Question

Decreasing acid strengths of $HI,HBr,HCl$ and $HF$ is:

• A. $HF>HCl>HBr>HI$
• B. $HI>HBr>HCl>HF$
• C. $HI>HCl>HBr>HF$
• D. $HI>HF>HCl>HBr$

Answer 1

Answer: (B)

$HI>HBr>HCl>HF$

As we know,

Acids have $A$ tendency to lose hydrogen ion. So, larger is the bond length, more is acidic nature (for halogen acids)

Bond	Bond Enthalpy ( $kJmo{l}^{-1}$)
H-F	+562
H-Cl	+431
H-Br	+366
H-I	+299

Because the fluorine atom is so small, the bond enthalpy (bond energy) of the hydrogen-fluorine bond is very high. In order for ions to form when the hydrogen fluoride reacts with water, the $H-F$ bond must be broken. It would seem reasonable to say that the relative reluctance of hydrogen fluoride to react with water is due to the large amount of energy needed to break that bond

So order is $HI>HBr>HCl>HF$

Option (B) is correct.