--%>

What are halogen oxoacids?

Fluorine yields only one oxyacid, hypofluorous acid (HOF). Chlorine, bromine and iodine form four series of acids with formulae: HOX, HXO2, HXO3 and HXO4, although many of these are known only in solutions or as salts.
    
The Hypohalous acids HOCl, HOBr and HOI are weak acids and are only formed in aqueous solutions by disproportionation of the halogen of the halogen water

X2 + H21402_Phosphorus trichloride.png  HOX + HX (X = Cl, Br, I)

Salts of these acids are known as hypohalites, e.g. bleaching powder, CaOCl2 is a common example of this category.
    
The halic acids HClO3 and HBrO3 are also known as solutions, but iodic acid HIO3 exists as a white solid. Thus, the stability of acids increases with increase in atomic number of the halogen. These acids act as strong oxidizing agents, e.g. these oxidize halides to give halogens in acid medium.

OX3- + 5X- + 6H+  1402_Phosphorus trichloride.png  3X2 + 3H2O

The salts of these are called halates. Amongst the halates, sodium chlorate (NaClO3and potassium chlorate (KClO3are prepared on industrial scale. It is also known as 'Berthelot salt'. NaClO3 is a powerful weed killer, whilst KClO3 is used in fireworks and matches.
    
Perhalic acid i.e. perchloric, periodic acids as well as their salts perchlorates and periodates are known to exist. The perhalates (MXO4)are prepared by the electrolytic oxidation of the corresponding halates, MXO3.

4ClO3 1402_Phosphorus trichloride.png  Cl+ 3ClO4-

The disproportionation of BrO3- to BrO4- is unfavorable, therefore per bromates are obtained only by oxidation of BrO3- by F2 in basic solution.

BrO3- + F2 + 2OH-  1402_Phosphorus trichloride.png  BrO4+ 2F- + H2O

Acidic character of oxyacids: the variation in the acidic character of the halogen acids in different oxidation states are summarized below:
    
The acid strength of oxyacid of the same halogen increases with the increase in oxidation number of the halogen. For example, among the different oxyacids of chlorine the acidic character follows the order

HOCl < HClO2 < HClO3 < HClO4

Reason: the acid strength can be explained on the basis Lowry-Bronsted concept that conjucate base of weak and is strong and conjugate base of strong acid is weaker.

   Related Questions in Chemistry

  • Q : Meaning of molality of a solution The

    The molality of a solution will be: (i) Number of moles of solute per 1000 ml of solvent (ii) Number of moles of solute per 1000 gm of solvent (iii) Number of moles of solute per 1000 ml of solution (iv) Number of gram equivalents of solute per 1000 m

  • Q : Problem on physical and thermodynamic

    The shells of marine organisms contain calcium carbonate CaCO3, largely in a crystalline form known as calcite. There is a second crystalline form of calcium carbonate known as aragonite. Physical and thermodynamic properties of calcite and aragonite at 298

  • Q : Problem on volumetric flow rate Methane

    Methane containing 4 mol% N2 is flowing through a pipeline at 105.1 kpa and 22 °C. To check this flow rate, N2 at the same temperature and pressure are introduced to the pipeline at the rate of 2.83 m3/min. At the end of the pipe (

  • Q : Explain oxygen and its preparation.

    Karl Scheele, the Swedish chemist, was

  • Q : BASIC CHARACTER OF AMINES IN GAS PHASE,

    IN GAS PHASE, BASICITIES OF THE AMINES IS JUST OPPOSITE TO BASICITY OF AMINES IN AQEUOUS PHASE .. EXPLAIN

  • Q : Explain vapour pressure of liquid

    Liquid solutions are obtained when the solvent is liquid. The solute can be a gas, liquid or a solid. In this section we will discuss the liquid solutions containing solid or liquid solutes. In such solutions the solute may or may not be volatile. We shall limit our d

  • Q : Molality of a glucose solution What

    What will be the molality of a solution containing 18g of glucose (having mol. wt. = 180) dissolved in 500g of water: (i) 1m  (ii) 0.5m  (iii) 0.2m  (iv) 2m

  • Q : Problem based on molarity Select the

    Select the right answer of the question. If 18 gm of glucose (C6H12O6) is present in 1000 gm of an aqueous solution of glucose, it is said to be: (a)1 molal (b)1.1 molal (c)0.5 molal (d)0.1 molal

  • Q : Vapour pressure over mercury Choose the

    Choose the right answer from following. At 300 K, when a solute is added to a solvent its vapour pressure over the mercury reduces from 50 mm to 45 mm. The value of mole fraction of solute will be: (a)0.005 (b)0.010 (c)0.100 (d)0.900

  • Q : Problem associated to vapour pressure

    Provide solution of this question. 60 gm of Urea (Mol. wt 60) was dissolved in 9.9 moles, of water. If the vapour pressure of pure water is P0 , the vapour pressure of solution is:(a) 0.10P0 (b) 1.10P0 (c) 0.90P0 (d) 0.99P0