Question

A hydrogen atom, unipositive helium, and dipositive lithium contain a single electron in the same shell. The radius of the shell:
A.Decreases
B.Increases
C.Remains unaffected
D.Cannot be predicted

Answer 1

Hint: For this question we must have the knowledge of the Bohr model for hydrogen and hydrogen-like species. We can calculate the radius of the shell by using the atomic number of their respective species in the formula.

Formula used: $${\text{R}}_{\text{n}}={\text{R}}_{\circ }\times {\displaystyle \frac{{\text{n}}^2}{\text{Z}}}$$ where $${\text{R}}_{\text{n}}$$ is radius of $${\text{R}}_{\text{n}}$$ orbital $${\text{R}}_{\text{n}}$$ is Bohr radius that is constant, n is number of orbit and s always a natural number and Z is atomic number of element.

Complete step by step solution:
We need to check the variation of radius for the given species. Let us calculate the radius of atom in each case:
The atomic number of hydrogen is 1. Hence $$\text{Z }=1$$ and we need to calculate for $$\text{n }=1$$. Putting the values in the formula we will get:
$${\text{R}}_{1(\text{H)}}={\text{R}}_{\circ }\times {\displaystyle \frac{1^2}{1}}={\text{R}}_{\circ }$$
The atomic number of Helium is 2. Helium ion is $$\text{H}{\text{e}}^{+}$$, but the atomic number remains the same whether it is an ion or atom of the same elements. Hence $$\text{Z }=2$$ and we need to calculate for $$\text{n }=1$$. Putting the values in the formula we will get:
$${\text{R}}_{1(\text{H}{\text{e}}^{+}\text{)}}={\text{R}}_{\circ }\times {\displaystyle \frac{1^2}{2}}={\displaystyle \frac{{\text{R}}_{\circ }}{2}}$$
The atomic number of Lithium is 3. Dipositive Lithium ion is $$\text{L}{\text{i}}^{2+}$$, but the atomic number remains same whether it is an ion or atom of same elements. Hence $$\text{Z }=3$$ and we need to calculate for $$\text{n }=1$$. Putting the values in the formula we will get:
$${\text{R}}_{1(\text{L}{\text{i}}^{+2}\text{)}}={\text{R}}_{\circ }\times {\displaystyle \frac{1^2}{3}}={\displaystyle \frac{{\text{R}}_{\circ }}{3}}$$
Hence the radius of $$\text{H, H}{\text{e}}^{+}\text{and L}{\text{i}}^{+2}$$ is $${\text{R}}_{\text{o}}\text{, }{\displaystyle \frac{{\text{R}}_{\text{o}}}{2}},{\displaystyle \frac{{\text{R}}_{\text{o}}}{3}}$$ respectively. Hence, we can clearly see that radius decreases.
The correct option is A.

Note: The given species $$\text{H, H}{\text{e}}^{+}\text{and L}{\text{i}}^{+2}$$ are iso-electronic in nature. That is all these species have same number of electron though have different atomic number. Hydrogen has 1 electron and 1 proton. Helium has atomic number 2 and have 2 electrons but in $$\text{H}{\text{e}}^{+}$$ has 1 less electron due to positive charge that makes 1 electron in $$\text{H}{\text{e}}^{+}$$ and similarly 1 electron is there in $$\text{L}{\text{i}}^{2+}$$ sue to 2 positive charge.