Assign oxidation number to the underlined elements in each of the following species

cbse
redoxreactions

#1

Assign oxidation number to the underlined elements in each of the following species


#2

(a) Let the oxidation no. of P in Na${{H}{2}}$P${{O}{4}}$ be x.

  • 1 + 2(+ 1) + x + 4(- 2) =0
  • 1 + 2 + x-8 =0
    x = + 5
    Oxidation number of P in Na${{H}{2}}$P${{O}{4}}$ = + 5.
    (b) Let the oxidation no. of S in NaHS${{O}_{4}}$ be x.
  • 1 + 1 + x + 4(- 2) =0
  • 2 + x - 8=0
    x = + 6
    Oxidation number of S in NaHS${{O}{4}}$ = + 6.
    © Let the oxidation number of P in ${{H}
    {4}}$${{P}{2}}$${{O}{7}}$ be x.
    4(+ 1) + 2x + 7(-2) =0
  • 4 + 2x -14 =0
    2x =10
    x = + 5
    Oxidation number of P in ${{H}{4}}$${{P}{2}}$${{O}{7}}$ = + 5.
    (d) Let the oxidation no. of Mn in ${{K}
    {2}}$Mn${{O}_{4}}$ be x.
    2(+ 1) + x + 4(- 2) =0
  • 2 + x-8 =0
    x = + 6
    Oxidation number of Mn in ${{K}{2}}$Mn${{O}{4}}$ = + 6.
    (e) Let the oxidation number of oxygen in Ca${{O}_{2}}$ be x.
  • 2 + 2x =0
    [oxidation no. of Ca = + 2]
    2x = -2
    x =-1
    Oxidation number of O in Ca${{O}{2}}$ =-1.
    (f) Let the oxidation no. of B in NaB${{H}
    {4}}$ be x.
  • l+x + 4(-l) =0
    [•.• oxidation no. of H = -1]
    x-3 =0
    x = + 3
    Oxidation number of B in NaB${{H}{4}}$ = + 3.
    (g) Let the Oxidation no. of S in ${{H}
    {2}}$${{S}{2}}$${{O}{7}}$ be x.
    2(+1)+ 2x + 7(-2) =0
  • 2 + 2x -14 =0
    2x =12
    x = + 6
    Oxidation number of S in ${{H}{2}}$${{S}{2}}$${{O}{7}}$ = + 6.
    (h) Let the oxidation no. of S in KAl(S${{({{O}
    {4}})}{2}}$).24${{H}{2}}$O = x
  • 1 + 3 + 2x + 8(- 2) + 24(+ 2) + 24(- 2) = 0
  • 4 + 2x -16 = 0
    [O. No. of K = 1,O. No. of Al = 3]
    2x =12
    x = +6
    Oxidation number of S in KAl(S${{({{O}{4}})}{2}}$).24${{H}_{2}}$O=+6