Which one forms the $KH{{X}_{2}}$ type compound?
1) HF
2) HCl
3) HI
4) HBr
Answer
558.3k+ views
Hint: X is the substituent term used for halogens. The electronegativity of halogens is responsible for the bonding of the halogens with hydrogen and thus forming compounds. The electronegativity decreases down the group and the most electronegative halogen is Fluorine that can also make hydrogen bonds.
Complete answer:
We have been given halogen acids, among them we have to find which one of them can form the compound of $KH{{X}_{2}}$ type.
$KH{{X}_{2}}$ compound is formed by the acid that has the smallest halogen. The smallest sized halogen is fluorine. It forms compound $KH{{X}_{2}}$ ztype because small size of fluorine is capable to form a molecule ${{H}_{2}}{{F}_{2}}$ as fluorine forms hydrogen bonds. It is more electronegative than hydrogen and have a small size thus can accommodate two hydrogen atoms forming ${{H}_{2}}{{F}_{2}}$, this compound forms $KH{{F}_{2}}$ by the reaction with potassium fluoride as
${{H}_{2}}{{F}_{2}}+KF\to KH{{F}_{2}}+HF$
Hence, hydrogen fluoride HF forms a compound $KH{{F}_{2}}$.
So option A is correct.
Additional information:
Electronegativity of any element is the ability to attract electrons from other atoms. Halogens have the tendency to accept electrons as they need 1 electron to complete their octet, therefore they are electronegative. Fluorine is the most electronegative in the periodic table.
Note:
The strength of halogen acids depends on the H – X bond strength. This bond strength decreases with increase in size of halogen. This makes hydrogen fluoride (HF) to be the weakest acid and hydrogen iodide (HI) to be the strongest acid due to the large size of iodide ion, the H – I bond weakens.
Complete answer:
We have been given halogen acids, among them we have to find which one of them can form the compound of $KH{{X}_{2}}$ type.
$KH{{X}_{2}}$ compound is formed by the acid that has the smallest halogen. The smallest sized halogen is fluorine. It forms compound $KH{{X}_{2}}$ ztype because small size of fluorine is capable to form a molecule ${{H}_{2}}{{F}_{2}}$ as fluorine forms hydrogen bonds. It is more electronegative than hydrogen and have a small size thus can accommodate two hydrogen atoms forming ${{H}_{2}}{{F}_{2}}$, this compound forms $KH{{F}_{2}}$ by the reaction with potassium fluoride as
${{H}_{2}}{{F}_{2}}+KF\to KH{{F}_{2}}+HF$
Hence, hydrogen fluoride HF forms a compound $KH{{F}_{2}}$.
So option A is correct.
Additional information:
Electronegativity of any element is the ability to attract electrons from other atoms. Halogens have the tendency to accept electrons as they need 1 electron to complete their octet, therefore they are electronegative. Fluorine is the most electronegative in the periodic table.
Note:
The strength of halogen acids depends on the H – X bond strength. This bond strength decreases with increase in size of halogen. This makes hydrogen fluoride (HF) to be the weakest acid and hydrogen iodide (HI) to be the strongest acid due to the large size of iodide ion, the H – I bond weakens.
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