--%>

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 : Problem related to molality Help me to

    Help me to solve this problem. What is the molality of a solution which contains 18 g of glucose (C6,H12, O6) in 250 g of water:  (a) 4.0 m (b) 0.4 m (c) 4.2 m (d) 0.8 m

  • Q : What is Spectroscopy? This is a very

    This is a very important aspect of Physical Chemistry in which knowledge of the size, shape, rigidity and electronic structure of molecules deduced from the experimental methods treated here goes hand in hand with the theoretical approaches of chemical reactions. Spec

  • Q : Amines why o-toluidine is a weaker base

    why o-toluidine is a weaker base than aniline?

  • Q : Concentration of an aqueous solution

    Give me answer of this question. The concentration of an aqueous solution of 0.01M CH3OH solution is very nearly equal to which of the following : (a) 0.01%CH3OH (b) 0.1%CH3OH (c) xCH3OH= 0.01 (d) 0.99MH2O (

  • Q : Question based on relative lowering of

    Give me answer of this question. When a non-volatile solute is dissolved in a solvent, the relative lowering of vapour pressure is equal to: (a) Mole fraction of solute (b) Mole fraction of solvent (c) Concentration of the solute in grams per litre

  • Q : Molarity of the final mixture Can

    Can someone please help me in getting through this problem. Two solutions of a substance (that is, non electrolyte) are mixed in the given manner 480 ml of 1.5M first solution + 520 ml of 1.2M second solution. Determine the molarity of the final mixture

  • Q : Oxoacids of halogens Why oxidising

    Why oxidising character of oxoacids of halogens decreases as oxidation number increases?

  • Q : Molal elevation constant of water The

    The boiling point of 0.1 molal aqueous solution of urea is 100.18oC  at 1 atm. The molal elevation constant of water is: (a) 1.8    (b) 0.18   (c) 18    (d) 18.6Answer: (a) Kb

  • Q : Molecular Symmetry Types The number of

    The number of molecular orbitals and molecular motions of each symmetry type can be deduced. Let us continue to use the C2v point group and the H2O molecule to illustrate how the procedure develop

  • Q : Problem on mole fraction of glucose

    Provide solution of this question. While 1.80gm glucose dissolve in 90 of H2O , the mole fraction of glucose is: (a) 0.00399 (b) 0.00199 (c) 0.0199 (d) 0.998