# Question 8530d

May 4, 2017

$S n C {l}_{4} \to S n = + 4 , \mathmr{and} C l = - 1$
$P b C {l}_{4} \to P b = + 4 \mathmr{and} C l = - 1$

#### Explanation:

Before getting started, take a look at this website for general rules for assigning oxidation numbers to atoms. Rules for Assigning Oxidation Numbers

$- - - - - - - - - - - - - - - - - -$

$\textcolor{b l u e}{\text{Step 1: Assign oxidation numbers for each atom. Multiply oxidation number by the number of atoms.}}$

$S n C {l}_{4}$

• Sn->"x" ("unknown")
• $C {l}_{4} \to - 1 \cdot \left(4\right)$

$\textcolor{b l u e}{\text{Step 2: Set up an equation to solve for the unknown.}}$

For polyatomic ions, the equation would be equal to the charge.
For neutral compounds and molecules, set equation equal to $0$

color(white)(aaaa)Sn->"x" ("unknown")color(white)(aaaa)Cl_4->-1*(4)#

• $\text{x} + \left(- 4\right) = 0$
• $\text{x} - 4 = 0$
• $\text{x} = + 4$

$x = + 4 \to \textcolor{m a \ge n t a}{\text{Sn = +4}}$

The oxidation numbers for each atom in $S n C {l}_{4}$ would be $S n = + 4 , \mathmr{and} C l = - 1$

You would repeat the same steps to solve for the oxidation number for $\text{Pb}$ in $P b C {l}_{4}$. If you did the steps correctly, the oxidation number for $P b$ would also be $+ 4$ and $C l$ would have an oxidation number of $- 1$.