Question

The radius of hydrogen atom in the ground state is 0.53 $\mathring{A}.$ The radius of $Li^{2+}$ ion (at. no. = 3) in a similar state is

• A. $0.17 \mathring{A}$
• B. $0.53 \mathring{A}$
• C. $0.265 \mathring{A}$
• D. $1.06 \mathring{A}$

Answer 1

The Correct Option is A

To determine the radius of the $Li^{2+}$ ion in its ground state, we need to compare it with the Bohr model of the hydrogen atom, which can be applied to hydrogen-like ions. The formula for the radius of an electron orbit in a hydrogen-like atom is given by:

r_n = \frac{n^2 \cdot a_0}{Z}

where:

• r_n is the radius of the nth orbit.
• a_0 is the Bohr radius, which is approximately 0.53 \mathring{A}.
• n is the principal quantum number.
• Z is the atomic number of the element.

For a ground state electron (n=1) in a hydrogen-like ion:

For hydrogen (H, with atomic number 1):

r_1^{\text{H}} = \frac{1^2 \cdot 0.53 \mathring{A}}{1} = 0.53 \mathring{A}

For Li^{2+} (with atomic number 3):

r_1^{\text{Li^{2+}}} = \frac{1^2 \cdot 0.53 \mathring{A}}{3} = \frac{0.53 \mathring{A}}{3} = 0.177 \mathring{A}

Upon rounding 0.177 \mathring{A}, we arrive at the nearest option, which is approximately 0.17 \mathring{A}.

Thus, the correct answer is $0.17 \mathring{A}$, which corresponds to option 1.