For electron in '2s' and '2p' orbitals, the orbital angular momentum values, respectively are :

Solution & Explanation

### Related Formula Orbital angular momentum (L) of an electron is determined exclusively by its azimuthal quantum number l: L = sqrtl(l+1) frach2pi where values of l are: - For s-orbitals: l=0 - For p-orbitals: l=1 - For d-orbitals: l=2 ### Core Logic Evaluate for both specified orbitals: - For '2s' orbital (l=0): L = sqrt0(0+1) frach2pi = 0 - For '2p' orbital (l=1): L = sqrt1(1+1) frach2pi = sqrt2 frach2pi ### Step 1: Match the values The orbital angular momentum values are 0 and sqrt2frach2pi respectively, corresponding to Option (4). ### Pattern Recognition Always note that orbital angular momentum does *not* depend on the principal quantum number n (the '2' in 2s and 2p is irrelevant). An electron in a 1s, 2s, or 3s orbital always has L=0. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

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Structure Of Atom

More Structure of Atom Previous-Year Questions — Page 5

Q66 jee_main_2024_29_jan_morning Quantum Numbers
The correct set of four quantum numbers for the valence electron of rubidium atom (Z = 37) is:
  • A. 5, 0, 0, +frac12
  • B. 5, 0, 1, +frac12
  • C. 5, 1, 0, +frac12
  • D. 5, 1, 1, +frac12

Solution

### Core Logic Rubidium (Rb) has the atomic number Z = 37. The noble gas core preceding it is Krypton (Kr, Z = 36). The electronic configuration is: Rb rightarrow [Kr] 5s^1 Thus, the valence electron resides in the 5s orbital. ### Step 1: Assigning Quantum Numbers For a 5s electron: - Principal quantum number, n = 5 (from the shell number). - Azimuthal quantum number, l = 0 (for an s-orbital). - Magnetic quantum number, m = 0 (since m ranges from -l to +l, and l=0). - Spin quantum number, s = +frac12 or -frac12. The correct set matching this is 5, 0, 0, +frac12. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom
Q90 jee_main_2024_30_january_evening Atomic Spectra
Number of spectral lines obtained in mathrmHe^+ spectra, when an electron makes transition from fifth excited state to first excited state will be
Numerical Answer. Answer: 10 to 10

Solution

### Related Formula textNumber of spectral lines = fracDelta n (Delta n + 1)2 where Delta n = n_2 - n_1 ### Core Logic Identify the principal quantum numbers (n) for the given states: - Fifth excited state means n_2 = 5 + 1 = 6. - First excited state means n_1 = 1 + 1 = 2. Calculate the difference: Delta n = n_2 - n_1 = 6 - 2 = 4 ### Step 1: Calculate Total Spectral Lines Substitute Delta n = 4 into the formula: textMaximum number of spectral lines = frac4(4 + 1)2 = frac4 times 52 = 10 ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom
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. textStatement-I is true but Statement-II is false
  • B. textBoth Statement-I and Statement-II are true.
  • C. textBoth Statement-I and Statement-II are false
  • D. textStatement-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 (1s^1), the energy of an orbital depends *only* on the principal quantum number 'n'. ### Step 1: Final conclusion In a hydrogen atom, energy of 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. Multi-electron atoms = Energy dictated by (n + l) rule. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom
Q77 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. text(1) n = 4, l = 2, m = -1, s = +frac12
  • B. text(2) n = 4, l = 0, m = 0, s = +frac12
  • C. text(3) n = 3, l = 0, m = 1, s = +frac12
  • D. text(4) n = 2, l = 0, m = 0, s = +frac12

Solution

### Core Logic The atomic number of potassium (K) is 19. Its electronic configuration is: 1s^2, 2s^2, 2p^6, 3s^2, 3p^6, 4s^1. The outermost orbital of potassium is the 4s orbital. For the 4s orbital: Principal quantum number, n = 4 Azimuthal quantum number, l = 0 (for s-subshell) Magnetic quantum number, m = 0 (since l = 0) Spin quantum number, s = pm frac12 ### Step 1: Final Matching Comparing these values with the options, option (2) provides n = 4, l = 0, m = 0, s = +frac12, which perfectly describes the outermost electron. ### Evaluation Rubric / Model Answer null ### 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 textkJ 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 = frac1240lambda (textnm) text eV E_texttotal = E times N_A ### Step 1: Energy per atom E = frac1240242 text eV = 5.12 text eV Converting eV to Joules: E = 5.12 times 1.6 times 10^-19 text J/atom E = 8.192 times 10^-19 text J/atom ### Step 2: Energy per mole To find the ionization energy per mole, multiply by Avogadro's number (N_A approx 6.022 times 10^23): IE = 8.192 times 10^-19 times 6.022 times 10^23 text J/mol IE approx 493.3 times 10^3 text J/mol approx 494 text kJ/mol ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom

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