
Arrange the hydra acids of the halogens in increasing order of acidity
A) HFB) HIC) HFD) HF
Answer
163.2k+ views
Hint: Usually the increase of atomic number results in the increase of acid strength because of the decrease in the force of interaction between nucleus and the electrons of the outermost or valence shell.
Complete step by step solution:Let's first discuss what an acid is. An acid is a chemical species that can donate hydrogen ions easily. The strength of an acid depends on the atomic size of the atom bonded with hydrogen. The smaller the size, the stronger the bond is and less acidity of the compound.
Let's discuss the acidity of the hydra acids of the halogens. In HF, the size of the atom of fluorine is lowest among the halogens. So, the attraction between the nucleus and electrons of the outermost shell is very high. This results in shorter bond length, so, the acidity of HF is least.
Let's discuss the acidity of HI. Iodine has the largest size among the given halogens. So, the attraction of the nucleus with the electrons of the valence shell is less. And this causes the longer bond length in HI. So, the acidity of HI is highest.
The order of increasing size of halogens is F, Cl, Br, I
So, the order of acidity of the hydra acid of halogens is HF
Therefore, option A is right.
Note: Always remember that, increase of bond length means breaking of bond is easy. And the decrease of bond length means the difficulty of breaking the bond. And the relation of bond length and bond order is inversely related.
Complete step by step solution:Let's first discuss what an acid is. An acid is a chemical species that can donate hydrogen ions easily. The strength of an acid depends on the atomic size of the atom bonded with hydrogen. The smaller the size, the stronger the bond is and less acidity of the compound.
Let's discuss the acidity of the hydra acids of the halogens. In HF, the size of the atom of fluorine is lowest among the halogens. So, the attraction between the nucleus and electrons of the outermost shell is very high. This results in shorter bond length, so, the acidity of HF is least.
Let's discuss the acidity of HI. Iodine has the largest size among the given halogens. So, the attraction of the nucleus with the electrons of the valence shell is less. And this causes the longer bond length in HI. So, the acidity of HI is highest.
The order of increasing size of halogens is F, Cl, Br, I
So, the order of acidity of the hydra acid of halogens is HF
Therefore, option A is right.
Note: Always remember that, increase of bond length means breaking of bond is easy. And the decrease of bond length means the difficulty of breaking the bond. And the relation of bond length and bond order is inversely related.
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