Question

Write chemical name of the following compound: $ {K_2}C{O_3} $

Answer 1

Hint: Potassium belongs to the Alkali metal metals family. It is the first group in the periodic table. The carbonates of the first group are quite stable except the carbonate of lithium. The first member of each group has some different behaviour.

Complete solution
The elements of group-1 are all metals. They are excellent conductors of electricity, heat. They are mostly soft in nature and known to be very reactive also.
The elements of group-1 belong to the s-block. This is because in the electronic configuration of these elements the last electron enters in the s- orbitals. Also as the group number is one, they have one electron in the s- orbital.
Like that we know that potassium belongs to the alkali metal family, i.e. group-1. The electronic configuration of potassium can be represented as ( $ K,Z = 19 $ )-
 $ 1{s^2}2{s^2}2{p^6}3{s^2}3{p^6}4{s^1} $
As the last electron enters in the $ 4s $ orbital the value of principal quantum number is four. Hence the period number of potassium is four.
The cohesive energy of group-1 metals are very low, thus the metals are not hard. Potassium and sodium metals are so soft they can be easily cut with a knife. They are also very reactive metals and for that they are kept in kerosene oils to avoid contact with air.
As we know that group-1 metals are electropositive in nature, they form very strong bases. The carbonates of group – 1 are very stable and they melt before their decomposition temperature.
Potassium forms both carbonate and bicarbonate salt which are quite stable.
 $ {K_2}C{O_3} $ Is the carbonate of potassium where the oxidation state of potassium is $ + 1 $ . The compound is chemically known as Potassium carbonate.

Note:
Lithium being the first member of group-1 has some anomalous behaviour. The carbonate of lithium is very unstable as on heating the compound gently it decomposes to lithium oxide and carbon dioxide. The reaction will be $ LiC{O_3}{ \to ^\Delta }LiO + C{O_2} \uparrow $ .