Question

Assign oxidation numbers to the atoms in each of the following species: ${\mathbf{SrBr}}_2,\mathbf{Zn}\mathbf{(}\mathbf{OH}{{\mathbf{)}}_4}^{2-},{\mathbf{SiH}}_4,{\mathbf{CaSiO}}_3\mathbf{,}{\mathbf{Cr}}_2{{\mathbf{O}}_7}^{2-}\mathbf{,}{\mathbf{Ca}}_5{\mathbf{(}\mathbf{P}{\mathbf{O}}_4\mathbf{)}}_3\mathbf{F}\mathbf{,}{\mathbf{KO}}_2\mathbf{,}\mathbf{CsH}.$

Short answer

Oxidation states of${\text{SrBr}}_{\text{2}}{{\text{,\hspace{0.17em}Zn(OH)}}_{\text{4}}}^{\text{2-}}{\text{,SiH}}_{\text{4}}{\text{,\hspace{0.17em}CaSiO}}_{\text{3}}{\text{,\hspace{0.17em}Cr}}_{\text{2}}{{\text{O}}_{\text{7}}}^{\text{2-}}{\text{,\hspace{0.17em}Ca}}_{\text{5}}{{\text{(PO}}_{\text{4}}\text{)}}_{\text{3}}{\text{F,\hspace{0.17em}KO}}_{\text{2}}\text{,\hspace{0.17em}CsH}$are +2, +2, +4,+4,+6, +2, +4, and +1 respectively.

Step-by-step solution

Introduction

An oxidation state of atoms is the number of electrons an atom has shared, gained, or lost while forming a chemical bond compared to a neutral atom.

To find an oxidation state:

• First, assign the oxidation states (constant) to all the other atoms bonded to the atom for which the oxidation number is to be found.
• Now, equate the sum of all the oxidation states to the net charge on the molecule or the ion given.

 SrBr2, Zn(OH)42−,SiH4, CaSiO3, Cr2O72-, Ca5(PO4)3F, KO2, CsH.

In the following manner, oxidation states are found:

$\begin{array}{c}{\text{SrBr}}_{\text{2}}\\ \mathrm{x}+2\times (-1)=0\\ \mathrm{x}=+2\end{array}$

$\begin{array}{c}\mathrm{Zn}{{\left(\mathrm{OH}\right)}_4}^{2-}\\ \mathrm{x}+4\times (-1)=-2\\ \mathrm{x}-4=-2\\ \mathrm{x}=+2\\ {\text{SiH}}_{\text{4}}\\ \mathrm{x}+4\times (-1)=0\\ \mathrm{x}=+4\\ {\text{CaSiO}}_{\text{3}}\\ 2+\mathrm{x}+3(-2)=0\\ 2+\mathrm{x}-6=0\\ \mathrm{x}=4\\ {\text{Cr}}_{\text{2}}{{\text{O}}_{\text{7}}}^{\text{2-}}\\ 2\times \mathrm{x}+7(-2)=-2\\ 2\mathrm{x}-14=-2\\ 2\mathrm{x}=12\\ \mathrm{x}=+6\\ {\text{Ca}}_{\text{5}}{\left({\text{PO}}_{\text{4}}\right)}_{\text{3}}\text{F}\\ 5\mathrm{x}+3\times (-3)+(-1)=0\\ 5\mathrm{x}-9-1=0\\ 5\mathrm{x}=10\\ \mathrm{x}=+2\\ {\text{KO}}_{\text{2}}\\ \mathrm{x}+2\times (-2)=0\\ \mathrm{x}-4=0\\ \mathrm{x}=+4\\ \text{CsH}\\ \text{x+\hspace{0.17em}}(-1)=0\\ \mathrm{x}=+1\end{array}$