Question

Arrange the following sets of orbits and species in the increasing order of the energies.

(A) H atom, n=2
(B) He⁺ ion, n=2
(C) Li⁺² ion, n=2
(D) Be⁺³ ion, n=2

• A. ADCB
• B. DCBA
• C. ABCD
• D. DBAC

Answer 1

The correct option is B

DCBA

The energy of nth orbit-

According to Bohr's model, the energy of the n^th orbit is-

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

Where Z is the nuclear charge (+1 for H, +2 for He)

The explanation of the correct option: (b)

A) H atom, n=2 : Now, Z = 1 and n = 2, thus the energy of the 2nd orbit is-

$$\begin{gathered} {\mathrm{E}}_2=-13.6\times \frac{1^2}{2^2}\mathrm{eV}=\frac{-13.6\mathrm{eV}}{4} \\ \therefore {\mathrm{E}}_2=-3.4eV \end{gathered}$$

B) He⁺ ion, n=2: For He⁺ , Z = 2 and n= 2, thus the energy of the 2nd orbit is-

$$\begin{gathered} {\mathrm{E}}_2=-13.6\times \frac{2^2}{2^2}\mathrm{eV} \\ \therefore {\mathrm{E}}_2=-13.6eV \end{gathered}$$

C) Li⁺² ion, n=2: For Li⁺² , Z = 3 and n= 2, thus the energy of the 2nd orbit is-

$$\begin{gathered} {\mathrm{E}}_2=-13.6\times \frac{3^2}{2^2}\mathrm{eV}=-13.6\times \frac{9}{4}\mathrm{eV} \\ \therefore {\mathrm{E}}_2=-20.4\mathrm{eV} \end{gathered}$$

D) Be⁺³ ion, n=2: For Be⁺³, Z = 4 and n= 2, thus the energy of the 2nd orbit is-

$$\begin{gathered} {\mathrm{E}}_2=-13.6\times \frac{4^2}{2^2}\mathrm{eV}=-13.6\times \frac{16}{4}\mathrm{eV} \\ \therefore {\mathrm{E}}_2=-54.4\mathrm{eV} \end{gathered}$$

2. Hence, the increasing order of energies is -D< C< B <A [More negative value means smaller energy]

The explanation for the incorrect option:

Option (a)

The energy of A is the largest, so it cannot come as the first alphabet.

Option (c)

The energy of A is the largest, so it cannot comes as the first alphabet. It is the reverse of the actual order.

Option (d)

The energy of D is the smallest but after this energy of B is still greater than the energy of C. So there should be no B and A after D.

The correct option is (b).