
Which of the hydrogen halides forms salts like $KH{{X}_{2}}$ (where X is a halogen atom)
A. $HF$
B. $HCl$
C. $HI$
D. $HBr$
Answer
161.4k+ views
Hint: Hydrogen Halides are the diatomic inorganic compounds which are dissolved in water to give acids which are known as hydrohalic acids. We have to answer the hydrogen halide that forms salts in the form of $KH{{X}_{2}}$.
Complete Step by Step Solution:
As we discuss hydrogen halides are the diatomic compounds with general $HX$ where $X\,\,be\,\,F,Cl,Br,I$. These exist as colourless gases at room temperature and are water soluble gases. It can be prepared easily by the direct reaction of hydrogen gas with the hydrogen molecule.
We know electronegativity is the tendency of an atom to attract the pair of electrons towards itself. As we go down the group, size of the atom goes on increasing and hence electronegativity goes on decreasing.
So the order of electronegativity in halogens is F > Cl > Br > I .
We know as the hydrogen of halide ions increases, the bond length also increases and more easily the ${{H}^{+}}$ ion is liberated. Hence $HI$forms the strongest hydrogen halide in the halide family.
Hence due to the highest electronegativity of fluorine the anion ${{[F--H-F]}^{-}}$ exist as a result of strong hydrogen bond by which ${{K}^{+}}$ associate to form $KH{{F}_{2}}$.
Thus, Option (A) is correct.
Note: $HI$ has low thermal stability due to this it exists as the strongest reducing agent. It can easily give the hydrogen atom to reduce the other compound. Thus it exists as the most stable compound in all the halogen family.
Complete Step by Step Solution:
As we discuss hydrogen halides are the diatomic compounds with general $HX$ where $X\,\,be\,\,F,Cl,Br,I$. These exist as colourless gases at room temperature and are water soluble gases. It can be prepared easily by the direct reaction of hydrogen gas with the hydrogen molecule.
We know electronegativity is the tendency of an atom to attract the pair of electrons towards itself. As we go down the group, size of the atom goes on increasing and hence electronegativity goes on decreasing.
So the order of electronegativity in halogens is F > Cl > Br > I .
We know as the hydrogen of halide ions increases, the bond length also increases and more easily the ${{H}^{+}}$ ion is liberated. Hence $HI$forms the strongest hydrogen halide in the halide family.
Hence due to the highest electronegativity of fluorine the anion ${{[F--H-F]}^{-}}$ exist as a result of strong hydrogen bond by which ${{K}^{+}}$ associate to form $KH{{F}_{2}}$.
Thus, Option (A) is correct.
Note: $HI$ has low thermal stability due to this it exists as the strongest reducing agent. It can easily give the hydrogen atom to reduce the other compound. Thus it exists as the most stable compound in all the halogen family.
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