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 , Mn₂O₇ , K₂Cr₂O₇ , KCrO₃ , MnCl₂ , MgO, MgCl₂ , Al₂O₃ ,AlCl₃

(Hint. Oxidation number 0 = -2, Cl = -1, K = +1)

Answer 1

MnO₂ Let x be the oxidation number of Mn

x = -2 x 2 = 0

x + -4 = 0

x = + 4

MnO₇ 

2Mn + -2 x 7 = 0

2Mn + -14 = 0

2Mn = 14

Mn = \(\frac{+14}{2}\) 

= +7

K₂Cr₂O₇ Let x be the oxidation number of Cr is x

+1 x 2 + 2x + -2 x 7 = 0

+ 2 + 2x + -2 x 7 = 0

2x + -12 = 0

2x = +12

x = \(\frac{+12}{2}\) = +6

KCiO₃

+1 + Cr + -2 x 3 = 0

+1 + Cr -6 = 0

Cr + -5 = 0

Cr = +5

MnCl₂ Let x be the oxidation

x + -1 x 2 = 0

x + -2 = 0

x = +2

MgO Let x be the oxidation number of Mg

x + -2 = 0

x = +2

MgCl₂ 

x + -1 x 2 = 0

x + -2 = 0

x = +2

Al₂O₃ Let x be the oxidation number of AI

2X x + -2 x 3 = 0

2x + -6 = 0

2 x = +6

x = \(\frac{+6}{2}\) = +3

AlCl₃

x + -1 x 3 = 0

x + -3 = 0

x = + 3