Question

The correct order of melting points of hydrides of group $16$ elements is :
(A) $H_{2}S>H_{2}Se>H_{2}Te$
(B) $H_{2}S<H_{2}Se<H_{2}Te$
(C) $H_{2}S>H_{2}Se<H_{2}Te$
(D) $H_{2}S<H_{2}Se>H_{2}Te$

Answer 1

Hint: To solve the given problem, we should have information about group $16$ elements, hydrides and melting point. Melting point of an element or compound is the temperature at which the solid state liquefies and exists in equilibrium and exists in equilibrium. Group $16$ elements start from oxygen and are collectively called Chalcogens as they form ores. They show catenation tendencies. Oxygen is the most reactive among the group $16$ . Hydride generally is the anion of hydrogen. They form covalent bonds with the elements of which hydrides are formed. Hydride is formed by the combination of hydrogen with other elements.

Complete answer:
Step-1 :
We know that hydrides form covalent bonds with the elements which they combine with. So, will be the case of Group $16$ hydrides like $H_{2}S,H_{2}Se$ and $H_{2}Te$ among which we have to find the order of melting point.

Step-2 :
In case of covalent hydride, the melting point is directly proportional to Vanderwaal’s force which is directly proportional to the molar mass. So, the hydride having high molar mass will have a high melting point.

Step-3 :
We know that going down the group, the molar mass of elements increases. So, $Te>Se>S$. So, the order of melting point is $H_{2}S<H_{2}Se<H_{2}Te$

Note:
We should remember that the hydrides are of different types such as covalent, ionic and metallic hydride. A hydride of oxygen and hydride of Carbon and hydride of Nitrogen.