Please scroll down to get to the study materials.
$$MnO_2+conc. H_2SO_4\longrightarrow MnSO_4+H_2O+[O]$$
$$[NaX+H_2SO_4\longrightarrow NaHSO_4+HX]\space X\space 2$$
These equatios results
Where X = Cl2, Br2, I2
Carnallite [KCl.MgCl2.6H2O] contains the small amount of KBr and MgBr2 as impurities. It is first dissolved in water and resulting solution is concentrated where less soluble chloride salt separates out leaving behind more soluble bromide in the mother liquor. The hot mother liquor is about to trickle down to tower packed with earthen balls with a current of air up which current of chlorine is passed. Chlorine liberates bromine from bromide.
Bromine thus liberated is converted to vapor. Bromine vapor travels up and then condenses in a spiral condenser to form liquid bromine. The vapor which is not condensed are absorbed by iron filings by forming ferrous - ferric bromide (Fe3Br8).
Bromine thus obtained contains chlorine and iodine as impurities which are removed by distillation with KBr and ZnO respectively.
Iodine is manufactured from deep seaweeds especially laminaria variety which contains iodine in the form of alkali iodide (NaI & KI). First of all, seaweeds are collected dried and burnt carefully so that no iodine is lost by decomposition and sublimation and ash called kelp is obtained. It contains 0.4 % to 1.3 % iodine in the form of iodide along with chloride and sulphate. The kelp is then treated with water and the aqueous solution is crystallized out where chloride and sulphate are crystallized whereas more soluble NaI and KI remains I mother liquor. It is mixed with MnO2 and concentrated H2SO4 and heated in the iron retort. Iodine vapors are liberated in a series of earthenwares condenser called aludels.
From caliche [crude chile salt pitre]
Crude chile salt pitre is NaNO3. It contains 0.3 % to 0.4 % iodine in form of iodide salt. The mineral is treated with water and then crystallized where crystal of NaNO3 separate out and NaIO3 remains in the mother liquor. The mother liquor is treated with sodium bisulphate which reduces sodium iodide to free iodine.
Chlorine and bromine react with water forming hydrohalide acid and hypohalous acid.
Where X2= Cl2 and Br2
HOX is unstable so it decomposes to give oxygen and hydrohalic acid in sunlight.
$$KI+I_2\longrightarrow 2KI_3 (Formation\space of\space potassium\space tri-iodide)$$
Colored matter + [O] → Colorless matter
pulse, Tracy. Introduction to chemistry. u.s.a: flex book, 2010.
Pathak, Sita Karki. The Text Book of Chemistry. Kathmandu: Vidhyarthi Pustak Bhandar, 2012.