Question

Excess ${F_2}\left( g \right)$ reacts at $150^\circ C$ and $1atm$ pressure with $B{r_2}\left( g \right)$ to give a compound$Br{F_n}$ . If $423mL$ of $B{r_2}\left( g \right)$ at the same temperature and pressure produced $4.2g$ of $Br{F_n}$, what is $n$?

Answer 1

Two gases combine under high temperature and pressure to produce a new compound. The stoichiometry of the combination of two gases determines the formula of the generated compound.

Complete step by step answer: Here the two gases undergoing chemical reaction are fluorine gas and bromine gas. The mole of the individual gases is to be evaluated first. The moles of the gas can be determined with the help of the ideal gas equation.
The ideal gas equation is,
$PV = mRT$ where,
$P$ is the pressure of the gas,
$V$ is the volume of the gas,
$m$ is the moles of the gas,
$R$ is the gas constant,
$T$ is the temperature in Kelvin.
The given entities for bromine gas are as follows:
$P = 1atm$ , $V = 423mL = 423 \times {10^{ - 3}}L$ , $R = 0.0821Latmmol{e^{ - 1}}{K^{ - 1}}$ ,$T = 150^\circ C = 150 + 273 = 423K$ .
Inserting in the equation,
$1 \times 423 \times {10^{ - 3}} = m \times 0.0821 \times 423$
$m = 0.0122$.
The probable equation of the reaction occurring between the gases is,
$\dfrac{n}{2}{F_2} + B{r_2} \to 2Br{F_n}$
The number of moles of $B{r_2}$ =$0.0122mole$. According to the equation, $1mole$ of $B{r_2}$ ​generates $2mole$ of $Br{F_n}$ ​. So $0.0122mole$ of $B{r_2}$ ​generates $2 \times 0.0122$ = $0.0244mole$ of $Br{F_n}$
The molar mass of $Br{F_n}$​= atomic mass of $Br$ atom + $n$ x atomic mass of $F$ atom.
=$80 + n \times 19$
=$80 + 19n$
The mole of a compound is the ratio of the weight taken and the molar mass.
Therefore,
$\dfrac{{4.2}}{{80 + 19n}} = 0.0244$
$80 + 19 n = \dfrac{{4.2}}{{0.0244}}$
$80 + 19n = 172.13$
$19n = 92.13$
$n = 4.85 \approx 5$
Thus the value of $n$ is $5$ , and the formula of the compound formed is $Br{F_5}$ .

Note: These types of compounds are known as interhalogen compounds. The availability of vacant inner $d$ and $f$ orbitals in halogen down the group allows them to form such types of hypervalent compounds. Common examples are $Br{F_3}$ , $Br{F_5}$ , $I{F_5}$ , $I{F_7}$ and so on.