The correct option is 2: C > D > B > A.
The acidity of hydrides generally increases down a group in the periodic table. In the case of Group 16 elements, which include oxygen (O), sulphur (S), selenium (Se), and tellurium (Te), the hydrides follow a trend of increasing acidity as you go down the group.
Here is a detailed explanation of the acidity trend for the given hydrides:
(A) \(H_2O\) (water):
Water is a unique molecule because it exhibits amphoteric behavior. It can act as both an acid and a base. However, when considering acidity, water is relatively weak compared to the other hydrides in this group. It can donate a proton (H⁺) but is not as acidic as the hydrides below it in the group.
(B) \(H_2S\) (hydrogen sulphide):
Hydrogen sulphide is more acidic than water. It can donate a proton to form \(HS^-\) ion. The larger atomic size of sulphur compared to oxygen allows for a more diffuse electron cloud, making the hydrogen atom more easily ionizable.
(D) \(H_2Se\) (hydrogen selenide):
Hydrogen selenide is more acidic than hydrogen sulfide. As you move down the group, the atomic size increases, and the ability of the element to stabilize a negative charge also increases. This makes \(H_2Se\) more acidic than \(H_2S\).
(C) \(H_2Te\) (hydrogen telluride):
Hydrogen telluride is the most acidic among the given hydrides. The larger size of tellurium compared to sulfur and selenium allows for even greater stabilization of the negative charge, making \(H_2Te\) the most acidic among the hydrides in this group.
Therefore, the correct order of decreasing acidity is C > D > B > A. |
