Question

The correct order of stability of hydrides of alkali metal is :
A. $LiH > NaH > KH > RbH$
B. $NaH > KH > RbH > LiH$
C. $RbH > KH > NaH > LiH$
D. $LiH > RbH > KH > NaH$

Answer 1

To answer this question the periodic properties of the elements from different groups should be known. A periodic table is a great source of information about the elements and their relation to each other. Changes of different periodic properties. The elements in periodic tables are arranged in a way that every element of each group shows similar physical characteristics and chemical characteristics.

Complete step by step answer:
Alkali metals are strongly reduced in nature; therefore, they cannot be extracted by reducing their oxides. And due to their strong electropositive alkali metals cannot be replaced by any other elements from its salt.
In the case of alkali metals, the hydrides of alkali metal are ionic in nature. The stability of the hydride of alkali metal depends upon the bond stability of the hydrides. The higher the bond strength, the higher will be the stability.
From top to bottom the size of the alkali metals increases, therefore the small size hydrogen cannot bind strongly with the bottom elements.
The bond strength of the bottom elements of the alkali in hydrides is very low as well as stability.
Therefore considering the size difference between the alkali metals and the hydrogen, the order of the stability is, $LiH > NaH > KH > RbH$.

So, the correct option is A.

Note:
-Due to their strong electropositive alkali metals cannot be replaced by any other elements from its salt. Due to this reason Alkali metals are prepared by electrolysis of their fused chlorides.
-But for electrolysis aqueous solution of their salt cannot be taken, due to their electropositive nature, hydrogen ions get reduced faster and produce hydrogen gas.
-For example, for the electrolysis of NaCl, sodium is formed in cathode by reduction, and chlorine gas is formed in the anode. The reactions are,
At cathode: $N{a^ + } + {e^ - } \to Na(s)$
At anode : $2C{l^ - } \to C{l_2}(g) + 2{e^ - }$
Total reaction Is $2N{a^ + } + 2C{l^ - } \to C{l_2}(g) + 2Na(s)$ .