Question

Complete the following reaction:
$
  {\text{ a) 2NaOH + C}}{{\text{l}}_2}{\text{ }} \to \\
  {\text{ (cold and dilute)}} \\
 $
$
  {\text{b) C}}{{\text{l}}_{\text{2}}}{\text{ + 3}}{{\text{F}}_{\text{2}}}{\text{ }} \to \\
  {\text{ (excess)}} \\
 $

Answer 1

Hint: Sodium chloride reacts with chlorine in different ways. For hot and concentrated there is different reaction and cold and dilute there is different reaction.
Similarly for the reaction of chlorine with fluorine there are different reactions. For example if chlorine is in excess then the product is different from the reaction when fluorine is in excess.

Complete step by step answer:
We know the reaction of the sodium hydroxide with chlorine yields different products under different conditions. For example: Cold dilute $NaOH$ and $C{l_2}$ reaction:
 \[{\text{2NaOH + C}}{{\text{l}}_{\text{2}}}{\text{}} \to {\text{NaCl + NaOCl + }}{{\text{H}}_{\text{2}}}{\text{O}}\]. Cold dilute sodium hydroxide and chlorine gas reactions are disproportionately. Chlorine gas is oxidized and reduced to \[OC{l^ - }\] ion and \[C{l^ - }\] ion respectively. As products sodium chloride, sodium chlorate(I) and water are given.
Hot concentrated Chlorine reacts with hot and concentrated sodium hydroxide solution to form sodium chloride and sodium chlorate.
\[{\text{3C}}{{\text{l}}_{\text{2}}}{\text{ + 6NaOH}} \to {\text{5NaCl + NaCl}}{{\text{O}}_{\text{3}}}{\text{ + 3}}{{\text{H}}_{\text{2}}}{\text{O}}\]
Hence the reaction will be :
\[{\text{2NaOH + C}}{{\text{l}}_{\text{2}}}{\text{}} \to {\text{NaCl + NaOCl + }}{{\text{H}}_{\text{2}}}{\text{O}}\].

We know that the reaction of chlorine with fluorine yields different products in different conditions. For example: When $C{l_2}$ reacts with an excess of ${F_2}$ then chlorine trifluoride instead of fluorine trichloride is formed.
Reason why chlorine trifluoride is formed instead of fluorine trichloride: In the first excited state chlorine atom can exhibit a covalency of three, hence cannot expand its octet due to the absence of empty d- orbitals in the $2nd$ energy shell. Hence, it cannot exhibit covalency more than $1$ therefore $FC{l_3}$ is not possible.
And when both chlorine and fluorine are in equal amounts then $ClF$ will form.
Hence the reaction will be:
$C{l_2} + 3{F_2} \to 2Cl{F_3}$.

Note:
Read the question carefully, In the question (i) there is a condition of cold and dilute and not hot and concentrated. Similarly for question (ii) there is a condition that fluorine is in excess and not chlorine.