Oxidation number solved problems

oxidation number

 

1. Calculate Oxidation number of Underlined element-
2. i) K₂CrO₄                                               ii)        K₂Cr₂O₇

2(+1)+x+4(-2)=0                                         2(+1)+2x+7(-2)=0

+2+x–8=0                                                     +2+2x–14 =0

x=+6                                                               2x–12=0

x=+6

O.No. of Cr = +6                                           O.No. of Cr =+6

iii)       H₂SO₄                                                 iv)       MnO₂

2(+1)+x+4(-2)=0                                         x+2(-2)=0

+2+x-8=0                                                      x=+4

x=+6

O.No. of S=+6                                               O.No. of Mn = +4

1. v) C₆H₁₂O₆ vi)       C₁₂H₂₂O₁₁

6x+12(+1)+6(-2)=0                         12(x)+22(+1)+11(-2)=0

6x+12-12=0                                                  12x+0=0

x=0                                                                 x=0

O.No. of C=0

vii)      K₄[Fe(CN)₆]                                      viii)     H₂S₂O₇

4(+1)+x+6(-1)=0                                         2(+1)+2x+7(-2)=0

+4+x-6=0                                                      +2+2x-14=0

x=+2                                                               x=+6

O.No. of Fe= +2                                            O.No. of S=+6

1. ix) Na₂S₄O₆ x)        (Ag(NH₃)₂)Cl

2(+1)+4x+6(-2)=0                                       x+2×0+(-1)=0

x=+10/4=+5/2                                              x=+1

O.No. of S=+5/2                                           O.No. of Ag= +1

1. xi) KClO₃                                     xii)      NH₄NO₃

+1+x+3 (-2)=0                                             [NH₄⁺)                        [NO₃^–]

x-5=0                                                             x+4(+1)=+1  x+3(-2)=-1

x=+5                                                               x+4=+1                      x-6=-1

O.No. of Cl = +5                                           x=-3                            x=+5

O.No. of N in O.No. of N

NH₄⁺ is -3                  inNO₃^– is+5

xiii)     Na₂[Fe(CN)₅NO]                              xiv)     Ni(CO)₄

2(+1)+x+5(-1)+1=0                                                x+4(0)=0

+2+x-5+1=0                                                 x=0

x-3+1=0                                                         O.No. of Ni=0

x-2=0

x=+2                                                               Note :  oxidation number of Metal in metal

O.No. of Fe = +2                                           caxbonyl is always zero

Q.2      Calculate oxidation number of underlined element in the Ion

1. i) PO₄^3–                                                  ii)        HSO₃^–

x+4(-2) = -3                                                  +1+x+3(-2)=-1

x-8 =-3                                                           +1+x-6=-1

x=-3+8                                                           x=+4

x=+5                                                               O.No. of 5 = +4

O.No. of P=+5

iii)       PtCl₆^2–                                                            iv)       PH₄⁺

x+6(-1)=-2                                                    x+4(+1)=+1

x-6=-2                                                            x=-3

x=+4                                                               O.No. of P=-3

O.No. of Pt=+4

1. v) S₄O₆^{– –}                                                            vi)       P₂O₇^4–

4x+6(-2)=-2                                                  2x+7(-2)=-4

4x=-2+12                                                      2x-14=-4

x= +5/2                                                          x=+10/2

O.No. of S= +5/2                                          = +5

O.No. of P=+5

vii)      (Cu(NH₃)₄)²⁺                                     viii)     CrO₄^{– –}

x+4×0=+2                                                     x+4(-2)=-2

x=+2                                                               x=-2+8

O.No. of Cu=+2                                            x=+6

O.No. of Cr=+6

1. ix) [Fe(H₂O)₆]³⁺ x)        [CuCl₄]^2–

x+6×0=+3                                                     x+4(-1)=-2

x=+3                                                               x-4=-2

O.No. of Fe=+3                                             x=-2+4

x=+2

O.No. of Cu= +2

1. xi) [Cu(CN)₄]^2– xii)      NO₃^–

x+4(-1)=-2                                                    x+3(-2)=-1

x=-2+4                                                           x=+5

x=+2                                                               O.No. of N=+5

O.No. of Cu=+2

Q.3      Calculate Oxidation number of underlined elements

1. i) H₂O₂                                                   ii)        H₂SO₅

2(+1)+2x=0                                                  2(+1)+x+3(-2)+2(-1)=0

x=-1                                                                +2+x-6-2=0

O.No. of O=-1                                               x=+6

O.No. of S=+6

iii)       H₂S₂O₈

2(+1)+2x+6(-2)+2(-1)=0

+2+2x+6(-2)+2(-1)=0

2x-12=0

x=+6

O.No. of S=+6

1. iv) CrO₅                                                               v)        Na₂O₂

x+4(-1)+1(-2)=0                                                      2(+1)+2x=0

x-4-2=0                                                                      2x=-2

x=+6                                                                           x=-1

O.No. of Cr= +6                                                        O.No. of O=-1

Q.4      Calculate O.No. of underlined element

1. i) N₃H                                         ii)        KIO₃

3x+1=0                                              +1+x+3(-2)=0

x=-1/3                                                x-5=0

O.No. of N=-1/3                               O.No.of I=+5

iii)       NaH₂PO₄                                iv)       O₃

+1+2(+1)+x+4(-2)=0                      O.No. of O in O₃=0 (Zero)

+1+2+x-8=0

x-5=0

x=+5

O.No. of P=+5

1. v) AlPO₄                                                 vi)       (NH₄)₂Cr₂O₇

3(+1)+x+4(–2)=0                                        2(+1)+2x+7(-2)=0

+3+x–8=0                                                     +2+2x–14=0

x=+5                                                               x=+6

O.No. of P=+5                                               O.No. of Cr=+6

vii)      K₃[Fe(CN)₆]                                      viii)     H₂PtCl₆

3(+1)+x+6(-1)=0                                         2(+1)+x+6(-1)=0

+3+x-6=0                                                      x-4=0

x= +3                                      x=+4

O.No. of Fe=+3                                 O.No.of Pt = +4

1. ix) N₂O₅                                       x)        Mn(OH)₃

2x+5(-2)=0                                       x+3(-1)=0

x=+5                                                   x=+3

O.No. of N=+5                                  O.No. of Mn=+3

1. xi) KMnO₄                                   xii)      K₂MnO₄

+1+x+4(-2)=0                                  2(+1)+x+4(-2)=0

x-7=0                                                 x-6=0

x=+7                                                   x=+6

O.No. of Mn=+7                               O.No. of Mn = +6

xiii)     Fe₂(SO₄)₃                               xiv)     Mn₂O₃

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

2x-6=0                                               2x-6=0

x=+3                                                   x=+3

O.No. of Mn=+3

1. Which of the following compounds has maximum Oxidation number of Mn?

KMnO₄,                      K₂MnO₄         MnO₂              Mn₂O₃

Ans     KMnO₄                                   K₂MnO₄                                 MnO₂

+1+x+4(-2)=0                      2(+1)+x+4(-2)=0                 x+2(-2)=0

x=+7                                       x=+6                                       x=+4

O.No. of Mn=+7                   O.No. of Mn=+6                   O.No. of Mn=+4

Mn₂O₃

2x+3(-2)=0

x=+3

O.No. of Mn=+3

KMnO₄ has maximum oxidation number of Mn i-e +7