For hydrogen like species, which of the following graphs provides the most appropriate representation of E vs Z plot for a constant n? [E: Energy of the stationary state, Z: atomic number, n: principal quantum number]

Core Logic

Statement-I defines degenerate orbitals correctly: Orbitals that share the exact same energy level are called degenerate orbitals. Statement-II claims 3p and 3d are not degenerate in a hydrogen atom. For a single-electron system like Hydrogen (1s¹$1s^1$), the energy of an orbital depends only on the principal quantum number 'n$n$'.

Step 1: Final conclusion

In a hydrogen atom, energy of 3s = 3p = 3d$3s = 3p = 3d$. Therefore, 3p and 3d are degenerate. Thus, Statement-II is false.

Pattern Recognition

Hydrogen atom (single electron species) = Energy dictated strictly by n$n$. Multi-electron atoms = Energy dictated by (n + l)$(n + l)$ rule.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Core Logic

The atomic number of potassium (K) is 19$19$. Its electronic configuration is: 1s², 2s², 2p⁶, 3s², 3p⁶, 4s¹$1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1$. The outermost orbital of potassium is the 4s$4s$ orbital.

For the 4s$4s$ orbital: Principal quantum number, n = 4$n = 4$ Azimuthal quantum number, l = 0$l = 0$ (for s-subshell) Magnetic quantum number, m = 0$m = 0$ (since l = 0$l = 0$) Spin quantum number, s = ± (1)/(2)$s = \pm \frac{1}{2}$

Step 1: Final Matching

Comparing these values with the options, option (2) provides n = 4, l = 0, m = 0, s = +(1)/(2)$n = 4, l = 0, m = 0, s = +\frac{1}{2}$, which perfectly describes the outermost electron.

Chapter Mix

Class 11 Chemistry: Structure of Atom

Related Formula

Step 1: Energy per atom

Converting eV to Joules:

Step 2: Energy per mole

To find the ionization energy per mole, multiply by Avogadro's number (NA ≈ 6.022 × 10²³$N_A \approx 6.022 \times 10^{23}$):

Chapter Mix

Class 11 Chemistry: Structure of Atom