Question

Hydrogen has three isotopes, the number of possible diatomic molecules will be
A. $2$
B. $6$
C. $9$
D. $12$

Answer 1

Hint: Hydrogen is the simplest chemical element with the symbol H. Naturally this element contains three stable isotopes named Protium, deuterium, and tritium. Here we have to place these elements together in such a way that more than one possible diatomic molecule can form.

Complete Step by Step Answer:
Hydrogen is the first and smallest element in the periodic table. It has three isotopes and each isotope is made up of only one proton; they only differ in the number of neutrons. There are three isotopes named Protium ($^{1}H$), Deuterium ($^{2}H$), and Tritium ($^{3}H$). Here the extra neutrons make the isotope Protium heavier. Deuterium is two times heavy and Tritium is three times heavier than Protium.
Hydrogen has three isotopes and here we have to find out the number of possible diatomic molecules.

The diatomic molecules which may be formed can be written in the following ways,
Selecting two Protium, then we get the diatomic molecule, $^{1}{{H}^{1}}H$
Selecting two deuteriums, then we get the diatomic molecule, $^{2}{{H}^{2}}H$
Selecting two Tritium, then we get the diatomic molecule, $^{3}{{H}^{3}}H$

Now taking one $^{1}H$ and one $^{2}H$ , then we get, $^{1}{{H}^{2}}H$ or $^{2}{{H}^{1}}H$
Taking one $^{1}H$ and one $^{3}H$ , then we get, $^{1}{{H}^{3}}H$ or $^{3}{{H}^{1}}H$
Finally taking one $^{2}H$ and $^{3}H$, then we get $^{2}{{H}^{3}}H$or $^{3}{{H}^{2}}H$ .
Therefore the number of possible diatomic molecules are $6$ and these are $^{1}{{H}^{1}}H{{,}^{2}}{{H}^{2}}H{{,}^{3}}{{H}^{3}}H{{,}^{1}}{{H}^{2}}H{{,}^{1}}{{H}^{3}}H,$and $^{2}{{H}^{3}}H$ .
Thus, option (B) is correct.

Note: To approach this type of question, we must be careful when possible arrangements are done. For example $^{1}{{H}^{2}}H$ and $^{2}{{H}^{1}}H$ are the same, here we have to consider them as one molecule instead of two molecules of hydrogen. Taking care of these facts, resist making calculation mistakes.