Question

Find the oxidation number of the elements which are underlined in the compounds given below. Among these find out the elements which show variable oxidation numbers. MnO2(Hint. oxidation number O=-2, Cl=1, K=+1)​

Answer 1

Answer:

MnO

2

Let x be the oxidation number of Mn

x+(−2)×2=0

x+(−4)=0

x=+4

MnO

7

2Mn+(−2)×7=0

2Mn+(−14)=0

2Mn=14

Mn=

2

+14

=+7

K

2

Cr

2

O

7

Let x be the oxidation number of Cr is x

+1×2+2x+(−2)×7=0

+2+2x+(−2)×7=0

2x+(−12)=0

2x=+12

x=

2

+12

=+6

KCrO

3

+1+Cr+(−2)×3=0

+1+Cr+(−6)=0

Cr+(−5)=0

Cr=+5

MnCl

2

Let x be the oxidation of Mn

x+(−1)×2=0

x+(−2)=0

x=+2

MgO Let x be the oxidation number of Mg

x+(−2)=0

x=+2

MgCl

2

x+(−1)×2=0

x+(−2)=0

x=+2

Al

2

O

3

Let x be the oxidation number of Al

2x+(−2)×3=0

2x+(−6)=0

2x=+6

x=

2

+6

=+3

AlCl

3

x+(−1)×3=0

x+(−3)=0

x=+3