Question

In the fifth group, ${(N{H_4})_2}C{O_3}$ is added to precipitate out the carbonates. We do not add $N{a_2}C{O_3}$ because
(A) $CaC{O_3}$ is soluble in $N{a_2}C{O_3}$
(B) $N{a_2}C{O_3}$ increases the solubility of fifth group carbonates
(C) $MgC{O_3}$ will be precipitated out in fifth group
(D) None of these

Answer 1

Hint: As we know that when we add ${(N{H_4})_2}C{O_3}$ with fifth group element of periodic table then the carbonates are got precipitated out but when we add $N{a_2}C{O_3}$ with fifth group elements then it gives a different precipitation which we are going to read below because every compound have different nature of reaction from this when react with different compound it gives a different output every time.

Complete Step by Step Solution:
As we know from the further study we did , when we add $N{a_2}C{O_3}$ then magnesium carbonate will be precipitated out from the solution.

That is the reason why we add ammonium carbonate in the place of sodium carbonate.
And we also know that sodium carbonate is a very strong electrolyte that can easily precipitate magnesium carbonate,

As comparison between ${(N{H_4})_2}C{O_3}$ and $N{a_2}C{O_3}$ we most of the time choose ${(N{H_4})_2}C{O_3}$ because it is a weak electrolyte.
As by taking all the conclusion above we get that why we don’t use $N{a_2}C{O_3}$ .
Hence, the correct option is (C).

Note: When we talk about the group system of periodic table so it is very necessary to know the format of left to right of elements and up to down of element because periodic table is based on many formats in which major ones are atomic number, atomic mass number and electrons present in the element etc. are the major parts which we must have to know in periodic table.