Question

The increasing order of reducing the power of the halogen acids is

• A. $\mathrm{HF}<\mathrm{HCI}<\mathrm{HBr}<\mathrm{HI}$
• B. $\mathrm{HI}<\mathrm{HBr}<\mathrm{HCI}<\mathrm{HF}$
• C. $\mathrm{HBr}<\mathrm{HCI}<\mathrm{HF}<\mathrm{HI}$
• D. HCI < HBr < HF < HI

Answer 1

The correct option is A

$\mathrm{HF}<\mathrm{HCI}<\mathrm{HBr}<\mathrm{HI}$

The explanation for the correct option

(A)$\mathrm{HF}<\mathrm{HCI}<\mathrm{HBr}<\mathrm{HI}$

Reducing the power of the halogen acids

• As the size of the halogen atom increases, the strength of the $\mathrm{H}-\mathrm{X}$ bond decreases.
• So, the tendency to donate hydrogen increases from $\mathrm{HF}$ to $\mathrm{HI}$.
• So, increasing the order of reducing the power of the halogen acids is as follows:-
• $\mathrm{HF}<\mathrm{HCI}<\mathrm{HBr}<\mathrm{HI}$.

The explanation for the incorrect option

• The increasing order of reducing the power of the halogen acids is $\mathrm{HF}<\mathrm{HCI}<\mathrm{HBr}<\mathrm{HI}$.
• Hence, options (B), (C), and (D) are incorrect.

Hence, option (A) is correct. The increasing order of reducing the power of the halogen acids is $\mathrm{HF}<\mathrm{HCI}<\mathrm{HBr}<\mathrm{HI}$.