Question

please please explain please​

Answer 1

Here the answer is option 3

CrO4^2- -------> Cr2O7^2-

In CrO4 ^ 2-

x + 4(-2) = -2

x = 8-2

x = 6

In Cr2O7 ^ 2-

2(x) + 7(-2) = -2

2x = 14 -2

x = 6

Here, Oxidation number of Cr remains the same on both sides that is 6

Extra information:

• Always remember to keep value of unknown element oxidation number as x or any alphabet
• Always equal sum of oxidation number to zero
• Oxidation number of certain elements like Oxygen, Hydrogen, Sodium and some exceptions should be kept in mind like Oxidation number of O = -2, Na = +1, H = +1 and so on

Here in first option

VO^-2 -----> V2O3

So, Here metal is V ( vanadium)

In VO ^-2

x -2 = -2

x = 0

In V2O3

2(x) + 3(-2) = 0

x = 6/2

x = 3

Here oxidation number of V is changed from 0 to 3

In second option

Na -------> Na+

When we take element in their free state like Na, O2, alone without any attachment to other compounds we consider their oxidation number as zero

So in Na oxidation number is zero and in Na+ Oxidation number is 1

Here oxidation number of Na changed from 0 to +1

In fourth option

Zn2+ -------> Zn

In Zn2+

Oxidation number is +2

In Zn

Oxidation number is zero as it is in free state

Here, oxidation number of zinc changed from +2 to 0