
Chlorine reacts with water according to the equation:
$ C{l_2} + {H_2}O \to HClO + {H^ + } + C{l^ - } $ . Given this, how do you write an equation to show how iodine reacts with water?
Answer
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Hint: Halogens reactivity depends on their electronegativity. Higher the electronegativity of the halogens, higher tendency of the halogens to protonate. So, they are more reactive towards hydrogen.
Complete answer:
Fluorine, chlorine, bromine and iodine are the elements of halogens. They belong to $ 17th $ group of the periodic table which is named as halogens. They have seven electrons in the valence shell. And need one electron to become stable thus they are more reactive towards group $ 1 $ and group $ 2 $ elements. As we know, hydrogen atoms are also placed in group $ 1 $ so they will show more reactivity toward hydrogen too.
Halogens react with hydrogen directly forming acids like hydrofluoric acid, hydrochloric acid, hydrobromic acid and hydroiodic acid.
In this question we are discussing Iodine. Iodine is strongly reactive toward hydrogen although it is less as compared to other halogens. Iodine can never be found as an element, but rather found as a molecule, i.e. $ {I_2} $ or as an ion which is $ {I^ - } $ or as an iodate that is a salt of iodic acid with $ IO_3^ - $ anion.
Thus, when iodine is added to water, the reaction we will obtain will be:
$ {I_2}\left( l \right) + {H_2}O\left( l \right) \to O{I^ - }\left( {aq} \right) + 2{H^ + }\left( {aq} \right) + {I^ - }\left( {aq} \right) $
Iodine molecules react with water and produce substances like hypoiodite $ O{I^ - } $ . The reaction can move both ways of the equilibrium which depend on the $ pH $ of the solution. Iodine can also occur as the following:
$ I_3^ - \left( {aq} \right),HIO\left( {aq} \right),I{O^ - }\left( {aq} \right),HI{O_3}\left( {aq} \right) $
Note:
When halogen combines with hydrogen they form a bond which is covalent. The order of acidic strength is opposite of the reactivity order. Thus $ HI $ is the stronger acid than other halogens as because of its larger size, the bond length between the $ H - I $ atoms increases and results in the decrease of bond strength.
Complete answer:
Fluorine, chlorine, bromine and iodine are the elements of halogens. They belong to $ 17th $ group of the periodic table which is named as halogens. They have seven electrons in the valence shell. And need one electron to become stable thus they are more reactive towards group $ 1 $ and group $ 2 $ elements. As we know, hydrogen atoms are also placed in group $ 1 $ so they will show more reactivity toward hydrogen too.
Halogens react with hydrogen directly forming acids like hydrofluoric acid, hydrochloric acid, hydrobromic acid and hydroiodic acid.
In this question we are discussing Iodine. Iodine is strongly reactive toward hydrogen although it is less as compared to other halogens. Iodine can never be found as an element, but rather found as a molecule, i.e. $ {I_2} $ or as an ion which is $ {I^ - } $ or as an iodate that is a salt of iodic acid with $ IO_3^ - $ anion.
Thus, when iodine is added to water, the reaction we will obtain will be:
$ {I_2}\left( l \right) + {H_2}O\left( l \right) \to O{I^ - }\left( {aq} \right) + 2{H^ + }\left( {aq} \right) + {I^ - }\left( {aq} \right) $
Iodine molecules react with water and produce substances like hypoiodite $ O{I^ - } $ . The reaction can move both ways of the equilibrium which depend on the $ pH $ of the solution. Iodine can also occur as the following:
$ I_3^ - \left( {aq} \right),HIO\left( {aq} \right),I{O^ - }\left( {aq} \right),HI{O_3}\left( {aq} \right) $
Note:
When halogen combines with hydrogen they form a bond which is covalent. The order of acidic strength is opposite of the reactivity order. Thus $ HI $ is the stronger acid than other halogens as because of its larger size, the bond length between the $ H - I $ atoms increases and results in the decrease of bond strength.
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