--%>

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 + H2O  1402_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 (NaClO3) and potassium chlorate (KClO3) are 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 : Moles of HCl present in .70 L of a .33

    Detail the moles of HCl which are present in .70 L of a .33 M HCl solution?

  • Q : Calculating Formulae Superphosphate has

    Superphosphate has the formula CaH4(PO4)2 H2O, what is the calculation to get the percentage of Phosphorus, I need to show the calculation. I know it is 30.9737622 u in weight and 2 atoms of the formula, but not sure how to work the calculation backwards.

  • Q : Question on Mole fraction Mole fraction

    Mole fraction of any solution is equavalent to: (a) No. of moles of solute/ volume of solution in litter (b) no. of gram equivalent of solute/volume of solution in litters (c) no. of  moles of solute/ Mass of solvent in kg (d) no. of moles of any

  • Q : Amount of glucose in blood What is the

    What is the normal amount of glucose in 100ml of blood (8–12 hrs after meal) is: (i) 8mg (ii) 80mg (iii) 200mg (iv) 800mg Choose the right answer from above.

  • Q : Depression in the freezing point When

    When 0.01 mole of sugar is dissolved in 100g of a solvent, the depression in freezing point is 0.40o. When 0.03 mole of glucose is dissolved in 50g of the same solvent, depression in the freezing point will be:(a) 0.60o  (b) 0.80o

  • Q : Decinormal concentration of Sulfuric

    Give me answer of this question. The volume of water to be added to 100cm3 of 0.5 N N H2SO4 to get decinormal concentration is : (a) 400 cm3 (b) 500cm3 (c) 450cm3 (d)100cm3

  • Q : Polymers comparison of biodegradable

    comparison of biodegradable and non-biodegradable polymers

  • Q : Chem Explain how dissolving the Group

    Explain how dissolving the Group IV carbonate precipitate with 6M CH3COOH, followed by the addition of extra acetic acid.

  • Q : Number of mlecules in methane Can

    Can someone please help me in getting through this problem. The total number of molecules in 16 gm of methane will be: (i) 3.1 x 1023 (ii) 6.02 x 1023 (iii) 16/6.02 x 1023 (iv) 16/3.0 x 1023

  • Q : Molarity 20mol of hcl solution requires

    20mol of hcl solution requires 19.85ml of 0.01 M NAOH solution for complete neutralisation. the molarity of hcl solution