Question

When ${H_2}S$ is passed through a mixture containing $C{u^{2 + }},N{i^{2 + }},Z{n^{2 + }}$ in acidic solution then ion will precipitate
(A) $C{u^{2 + }},N{i^{2 + }}$
(B) $N{i^{2 + }}$
(C) $C{u^{2 + }},Z{n^{2 + }}$
(D) $C{u^{2 + }}$

Answer 1

Hint: We know that, when ${H_2}S$ is in the presence of an acidified water solution it will precipitate into the form of sulphides. ${H_2}S$ Is a colourless gas as well as it is a very explosive and poisonous gas the hybridization of Hydrogen sulphide gas is $s{p^3}$ . Hydrogen sulphide can be separated from water as from the process named as aeration.

Complete Step by Step Solution:
As we know, when ${H_2}S$ gas is through any mixture containing the given ions there are many types of factors that can affect which ion will be precipitated.

We get that, the nature of ${H_2}S$ is very reactive from which we can say that, as per seeing all the ions, only $C{u^{2 + }}$ get bonded first as per the reactivity order.

From getting reacted it can be converted as a precipitate most likely as comparison from the other ions.

From getting all the conclusion we say that, $C{u^{2 + }}$ ion will precipitate when hydrogen sulphide is passed through a mixture containing the $C{u^{2 + }},N{i^{2 + }},Z{n^{2 + }}$ ions.
Hence, the correct option is (D).

Note: It is very important to know the reactivity order of any element from which we can easily classify the product of any reaction. As per talking about most reactive, the halogens are the most reactive elements in the periodic table as a formula is there to easily learn the reactivity order of elements the learning formula is F > O > N > C > H. Here all the terms stand for fluorine (F), oxygen (O) nitrogen (N), carbon (C) and hydrogen (H).