Question

$A{l^{3 + }},F{e^{3 + }},Z{n^{2 + }}$ and $N{i^{2 + }}$ ions are present in an acidic solution. Excess ammonium chloride solution is added followed by addition of ammonium hydroxide solution. The available precipitate
(A) $Zn{(OH)_2}$ and $Ni{(OH)_2}$
(B) $Al{(OH)_3}$ and $Fe{(OH)_3}$
(C) $Zn{(OH)_2}$ and $Al{(OH)_3}$
(D) $Ni{(OH)_2}$ and $Fe{(OH)_3}$

Answer 1

Hint: There are given many types of ions which are present in a acidic solution as a mixture of ammonium chloride which are followed by an addition of ammonium hydroxide solution as in this we have to find the precipitate made in this reaction. As for doing this question firstly we have to know the nature and reactivity of every element from which we can easily solve this question.

Complete Step by Step Solution:
Firstly, we can classify all the radicals and denotes to which group,
Third group radicals $ = A{l^{3 + }},C{r^{3 + }},F{e^{3 + }}$
Fourth group radicals $ = Z{n^{2 + }},N{i^{2 + }},M{n^{2 + }},C{o^{2 + }}$
Fifth group radicals $ = C{a^{2 + }},B{a^{2 + }},S{r^{2 + }}$
Sixth group radicals $ = M{g^{ + 2}} \ and \ (N{a^ + },{K^ + })$ also included

As we know that,
Excess ammonium chloride + ammonium hydroxide, can form Precipitate,
And we also know that only Third group radicals can form the precipitate,
Now applying all the data to find the answer from the options we get,
As we can see that, $Al{(OH)_3}$ and $Fe{(OH)_3}$ both are present in third group radicals,
From which we can say that $Al{(OH)_3}$ and $Fe{(OH)_3}$ will contain as precipitate.
Hence, the correct option is (B).

Note: Due to common ion effects precipitate to be formed from third group radicals because the product of solubility of third group radicals is less as compared to the forth, fifth and sixth group of radicals. As according to the question that is asking about precipitate only but originally this is a radical based question.