Answered By Leonardo F

For the first chemical compound,   $I_{2\left(s\right)}$I_2(s) , we have to remember that the oxidation number of each atom in a simple substance (only one chemical species) is 0. Hence, the oxidation number of I is 0.

For the second species,  $HS^-$HS⁻ , we know that in complex ions, the sum of the oxidation numbers of all the atoms is equal to the charge of the ion. The H atom has oxidation number of +1. Hence, we have:

 $+1+x=-1$+1+x=−1 

In which x is the oxidation number of S. Solving, we have: x=-2

Hence, oxidation number of S is -2.

Applying the same concept to the third compound, the oxidation number of F is -1. The sum of all the oxidation numbers in a molecule is 0. Hence:

 $x+6\left(-1\right)=0\rightarrow x=+6$x+6(−1)=0→x=+6 

The oxidation number of Xe is +6.

In a simple ion, the oxidation number is always equal to the charge of the ion. Hence, for the  $Sn^{4+}$Sn⁴⁺ , the oxidation number of Sn is +4.

For  $BrO^{3-}$BrO³⁻ , the oxidation number of O is -2. Hence:

 $x+3\left(-2\right)=-1\rightarrow x=+5$x+3(−2)=−1→x=+5 

The oxidation number of Br is +5. For  $N_2O_5$N₂O₅ , the oxidation number of O is -2. Hence:

 $2x+5\left(-2\right)=0\rightarrow x=+5$2x+5(−2)=0→x=+5 

The oxidation number of N is +5. For  $Pb\left(SO_4\right)_2$Pb(SO₄)₂ , the oxidation number of the anion sulfate ( $SO_4^{2-}$SO₄²⁻ ) is -2. Hence:

 $x+2\left(-2\right)=0\rightarrow x=+4$x+2(−2)=0→x=+4 

The oxidation number of Pb is +4. For  $Na_2O_2$Na₂O₂ , the O, in peroxides, has always an oxidation number of -1.

For  $CaH_2$CaH₂ , this is a metal hydride. The hydrogen has an oxidation number of -1.

For  $S_2O_3^{2-}$S₂O₃²⁻ , the oxygen has an oxidation number of -2. Hence:

 $2x+3\left(-2\right)=-2\rightarrow x=+2$2x+3(−2)=−2→x=+2 

The oxidation number of S is +2.

Putting all the answers together:

  $I$I in  $I_2$I₂ : 0

 $S$S in  $HS^-$HS⁻ : -2

 $Xe$Xe in  $XeF_6$XeF₆ : +6

 $Sn$Sn in  $Sn^{4+}$Sn⁴⁺ : +4

 $Br$Br in  $BrO_3^-$BrO₃⁻ : +5

 $N$N in  $N_2O_5$N₂O₅ : +5

 $Pb$Pb in  $Pb\left(SO_4\right)_2$Pb(SO₄)₂ : +4

 $O$O in  $Na_2O_2$Na₂O₂ : -1

 $H$H in  $CaH_2$CaH₂ : -1

 $S$S in  $S_2O_3^{2-}$S₂O₃²⁻ : +2