Q 14.126P - Chemistry: Molecular Nature Of Matter And Change

Question

(a) A halogen  $\left(X_2\right)$ disproportionates in base in several steps to $X^{-}$  and $X{O}_3^{-}$ . Write the overall equation for the disproportionation of $B{r}_2$  to $B{r}^{-}$  and $Br{O}_3^{-}$ . (b) Write a balanced equation for the reaction of $Cl{F}_5$  with aqueous base (see the reaction of $Br{F}_5$  shown on p. 614).

Step-by-Step Solution

Answer

(a)   $3B{r}_2\left(l\right)+6O{H}^{-}\left(aq\right)\to 5B{r}^{-}\left(aq\right)+Br{O}_3^{-}\left(aq\right)+3H_{2}O\left(l\right)$

(b)  $Cl{F}_5\left(g\right)+6NaOH\left(aq\right)\to 5NaF\left(aq\right)+NaCl{O}_3\left(aq\right)+3H_{2}O\left(l\right)$

Step 1: Disproportionation reaction

A reaction in which a substance undergoes oxidation and reduction simultaneously is called a disproportionation reaction.

Therefore, the element in the reaction can exist in three oxidation states. The element will be in an intermediate oxidation state in the reactant.

Disproportionation reactions are identified from the element's oxidation state in the reactant and the product side. The disproportionation reaction is an example of a redox reaction in which oxidation and reduction occur simultaneously.

 

Step 2: Interhalogen compounds

Compounds formed between two halogens and have the general formula  , where X and Y are halogens are called interhalogen compounds. ‘n’ is an integer with values 1, 3, 5 or 7. X will be less electronegative than Y and will be in its positive oxidation state.

These compounds are more reactive than halogens and, therefore, undergo decomposition easily. They are also diamagnetic and contain an even number of atoms in them.

Step 3: Balanced equation for the given reactions

(a) The disproportionation of bromine occurs in a basic medium. Bromine is a liquid, and when mixed with an aqueous solution of a base, it forms bromide and bromate ions. Water is formed as a by-product.

The oxidation state of bromine changes from 0 in bromine to -1 in bromide ion and +5 in bromate ion. Therefore, bromine undergoes both oxidation and reduction. It is a disproportionation reaction.

The reaction can be shown as

 $3B{r}_2\left(l\right)+6O{H}^{-}\left(aq\right)\to 5B{r}^{-}\left(aq\right)+Br{O}_3^{-}\left(aq\right)+3H_{2}O\left(l\right)$

 

(b) Chlorine hexafluoride is a gaseous interhalogen compound and is less stable than chlorine or fluorine. Therefore, in a basic solution, it undergoes decomposition to form fluoride and chlorate ion. 

If the base used is sodium hydroxide, the products formed are sodium fluoride and sodium chlorate. Water is also formed as a by-product in the reaction, which dissolves sodium fluoride and sodium chlorate.

The balanced equation for the reaction is

 $Cl{F}_5\left(g\right)+6NaOH\left(aq\right)\to 5NaF\left(aq\right)+NaCl{O}_3\left(aq\right)+3H_{2}O\left(l\right)$