Question

Assign oxidation numbers to the atoms in each of the followingspecies:

${\mathbf{NH}}_4{\mathbf{NO}}_3$, ${\mathbf{CaMgSiO}}_4$, $\mathbf{Fe}{\mathbf{\left(}\mathbf{CN}\mathbf{\right)}}_6^{4-}$, ${\mathbf{B}}_2{\mathbf{H}}_6$, ${\mathbf{BaH}}_2$,${\mathbf{PbCl}}_2$ , ${\mathbf{Cu}}_2\mathbf{O}\mathbf{\left(}{\mathbf{SO}}_4\mathbf{\right)}$, ${\mathbf{S}}_4{\mathbf{O}}_6$ .

Short answer

$\stackrel{\text{+3}}{\text{N}}\stackrel{\text{+1}}{{\text{H}}_{\text{4}}}\stackrel{\text{+5}}{\text{N}}\stackrel{\text{-2}}{{\text{O}}_{\text{3}}^{\text{-}}}$,$\stackrel{\text{+2}}{\text{Ca}}\stackrel{\text{+2}}{\text{Mg}}\stackrel{\text{+4}}{\text{Si}}\stackrel{\text{-2}}{{\text{O}}_{\text{4}}}$,$\stackrel{\text{+2}}{\text{Fe}}\stackrel{\text{-1}}{{\left(\text{CN}\right)}_{\text{6}}}$,${\stackrel{+3}{\text{B}}}_{\text{2}}{\stackrel{-1}{\text{H}}}_{\text{6}}$,$\stackrel{\text{+2}}{\text{Ba}}{\stackrel{\text{-1}}{\text{H}}}_{\text{2}}$,$\stackrel{+2}{\text{Pb}}{\stackrel{-1}{\text{Cl}}}_{\text{2}}$,${\stackrel{\text{+2}}{\text{Cu}}}_{\text{2}}\stackrel{\text{-2}}{\text{O}}\left(\stackrel{\text{+6}}{\text{S}}{\stackrel{\text{-2}}{\text{O}}}_{\text{4}}\right)$,${\stackrel{\text{+3}}{\text{Si}}}_{\text{4}}{\stackrel{\text{-2}}{\text{O}}}_{\text{6}}$

Step-by-step solution

Assigning Oxidation number

The compound ${\text{NH}}_{\text{4}}{\text{NO}}_{\text{3}}$ has two ions: ${\mathbf{NH}}_4^{+}$ ion nitrogen has +3 oxidation number and hydrogen has +1 oxidation number, and in the ${\mathbf{NO}}_3^{-}$ion, nitrogen has +5 oxidation number, and oxygen has -2 oxidation number

Oxidation numbers in ${\text{CaMgSiO}}_{\text{4}}$ thecalcium and magnesium are group II metal having +2 oxidation state, oxygen having -2 oxidation state, and silicon oxidation state calculated as follows:

$\begin{array}{c}\text{x+}\left(\text{6×-1}\right)\text{=\hspace{0.17em}-4}\\ \text{x\hspace{0.17em}=+2}\end{array}$

In the coordination complex$\text{Fe}{\left(\text{CN}\right)}_{\text{6}}^{\text{4-}}$cyanide, the ion is in a -1 oxidation state, and the iron oxidation state is calculated as follows:

$\begin{array}{c}\text{x+}\left(\text{6×-1}\right)\text{=\hspace{0.17em}-4}\\ \text{x\hspace{0.17em}=+2}\end{array}$

The oxidation state of iron is +2.

Assigning Oxidation number 

In the ${\text{B}}_{\text{2}}{\text{H}}_{\text{6}}$ hydrogen oxidation state is -1(present in a hydride form), theboron oxidation state is calculated as follows:

$\begin{array}{c}\text{2x+}\left(\text{6×-1}\right)\text{=\hspace{0.17em}}0\\ \text{x\hspace{0.17em}=+3}\end{array}$

The Boron is in a +3 oxidation state.

In the compound, ${\text{BaH}}_{\text{2}}$hydrogen is in -1 because it is present with metal barium, whose oxidation state is calculated as follows:

$\begin{array}{c}\text{x+}\left(\text{2×-1}\right)\text{=\hspace{0.17em}}0\\ \text{x\hspace{0.17em}=+2}\end{array}$

The barium oxidation state is +2.

In the compound,${\text{PbCl}}_{\text{2}}$ the chlorine atom is in a -1 oxidation state, and the oxidation state of Pb is calculated as follows:

$\begin{array}{c}\text{x+}\left(\text{2×-1}\right)\text{=\hspace{0.17em}0}\\ \text{x\hspace{0.17em}=\hspace{0.17em}+2}\end{array}$

Lead (Pb) oxidation state is +2.

Assigning Oxidation number

In the compound,${\text{Cu}}_{\text{2}}\text{O}\left({\text{SO}}_{\text{4}}\right)$ the oxygen atoms are in a -2, sulfur in a +6 oxidation state, and the oxidation state of copper is calculated as follows:

$\begin{array}{c}\text{2x-2}\left(\text{6+4×-2}\right)\text{=\hspace{0.17em}0}\\ \text{x\hspace{0.17em}=\hspace{0.17em}+2}\end{array}$

The copper (Cu) oxidation state is +2.

In the compound, ${\text{S}}_{\text{4}}{\text{O}}_{\text{6}}$the oxygen atoms are in a -2, and the oxidation state of sulfur is calculated as follows:

$\begin{array}{c}\text{4x+}(6\times -2)\text{=\hspace{0.17em}0}\\ \text{x\hspace{0.17em}=\hspace{0.17em}+3}\end{array}$

The sulfur (S) oxidation state is +3.