Question

Hydrogen does not combine with:
A.Sb
B.Na
C.He
D.Zn

Answer 1

Hint: For this question, we need to check the electronic configuration of the elements given in the options. The one with the most stable electronic configuration will not react with hydrogen. So just write the electronic configuration of the element and check which has fully filled valence shell configuration, that element will not react with hydrogen.

 Complete step by step answer:
We will start the answer by writing the electronic configuration of each of the elements even in the question.
First of all, in option A, we have been given Antimony (Sb). The Electronic configuration of Sb is:
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{s}}^{\text{2}}}{\text{4}}{{\text{p}}^{\text{6}}}{\text{4}}{{\text{d}}^{{\text{10}}}}{\text{5}}{{\text{s}}^{\text{6}}}{\text{5}}{{\text{p}}^{\text{3}}}$
The valence shell is not fully filled, because in p-orbital 6 electrons can accommodate but only 3 electrons are there in Sb. So, Sb can react with hydrogen.
In option B, Sodium (Na) is there. So, writing the electronic configuration of sodium as:
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^1}$
s-orbital can accommodate 2 electrons but sodium has only 1 electron in the valence shell. So, it will also react with Hydrogen.
Moreover, sodium is a well known alkali metal, having one electron in the valence shell and is very reactive because after losing this electron Na will acquire a stable fully filled electronic configuration. Hence, Na readily reacts with Hydrogen because it wants to lose the electron to get stable.
Now, coming to option C, Helium (He) is given to us. Again, writing electronic configuration for helium:
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}$
Clearly, we can see in helium, there are two electrons in s orbital and s orbital can have a maximum of two electrons that means Helium has a fully filled valence shell electronic configuration. So, helium is very stable and it does not want to react with hydrogen.
Also, Helium is a noble gas or inert gas which does not react with other elements and so it would not react with Hydrogen.
Let us check the last option that is option D which is Zinc (Zn) writing the electronic configuration of zinc,
${\text{1}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{s}}^{\text{2}}}{\text{2}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{s}}^{\text{2}}}{\text{3}}{{\text{p}}^{\text{6}}}{\text{3}}{{\text{d}}^{{\text{10}}}}{\text{4}}{{\text{s}}^{\text{2}}}$
No doubt zinc has a fully filled electronic configuration but being metal zinc reacts with hydrogen and forms zinc hydride but the zinc hydride is less stable and readily decomposes at room temperature.
From the above discussion, it is clear that Hydrogen does not react with Helium.

So, the correct answer is Option C .

Note:
To save the time and to get the answer quickly, have a look at the options carefully as you can see in options A, B, and D, all are metals but in C, there is helium which is an inert gas. So, option C is the odd one among the given options and hence, it will be the answer. The zinc hydride is unstable at room temperature and breaks down into zinc and hydrogen, but we will consider it because it at least forms a compound with hydrogen, helium does not even form any compound because of its inert nature.