Question

Lead forms compounds in the $\mathbf{+}\mathbf{2}$and $\mathbf{+}\mathbf{4}$oxidation states. All lead (ll) halides are known (and are known to be ionic). Only${\mathbf{\text{PbF}}}_{\mathbf{\text{4}}}$ and${\mathbf{\text{PbCl}}}_{\mathbf{\text{4}}}$ are known among the possible lead (IV) halides. Presumably lead (IV) oxidizes bromide and iodide ions, producing the lead (II) halide and the free halogen:

${\mathbf{\text{PbX}}}_{\mathbf{\text{4}}}\mathbf{\to }{\mathbf{\text{PbX}}}_{\mathbf{\text{2}}}{\mathbf{\text{+X}}}_{\mathbf{\text{2}}}$

Suppose$\mathbf{\text{25.00g}}$ of a lead (IV) halide reacts to form$\mathbf{\text{16.12g}}$ of a lead (II) halide and the free halogen. Identify the halogen.

Short answer

The unknown element (free halogen) with a mass of$\text{Ar}\left(x\right)={\text{127.00gmol}}^{\text{-1}}$ is Iodine.

Step-by-step solution

Concept Introduction:

Chemical compounds known as compounds are made up of two or more elements chemically combined in a predetermined ratio.

Information Provided:

Mass of lead (IV) halide is $\text{25.00g}$.

Mass of lead (II) halide is $\text{16.12g}$.

Let the unknown halogen be $x$.

Identifying the halogen:

The molarity of reactant and product is the same. This will be the starting point.

$$\begin{gathered} n_1=n_2 \\ \frac{m_1}{M_1}=\frac{m_2}{M_2} \\ {M}_1\cdot m_2=M_2\cdot m_1 \\ {M}_1\cdot m_2-M_2\cdot m_1=0 \end{gathered}$$

Then, substituting the elements and you get,

$$\begin{gathered} \left[\text{Ar}\left(\text{Pb}\right)\text{+4Ar}\left(\text{x}\right)\right]\cdot {\text{m}}_{\text{2}}\text{-}\left[\text{Ar}\left(\text{Pb}\right)\text{+2Ar}\left(\text{x}\right)\right]\cdot {\text{m}}_{\text{1}}=\text{0} \\ \text{Ar}\left(\text{Pb}\right){\text{m}}_{\text{2}}\text{+4Ar}\left(\text{x}\right){\text{m}}_{\text{2}}\text{-Ar}\left(\text{Pb}\right){\text{m}}_{\text{1}}\text{-2Ar}\left(\text{x}\right){\text{m}}_{\text{1}}=\text{0} \\ \text{Ar}\left(\text{x}\right)\left({\text{4m}}_{\text{2}}{\text{-2m}}_{\text{1}}\right)=\text{Ar}\left(\text{Pb}\right){\text{m}}_{\text{1}}\text{-Ar}\left(\text{Pb}\right){\text{m}}_{\text{2}} \end{gathered}$$

On substituting the mass and on further calculation, you get

localid="1663853371684" $\begin{array}{rcl}\text{Ar}\left(x\right)& =& \frac{\text{Ar}\left(\text{Pb}\right){\text{m}}_{\text{1}}\text{-Ar}\left(\text{Pb}\right){\text{m}}_{\text{2}}}{{\text{4m}}_{\text{2}}{\text{-2m}}_{\text{1}}}\\ & =& \frac{{\text{207.2gmol}}^{\text{-1}}\cdot {\text{25g-207.2gmol}}^{\text{-1}}\cdot \text{16.12g}}{\text{4}\cdot \text{16.12g-2}\cdot \text{25g}}\\ & =& {\text{127.00gmol}}^{\text{-1}}\end{array}$

Hence, according to the value of $x$the halogen is Iodine.