In a multielectron atom, which of the following orbitals described by three quantum numbers with have same energy in…

Year	2026	2025	2024	Total
Questions	14	15	9	38

Year

2026

2025

2024

Total

Questions

Solution & Explanation

Core Logic

In multi-electron systems, the energy profile depends on both the primary shell ( $n$ $n$) and azimuthal subshell ( $l$ $l$) quantum values via the $(n+l)$ $(n+l)$ rule. Orbitals sharing the exact same $n$ $n$ and $l$ $l$ values are energy-degenerate as long as external field metrics are zero. Let us map each set:

A: $1s$ $1s$
B: $2s$ $2s$
C: $2p$ $2p$
D: $3d$ $3d$ ( $ml = 1$ $m_l = 1$)
E: $3d$ $3d$ ( $ml = 0$ $m_l = 0$)

Since both D and E describe subcomponents of the same $3d$ $3d$ subshell ( $n=3, l=2$ $n=3, l=2$), they share identical energies, establishing perfect degeneracy.

Pattern Recognition

Sees: Orbitals matching energy level in multi-electron space. Trap: Confusing single-electron hydrogen atoms (where energy depends solely on $n$ $n$) with multielectron atoms.

Chapter Mix

Class 11 Chemistry: Structure of Atom

More Structure of Atom Previous-Year Questions — Page 8

Q68 jee_main_2024_30_jan_morning Quantum Mechanical Model

Given below are two statements: Statement-I: The orbitals having same energy are called as degenerate orbitals. Statement-II: In hydrogen atom, 3p and 3d orbitals are not degenerate orbitals. In the light of the above statements, choose the most appropriate answer from the options given

A. $Statement-I is true but Statement-II is false$
B. $Both Statement-I and Statement-II are true.$
C. $Both Statement-I and Statement-II are false$
D. $Statement-I is false but Statement-II is true$

Solution

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

Q jee_main_2024_31_jan_evening Quantum Numbers

The four quantum numbers for the electron in the outer most orbital of potassium (atomic no. 19) are

A. $n = 4, l = 2, m = -1, s = +(1)/(2)$
B. $n = 4, l = 0, m = 0, s = +(1)/(2)$
C. $n = 3, l = 0, m = 1, s = +(1)/(2)$
D. $n = 2, l = 0, m = 0, s = +(1)/(2)$

Solution

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, $s = \pm (1)/(2)$ $s = \pm \frac{1}{2}$

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

Q89 jee_main_2024_31_jan_morning Photoelectric Effect and Ionization Energy

The ionization energy of sodium in

kJ mol⁻¹

. If electromagnetic radiation of wavelength 242 nm is just sufficient to ionize sodium atom is ________

Numerical Answer. Answer: 493 to 495

Solution

Related Formula

E = 1240λ (nm) eV

Etotal = E \times NA

Step 1: Energy per atom

E = (1240)/(242) eV = 5.12 eV

Converting eV to Joules:

E = 5.12 \times 1.6 \times 10⁻¹⁹ J/atom

E = 8.192 \times 10⁻¹⁹ J/atom

Step 2: Energy per mole

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

IE = 8.192 \times 10⁻¹⁹ \times 6.022 \times 10²³ J/mol

IE \approx 493.3 \times 10³ J/mol \approx 494 kJ/mol

Chapter Mix

Class 11 Chemistry: Structure of Atom

Solution & Explanation

Core Logic

Pattern Recognition

Chapter Mix

Reference Study Guides

More Structure of Atom Previous-Year Questions — Page 8

Solution

Core Logic

Step 1: Final conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Matching

Chapter Mix

Solution

Related Formula

Step 1: Energy per atom

Step 2: Energy per mole

Chapter Mix

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Solution & Explanation

Core Logic

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Reference Study Guides

More Structure of Atom Previous-Year Questions — Page 8

Solution

Core Logic

Step 1: Final conclusion

Pattern Recognition

Chapter Mix

Solution

Core Logic

Step 1: Final Matching

Chapter Mix

Solution

Related Formula

Step 1: Energy per atom

Step 2: Energy per mole

Chapter Mix

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