{{P}_{2}}{{O}_{6}}$ .Keep in mind the valency of the P atom to avoid any mistakes. Try calculating oxidation state of P by trying to balance the positive and
the negative charge in the compound.

Complete step by step solution:
 We will draw the structure for the three compounds given above to find the oxidation state.
Structure of $POC{{l}_{3}}$:

There are 5 bonds around the central P atom, hence the oxidation state of P is 5.
Structure of ${{H}_{2}}P{{O}_{3}}$:



There are 4 bonds around the central P atom, hence the oxidation state of P is 4.
Structure of ${{H}_{4}}{{P}_{2}}{{O}_{6}}$:
In the above case, we consider the number of bonds around each P excluding the P-P bond as it is not counted when calculating the oxidation state. There are 4 bonds around each of the P atoms, hence the oxidation state is 4.

Therefore, the correct answer is option (A).

Additional information:
While calculating the oxidation state of an atom we count the number of bonds around the atom. In reality, we mean to count the number of heterogeneous bonds i.e. both the atoms are not identical. Homogenous bonds like P-P do not contribute to the oxidation state of the atom and hence it is not counted.

Note: While writing the structure of the compound, keep in mind the maximum valency the central atom can show and make the bonds accordingly. Do not count homogenous bonds while calculating the oxidation state of the central atom.