Question

a. What is the oxidation state of the following atoms? • O in O2 _____ • S in S8 _____ b. What is the oxidation state of Cr in the following compounds? • CrPO4 _____ • Cr3(PO4)2 _____

Answer 1

(a) The oxidation state of (O) in
`O_2` is, (0) and the oxidation state of (O) in
`O_2` is, (0)

(b) The oxidation state of (Cr) in
`CrPO_4` and
`Cr_3(PO_4)_2` are, (+3) and (+2) respectively.

Explanation :

Oxidation number or oxidation state : It represent the number of electrons lost or gained by the atoms of an element in a compound.

Oxidation numbers are generally written with the sign (+) and (-) first and then the magnitude.

When the atoms are present in their elemental state then the oxidation number will be zero.

The oxidation number of oxygen (O) in compounds is usually -2.

The sum of the oxidation numbers in a polyatomic ion is equal to the charge of the ion.

(a) The given molecule is,
`O_2`

Let the oxidation state of O be, 'x'

`2x=0\\\\x=0`

Therefore, the oxidation number of oxygen (O) is, (0)

The given molecule is,
`S_8`

Let the oxidation state of S be, 'x'

`8x=0\\\\x=0`

Therefore, the oxidation number of sulfur (S) is, (0)

(b) The given molecule is,
`CrPO_4`

Let the oxidation state of Cr be, 'x'

As we know that the charge on polyatomic ion phosphate ion is, (-3).

`x+(-3)=0\\\\x-3=0\\\\x=+3`

Therefore, the oxidation number of (Cr) in
`CrPO_4` is, (+3)

The given molecule is,
`Cr_3(PO_4)_2`

Let the oxidation state of Cr be, 'x'

As we know that the charge on polyatomic ion phosphate ion is, (-3).

`3x+2(-3)=0\\\\3x-6=0\\\\x=+2`

Therefore, the oxidation number of (Cr) in
`Cr_3(PO_4)_2` is, (+2)