Question

Complete the following reaction equations:
$Xe{{F}_{2}}+P{{F}_{5}}\to$
$C{{l}_{2(g)}}+NaO{{H}_{(aq)}}\to$

Answer 1

We know that $Xe{{F}_{6}}$ is an example of a noble gas compound. Generally noble gases are inert in nature due to their stable electron configuration but in certain circumstances heavier noble gases react with highly reactive elements. Sodium chloride reacts with chlorine in different ways. For hot and concentrated there is a different reaction. Similarly for the reaction of chlorine with different reactions. For example if chlorine is in excess then the product is different from the reaction.

Complete step by step answer:
$Xe{{F}_{2}}$ is a fluorinating agent and $P{{F}_{5}}$ is a Lewis acid. $Xe{{F}_{2}}$ donates a fluoride ion to $P{{F}_{5}}$ which results in the formation of ${{\left[ XeF \right]}^{+}}{{\left[ P{{F}_{6}} \right]}^{-}}.$ This is possible because of the availability of vacant d-orbitals in phosphorus. The reaction is given by; $Xe{{F}_{2}}+P{{F}_{5}}\to {{\left[ XeF \right]}^{+}}{{\left[ P{{F}_{6}} \right]}^{-}}.$
We know the reaction of the sodium hydroxide with chlorine yields different products under different conditions. For example: Hot and concentrated $NaO{{H}_{(aq)}}$ and $C{{l}_{2(g)}}$, Hot concentrated Chlorine reacts with hot and concentrated sodium hydroxide solution to form sodium chloride and sodium chlorate. The reaction is given by; $3C{{l}_{2}}+6NaOH\to 5NaCl+NaCl{{O}_{3}}+3{{H}_{2}}O\text{ }\text{.}$
Also keep note that the reaction of chlorine with fluorine yields different and different or example: When reacts with an excess of then chlorine trifluoride instead of fluorine trichloride and reason why chlorine trifluoride is formed instead of fluorine trichloride. In the first excited state chlorine atom can exhibit a covalence of three, hence cannot expand its octet due to the absence of empty d- orbitals in the energy shell.
Therefore the complete reaction is given by;
$Xe{{F}_{2}}+P{{F}_{5}}\to {{\left[ XeF \right]}^{+}}{{\left[ P{{F}_{6}} \right]}^{-}}.$
$3C{{l}_{2}}+6NaOH\to 5NaCl+NaCl{{O}_{3}}+3{{H}_{2}}O\text{ }\text{.}$

Note: Read the question carefully, in the question (i) $Xe{{F}_{2}}$ is not the only xenon compound that can be synthesized: there are several oxides, fluorides and oxofluorides of xenon. Similarly for question (ii) there is a condition of Hot and concentrated and not cold and dilute.