Question

$ KH{F_2} $ is well known whereas $ KHC{l_2} $ and $ KHB{r_2} $ does not exist.

Answer 1

Hint :The halogens are a category of five chemically associated elements in the periodic table: fluorine $ \left( F \right) $ , chlorine $ \left( {Cl} \right) $ , bromine $ \left( {Br} \right) $ , iodine $ \left( I \right) $ , and astatine $ \left( {As} \right) $ . Tennessine $ \left( {Ts} \right) $ , an artificially formed element, may also be a halogen.

Complete Step By Step Answer:
The colourless and inorganic potassium bifluoride $ \left( {KH{F_2}} \right) $ is used as a glass etchant. The electrolysis of molten $ KH{F_2} $ and $ K{H_2}{F_3} $ is used in the industrial processing of fluorine. Henri Moissan was the first to use $ KH{F_2} $ electrolysis in $ 1886 $ .
Fluorine has a higher electronegative charge than $ Cl $ and $ Br $ . An atom's ability to draw shared electrons or electron density to itself is known as electronegativity. The electronegativity of an atom is determined by its atomic number as well as the distance between its valence electrons and the charged nucleus. The higher the related electronegativity, the more electrons are attracted to an atom or a substituent group.
Stronger hydrogen bonding is associated with higher electronegativity, so $ F $ binds with $ H $ to form the $ HF_2^ - $ ion. Other halides lack this tendency due to the absence of such intense intermolecular attraction also.
As a result, the $ {K^ + } $ cation joins with the $ HF_2^ - $ anion to form the ionic compound Potassium bifluoride $ \left( {KH{F_2}} \right) $ . $ Cl $ and $ Br $ cannot form $ HCl_2^ - $ and $ HBr_2^ - $ ions due to their lower electronegativity.
We can also find that there is no hydrogen bonding in hydrochloric acid $ \left( {HCl} \right) $ and hydrogen bromide $ \left( {HBr} \right) $ ; both occur in monomeric form. As a result, it is unable to form a salt of the type $ KHC{l_2} $ and $ KHB{r_2} $ .

Note :
Owing to the presence of hydrogen bonding, hydrofluoric acid occurs in dimeric form $ \left( {{H_2}{F_2}} \right) $ and has a dibasic composition. As it reacts with a base like $ KOH $ , it can form an acid salt that is, $ KH{F_2} $ .