Question

The ratio of atomic numbers of two elements ‘ P’ and ' Q ' is 2:7. The energy of the fourth orbit of single-electron species of P is $-13.6\mathrm{e}/\mathrm{v}.$ Calculate the difference in the number of electrons present in the valence and penultimate shells of Q.

• A. 2
• B. 4
• C. 3
• D. 1

Answer 1

The correct option is B

4

The explanation for the correct option:

1. Let the atomic number of P and Q be 2x and 7x as their ratio is 2:7.
2. According to Bohr's model, the energy of the n^th orbit is-

${\mathrm{E}}_{\mathrm{n}}=-13.6\times \frac{{\mathrm{Z}}^2}{{\mathrm{n}}^2}\mathrm{eV}$

3. Now, the Atomic number of P, Z = 2x and n = 4, and the energy of the 4th orbit = -13.6 eV.

Thus,

$$\begin{gathered} -13.6\mathrm{eV}=-13.6\times \frac{{\left(2\mathrm{x}\right)}^2}{4^2}\mathrm{eV} \\ {\left(2\mathrm{x}\right)}^2=16\mathrm{Taking}\mathrm{square}\mathrm{root}\mathrm{of}\mathrm{both}\mathrm{the}\mathrm{sides}, \\ 2\mathrm{x}=4 \\ \therefore \mathrm{x}=2 \end{gathered}$$

4. Hence, the atomic number of P = 2x = 4, and atomic number of Q = 7x = 14.

5. Now, the electronic configuration of Q is written as-

$$\begin{gathered} \mathrm{K}\mathrm{L}\mathrm{M}\mathrm{N} \\ 284 \end{gathered}$$

6. In Q, the valence shell M has 4 electrons and the penultimate shell L has 8 electrons.

7. Hence, the difference in the number of electrons present in the valence and penultimate shells of Q = 8-4 = 4 electrons.

The correct option is b).