Answer
Please see the answer below.
Work Step by Step
a)
$Cs_{2}O$
The oxidation number of oxygen in all cases except in peroxides and superoxides is -2.
Let the oxidation number of Cs be ‘x’.
$(2\times x)+ - 2 = 0$
Then x = +1
Hence oxidation number of Cs in $Cs_{2}O$ =+1.
b)
$CaI_{2}$
The oxidation number of Calcium in all compounds is +2.
Let the oxidation number of I be ‘x’.
$+2 +(2\times x) = 0$
Then x = - 1
Hence oxidation number of I in $CaI_{2}$ = -1.
c)
$Al_{2}O_{3}$
The oxidation number of oxygen in all cases except in peroxides and super oxides is -2.
Let the oxidation number of Al be ‘x’.
$(2\times x) + +(3\times -2)+ x = 0$
Then x = +3
Hence oxidation number ofAl in $Al_{2}O_{3}$ = +3.
d)
$H_{3}AsO_{3}$
The oxidation number of hydrogen in all cases except in hydrides is +1.
The oxidation number of oxygen in all cases except in peroxides and superoxides is -2.
Let the oxidation number of As be ‘x’.
$(3\times+1) +x +(3\times -2)+ x = 0$
Then x = +3
Hence oxidation number of As in $H_{3}AsO_{3}$ = +3.
e)
$TiO_{2}$
The oxidation number of oxygen in all cases except in peroxides and superoxides is -2.
Let the oxidation number of Ti be ‘x’.
$(2\times -2)+ x = 0$
Then x = +4
Hence oxidation number of Ti in $TiO_{2}$ =+4.
f)
$ MoO_{4}^{2-} $
The oxidation number of oxygen in all cases except in peroxides and superoxides is -2.
Let the oxidation number of Mo be ‘x’.
$ (4\times-2) + x = -2$
Then x = +6
Hence oxidation number of Mo in $MoO_{4}^{2-}$ = +6.
g) $ PtCl_{4}^{2-} $
The oxidation number of Chlorine in all cases except in the compounds with Fluorine and Oxygen is -1.
Let the oxidation number of Pt be ‘x’.
$ (4\times-1) + x = -2$
Then x = +2
Hence oxidation number of Pt in $PtCl_{4}^{2-}$ = +2.
h) $ PtCl_{6}^{2-} $
The oxidation number of Chlorine in all cases except in the compounds with Fluorine and oxygen is -1.
Let the oxidation number of Pt be ‘x’.
$ (6\times-1) + x = -2$
Then x = +4
Hence oxidation number of Pt in $PtCl_{6}^{2-}$ = +4.
i)
$ SnF_{2} $
The oxidation number of Flourine in all compounds is -1.
Let the oxidation number of Sn be ‘x’.
$ (2\times-1) + x = 0$
Then x = +2
Hence oxidation number of Sn in $SnF_{2}$ = +2
j)
$ ClF_{3} $
The oxidation number of Flourine in all compounds is -1.
Let the oxidation number of Cl be ‘x’.
$ (3\times-1) + x = 0$
Then x = +3
Hence oxidation number of Cl in $ClF_{3}$ = +3
k)
$ SbF_{6}^{-} $
The oxidation number of Flourine in all compounds is -1.
Let the oxidation number of Sb be ‘x’.
$ (6\times-1) + x =-1$
Then x = +5
Hence oxidation number of Sb in $SbF_{6}^{-}$ = +5.
