Question
How are interhalogen compounds prepared?

Answer

All interhalogen compounds are prepared by direct reaction of halogens or reaction of halogen on lower interhalogen compounds. The compound formed depends upon specific conditions of the reactions,
$(a)$ By direct combination :
  • Chlorine monofluoride, $\text{ClF}$ : It is a colourless gas obtained by reacting equal volumes of $Cl_2$ and $F_2.$
    $Cl _2+ F _2 \longrightarrow 2 ClF$
    $($equal volumes$)$
  • Chlorine trifluoride, $ClF_3 :$ It is a colourless gas, obtained by reacting $Cl_2$ with excess of $F_2.$
    $Cl _2+3 F _2 \longrightarrow 2 ClF _3$
    excess
  • Bromine trifluoride, $BrF_3 :$ It is a yellow green liquid obtained by reacting $Br_2.$
    $Br _2+3 F _2 \longrightarrow 2 BrF _3$
    $($diluted with water$)$
  • Iodine trichloride, $ICl_3 :$ It is an orange solid obtained by reacting $I_2$ with excess of $Cl.$ This orange solid dimerises to form$ I_2Cl_6$ having $Cl-$bridges.
    $I _2+3 Cl _2 \longrightarrow 2 ICl _3$
    excess
  • Iodine chloride, $ICl :$ It is obtained by reaction between equimolar amounts of $I_2$ and $Cl_2. a-$ form of it is a ruby red solid while $1-$form is a brown$-$red solid.
    $\underset{\text { (equimolar) }}{ I _2+ Cl _2} \longrightarrow 2 ICl$
  • Bromine pentafluoride, $BrF_5 :$ It is a colourless liquid and obtained by the reaction of bromine with the excess of fluorine.
    $Br _2+\underset{\text { (excess) }}{5 F _2 \longrightarrow} 2 BrF _5$
  • Bromine chloride $BrCl :$ It is a gas, obtained by the reaction of bromine with chlorine taken in equal volumes.
    $Br _2+ Cl _2 \longrightarrow 2 BrCI$
    $($equal volumes$)$
$(b)$ By the reaction of halogen with interhalogen compound :
  • Bromine fluoride $(BrF) :$ It is a pale brown coloured gas prepared by the action of $Br_2$ on $BrF_3.$ $Br_2 + BrF_3 \rightarrow 3BrF$
  • Bromine trifluoride $( BrF_3) : BrF_3$ can also be prepared by the action of $Br_2$ on $ClF’_3.$
  • $Br_2 + ClF_3 \rightarrow 2BrF_3 + BrCl$
$(c)$ Special reaction for the preparation of $ICl :$
  • $ICl($ Wij’s solution$)$ can be prepared by the action of Iodinef $(I_2)$ on potassium chlorate $(KClO_3)I_2 + KClO_3 \rightarrow ICl + KlO_3$

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