Question

Oxidation states of P in ${{H}_{4}}{{P}_{2}}{{O}_{5}}$,${{H}_{4}}{{P}_{2}}{{O}_{6}}$,${{H}_{4}}{{P}_{2}}{{O}_{7}}$are respectively:
A) +3, +5, +4
B) +5, +3, +4
C) +5, +4,+3
D) +3, +4, +5

Answer 1

When the two hydroxyl groups are attached to the central atom then it is known as dibasic acid. The number of hydrogens attached directly to the central atom will help us to determine the reducing property of the compound. The oxidation state helps to determine the number of electrons lost or gained by the atom.

Complete step by step answer:
The compound ${{H}_{4}}{{P}_{2}}{{O}_{5}}$is known as pyrophosphorous acid. We can calculate the oxidation state of P in this acid by knowing the formal charge of hydrogen and oxygen which are +1 and -2 respectively. Let the oxidation state of P be x. so now,

& 4(+1)+2x+5(-2)=0 \\\ & 2x=6 \\\ & x=+3 \\\ \end{aligned}$$ So the oxidation of P in $${{H}_{4}}{{P}_{2}}{{O}_{5}}$$is +3. Now we will calculate the oxidation state of P in$${{H}_{4}}{{P}_{2}}{{O}_{6}}$$. So now the formal charge of hydrogen and oxygen is +1 and -2 respectively. Let the oxidation state of P be x. so now, $$\begin{aligned} & 4(+1)+2x+6(-2)=0 \\\ & 2x=8 \\\ & x=+4 \\\ \end{aligned}$$ So the oxidation state of P in $${{H}_{4}}{{P}_{2}}{{O}_{6}}$$is +4. Now we will calculate the oxidation state in $${{H}_{4}}{{P}_{2}}{{O}_{7}}$$. So now the formal charge of hydrogen and oxygen is +1 and -2 respectively. Let the oxidation state of P be x. so now, $$\begin{aligned} & 4(+1)+2x+7(-2)=0 \\\ & 2x=10 \\\ & x=+5 \\\ \end{aligned}$$ So the oxidation state of P in $${{H}_{4}}{{P}_{2}}{{O}_{7}}$$is +5. **So the correct answer is option D.** **Note:** Phosphorus is used in the manufacturing of steel and fertilizers. It is used in producing fine chinaware and special glasses. It is white colour in pure form. It is considered as a critical plant nutrient.