Question

Phosphorus has the oxidation state of +3 in:
(A) Orthophosphoric acid
(B) Phosphorous acid
(C) Meta phosphoric acid
(D) Pyrophosphoric acid

Answer 1

Hint: The hypothetical charge that an atom would possess if all bonds to atoms of different elements were 100% ionic with no covalent compound is known as the oxidation state of an atom. The oxidation state describes the degree of oxidation that is the loss of electrons of an atom in a chemical compound.

Complete step by step solution:
-The first element whose discovery can be traced to a single individual was Phosphorus. Elemental phosphorus exists in two major forms, that is white phosphorous and the red phosphorous.
-Phosphorous being highly reacted reacts vigorously with air in its elemental form and bursts into flames.
-Phosphorus often forms compounds with the same oxidation states but with different formulas.
-For calculating the oxidation state of a compound, we need to remember the following rules-
(i) The oxidation number of a free element is always zero.
(ii) The oxidation number in the case of a monatomic ion is equal to the charge of the ion.
(iii) The oxidation state of hydrogen is generally +1, but when it combines with less electronegative elements, it exhibits -1 oxidation state.
(iv) The oxidation state of oxygen in compounds is usually -2. In the case of peroxides, the oxidation state of oxygen is -1.
(v) For Group 1 elements, the oxidation number in a compound is +1.
(vi) For Group 2 elements, the oxidation number is +2.
(vii) For the oxidation of Group 17 elements, in a binary compound id -1.
(viii) The sum of the oxidation numbers of all the atoms and ions in a neutral compound is zero.
(ix) The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

-Following the above rules, let us now calculate the oxidation states of the compounds given in the options.
-For Orthophosphoric acid, the chemical formula is $H_{3}P{O}_4$. Let us consider the oxidation state of P in this complex be x.
According to rule (iii), the hydrogen will have a +1 oxidation state.
According to rule (iv), the oxygen will have -2 oxidation state.
The overall charge on the complex is neutral that is equal to zero.
Therefore, the oxidation state of Phosphorus in $H_{3}P{O}_4=$ 3(1) + x + 4(-2) = 0
$\Rightarrow x=+5$
-For phosphorous acid, the chemical formula is $H_{3}P{O}_3$. Let us consider the oxidation state of P in this complex be x.
According to rule (iii), the hydrogen will have a +1 oxidation state.
According to rule (iv), the oxygen will have -2 oxidation state.
The overall charge on the complex is neutral that is equal to zero.
Therefore, the oxidation state of Phosphorus in $H_{3}P{O}_3=$3(1) + x + 3(-2)=0
$\Rightarrow x=+3$
-For metaphosphoric acid, the chemical formula is$HP{O}_3$. Let us consider the oxidation state of P in this complex be x.
According to rule (iii), the hydrogen will have a +1 oxidation state.
According to rule (iv), the oxygen will have -2 oxidation state.
The overall charge on the complex is neutral that is equal to zero.
Therefore, the oxidation state of Phosphorus in $HP{O}_3=$ 1 + x + 3(-2) = 0
$\Rightarrow x=+5$
-For pyrophosphoric acid, the chemical formula is $H_4{P}_2{O}_7$. Let us consider the oxidation state of P in this complex be x.
According to rule (iii), the hydrogen will have a +1 oxidation state.
According to rule (iv), the oxygen will have -2 oxidation state.
The overall charge on the complex is neutral that is equal to zero.
Therefore, the oxidation state of Phosphorus in $H_4{P}_2{O}_7=$ 4(1) + 2x + 7(-2) = 0
$\Rightarrow x=+5$ .

So, the correct answer only options B.

Note: Phosphorous being a third-row element has five empty 2d-orbitals which can be used for making p-d bonding in a fashion similar to pi bonding, hence the phosphorus may expand its valence shell by two electrons. Organophosphorus compounds have phosphorus in oxidation states ranging from -3 to +5.