
Order of reactivity of halogens towards hydrogen is:
[A]${{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}$
[B]${{I}_{2}}>B{{r}_{2}}>C{{l}_{2}}>{{F}_{2}}$
[C]$C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}>{{F}_{2}}$
[D]$B{{r}_{2}}>C{{l}_{2}}>{{F}_{2}}>{{I}_{2}}$
Answer
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Hint: Reactivity of the halogens depends on their electronegativity. Higher the electronegativity, higher will be the tendency of the halogens to protonate, thus it will have a higher reactivity towards hydrogen than other halogens.
Complete step by step answer:
Fluorine, chlorine, bromine and iodine are known as the halogens. They belong to Group 17 which we also know by the name halogen group. These halogens exist as a diatomic molecule.
Halogens are highly electronegative and they have 7 electrons in their valence shells. Gaining one electron gives them the stable noble gas configuration thus, the react readily with the Group 1 and 2 elements.
As mentioned above, halogens react readily with Group 1 and 2 elements and as hydrogen sits in Group 1, they react with hydrogen too.
Halogens react directly with hydrogen forming acids like hydrochloric acid, hydrobromic acid, hydroiodic acid and hydrofluoric acid.
As we know, the atomic radius increases down the group and due to this, the nuclear attraction faced by the valence shell electrons of the atom decreases, also down the group electronegativity decreases as, a smaller negative ion will have higher tendency to attract a proton towards itself than a bigger atom. Therefore on moving down the group, reactivity decreases.
We know that fluorine lies on the top in Group 17 followed by chlorine, bromine and iodine. Therefore, fluorine is the most reactive and iodine is the least.
Therefore, the correct order of reactivity of halogens towards hydrogen will be ${{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}$.
Hence, the correct answer is option A.
Note:
The hydrogen halides are formed through covalent bonding. The order of acidity of the acids thus formed is the opposite of the reactivity order. HI is the strongest acid among the halogens. Due to its larger size, the bond length between the H-I atoms increases and results in the decrease of bond strength. Hydroiodic acid will lose its proton readily, thus is a stronger acid than the other halogen acids.
Complete step by step answer:
Fluorine, chlorine, bromine and iodine are known as the halogens. They belong to Group 17 which we also know by the name halogen group. These halogens exist as a diatomic molecule.
Halogens are highly electronegative and they have 7 electrons in their valence shells. Gaining one electron gives them the stable noble gas configuration thus, the react readily with the Group 1 and 2 elements.
As mentioned above, halogens react readily with Group 1 and 2 elements and as hydrogen sits in Group 1, they react with hydrogen too.
Halogens react directly with hydrogen forming acids like hydrochloric acid, hydrobromic acid, hydroiodic acid and hydrofluoric acid.
As we know, the atomic radius increases down the group and due to this, the nuclear attraction faced by the valence shell electrons of the atom decreases, also down the group electronegativity decreases as, a smaller negative ion will have higher tendency to attract a proton towards itself than a bigger atom. Therefore on moving down the group, reactivity decreases.
We know that fluorine lies on the top in Group 17 followed by chlorine, bromine and iodine. Therefore, fluorine is the most reactive and iodine is the least.
Therefore, the correct order of reactivity of halogens towards hydrogen will be ${{F}_{2}}>C{{l}_{2}}>B{{r}_{2}}>{{I}_{2}}$.
Hence, the correct answer is option A.
Note:
The hydrogen halides are formed through covalent bonding. The order of acidity of the acids thus formed is the opposite of the reactivity order. HI is the strongest acid among the halogens. Due to its larger size, the bond length between the H-I atoms increases and results in the decrease of bond strength. Hydroiodic acid will lose its proton readily, thus is a stronger acid than the other halogen acids.
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