Question

Again back with a Science Question !!!


Give the oxidation state of compound in given formulas :


1. H2SO4 ( for sulphur )

2. H3PO4 ( of phosphorus )

3/. H2PO3 ( of phosphorus ) ​

Answer 1

${ \textbf{ \underline{ \underline{Oxidation Number}} }} \implies$

•It is an imaginary number assigned to an element when it is present in combined state with other elements .

Before coming to the question , let's the know the oxidation state of two basic elements to be used while calculation :-

★ Hydrogen : +1

★ Oxygen : -2

Let's come to the question ,

$|1.| \: { \sf{H_2SO_4}} \:$

✔let oxidation state of sulphur be x.

Using the oxidation states of Oxygen and Hydrogen and equating the sum to zero ( as its a neutral compound ) we have ,

=> 2 + x + 4 × (-2) = 0

=> 2 + x - 8 = 0

${ \red{ \bold{\boxed{x = 6}}}}$

${ \underline{ \textbf{Hence oxidation state of Sulphur is +6}}}$

$|2.| \: { \sf{H_3 PO_4}} \:$

✔ Let oxidation state of Phosphorus be x

=> 3 + x + 4 × (-2) = 0

=> 3 + x - 8 = 0

${ \red{ \bold{\boxed{x = 5}}}}$

${ \underline{ \textbf{Hence oxidation state of \: phosphorous is +5}}}$

$|2.| \: { \sf{H_2 PO_3}} \:$

✔let oxidation state of Phosphorus be x

=> 2 + x + 3 × (-2) = 0

=> 2 + x - 6 = 0

${ \red{ \bold{\boxed{x = 4}}}}$

${ \underline{ \textbf{Hence oxidation state of \: phosphorous is +4}}}$

Answer 2

$\LARGE{\mathfrak{\underline{\underline{\green{Solution:-}}}}}$

Oxidation Number / oxidation state :- A number given to a element in combination state which represents electron gained or lost .

$\rule{200}{2}$

A.T.Q,

Oxidation number

➭ Hydrogen = +1

➭ Oxygen = -2

_______________________

(1) H2 So4 (for sulphur)

$\large\bf{H_{2}^{+1} S^{x} O_{4}^{-2}}$

Take charges

➭ 2(+1) + x + 4(-2) = 0

⇒ 2 + x - 8 = 0

⇒ x -6 = 0

⇒ x = 6

Oxidation number of Sulphur is +6.

________________________

(2) H3 PO4 (of phosphorus)

Take charges

$\large\bf{H_{3}^{+1}P^{x}O_{4}^{-2}}$

⇒ 3(+1) + x + 4(-2) = 0

⇒ 3 + x -8 = 0

⇒ x - 5 = 0

⇒ x = 5

Oxidation number of Phosphorus is +5.

_________________________

(3) H2PO3 (of Phosphorus)

Take charges

$\large\bf{H_{2}^{+1}P^{x}O_{3}^{-2}}$

⇒ 2(+1) + x + 3(-2) = 0

⇒ 2 + x -6 = 0

⇒ x - 4 = 0

⇒ x = 4

Oxidation number of Phosphorus is +4