To determine the order of the oxidation state of the phosphorus atom in the given compounds, we need to calculate the oxidation state of phosphorus (P) in each molecule. We will use the standard rules for assigning oxidation states:

1. The oxidation state of hydrogen (H) is +1 when bonded to a non-metal.
2. The oxidation state of oxygen (O) is -2 in most compounds.
3. The sum of the oxidation states of all atoms in a neutral molecule is zero.

Let the oxidation state of phosphorus be $x$.

Step 1: Calculate the oxidation state of P in ${H_3}P{O_2}$ (Hypophosphorous acid)

We set up the equation for the sum of oxidation states: $(3 \times \text{Oxidation state of H}) + (1 \times \text{Oxidation state of P}) + (2 \times \text{Oxidation state of O}) = 0$ $(3 \times (+1)) + (1 \times x) + (2 \times (-2)) = 0$ $+3 + x - 4 = 0$ $x - 1 = 0$ $x = +1$ So, the oxidation state of P in ${H_3}P{O_2}$ is +1.

Step 2: Calculate the oxidation state of P in ${H_3}P{O_4}$ (Phosphoric acid)

We set up the equation for the sum of oxidation states: $(3 \times \text{Oxidation state of H}) + (1 \times \text{Oxidation state of P}) + (4 \times \text{Oxidation state of O}) = 0$ $(3 \times (+1)) + (1 \times x) + (4 \times (-2)) = 0$ $+3 + x - 8 = 0$ $x - 5 = 0$ $x = +5$ So, the oxidation state of P in ${H_3}P{O_4}$ is +5.

Step 3: Calculate the oxidation state of P in ${H_3}P{O_3}$ (Phosphorous acid)

We set up the equation for the sum of oxidation states: $(3 \times \text{Oxidation state of H}) + (1 \times \text{Oxidation state of P}) + (3 \times \text{Oxidation state of O}) = 0$ $(3 \times (+1)) + (1 \times x) + (3 \times (-2)) = 0$ $+3 + x - 6 = 0$ $x - 3 = 0$ $x = +3$ So, the oxidation state of P in ${H_3}P{O_3}$ is +3.

Step 4: Calculate the oxidation state of P in ${H_4}{P_2}{O_6}$ (Hypophosphoric acid)

We set up the equation for the sum of oxidation states: $(4 \times \text{Oxidation state of H}) + (2 \times \text{Oxidation state of P}) + (6 \times \text{Oxidation state of O}) = 0$ $(4 \times (+1)) + (2 \times x) + (6 \times (-2)) = 0$ $+4 + 2x - 12 = 0$ $2x - 8 = 0$ $2x = 8$ $x = +4$ So, the oxidation state of P in ${H_4}{P_2}{O_6}$ is +4.

Step 5: Arrange the compounds in descending order of the oxidation state of P

We have the following oxidation states:

• ${H_3}P{O_4}$: +5
• ${H_4}{P_2}{O_6}$: +4
• ${H_3}P{O_3}$: +3
• ${H_3}P{O_2}$: +1

The descending order is +5 > +4 > +3 > +1.

Therefore, the order of the compounds is: ${H_3}P{O_4} \gt {H_4}{P_2}{O_6} \gt {H_3}P{O_3} \gt {H_3}P{O_2}$

This corresponds to option C.