Question

The radius of hydrogen atom in ground state is 0.53 angstrom. What will be the radius of $$L{i}^{2+}$$ in the ground state?
a.) 1.06 angstrom
b.) 0.265 angstrom
c.) 0.17 angstrom
d.) 0.53 angstrom

Answer 1

Hint: This question can be solved by considering the Bohr’s atomic model. We can solve the given question by keeping the following formula in mind:
$$r={\displaystyle \frac{n^2}{z}}{r}_H$$
Where r is the radius of the atom in consideration, n is the principal quantum number of the element, z is the atomic number of the element and $$r_H$$ is the radius of hydrogen atom.

Complete step by step answer:
The Bohr atomic model is only valid for hydrogen or hydrogen like single electronic species. As we can see that $$L{i}^{2+}$$is also a single electronic species we can apply the mentioned model to this ion.
$$r\propto {\displaystyle \frac{n^2}{z}}$$
$$r={\displaystyle \frac{n^2}{z}}{r}_H$$

As given in the question the $$L{i}^{2+}$$ is in ground state so n=1 , for $$L{i}^{2+}$$ atomic number= z= 3 and $$r_H$$ is 0.53, substituting the given values in the above equation we get,
$$r_{Li}={\displaystyle \frac{r_H}{3}}=0.17$$ angstrom
So, the correct answer is “Option C”.

Note: Niels Bohr first introduced the atomic Hydrogen model in 1913. He described the hydrogen atom as a positively charged nucleus, consisting of protons and neutrons and surrounded by a negatively charged electron cloud. The atom is held together by the strong electrostatic forces of attraction between the positive nucleus and negative surroundings. The model is only valid for hydrogen or hydrogen like single electronic species. A hydrogen-like species is any atomic nucleus bound to one electron and thus is isoelectronic with hydrogen. These atoms or ions can carry a positive charge.