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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

Solution & Explanation

### 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

Reference Study Guides

Structure Of Atom

More Structure of Atom Previous-Year Questions — Page 4

Q jee_main_2025_29_jan_morning Bohr's Model and de-Broglie Wavelength
If a_0 is denoted as the Bohr radius of hydrogen atom, then what is the de-Broglie wavelength ( lambda ) of the electron present in the second orbit of hydrogen atom? [n: any integer]
  • A. frac2mathrma_0mathrmnpi
  • B. frac8pi a_0n
  • C. frac4 pi a0n
  • D. frac4mathrmnpimathrma0

Solution

### Related Formula 2pi r_n = nlambda r_n = a_0 cdot n^2 ### Core Logic According to Bohr's quantization postulate condition coupled with de-Broglie's hypothesis : 2pi r_n = nlambda For the second orbit (n = 2), the radius is: r_2 = a_0 cdot (2)^2 = 4a_0 Substituting into the wave perimeter formula : 2pi (4a_0) = nlambda lambda = frac8pi a_0n This strictly matches option (2). ### Pattern Recognition The circumference of an electron's orbit must encompass an exact integral count of complete standing de-Broglie wave wavelengths. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom
Q66 jee_main_2024_01_february_morning Dual Behaviour of Matter
According to the wave-particle duality of matter by de-Broglie, which of the following graph plot presents most appropriate relationship between wavelength of electron (lambda) and momentum of electron (p)?
  • A. Graph 1
  • B. Graph 2
  • C. Graph 3
  • D. Graph 4

Solution

### Related Formula lambda = frachp ### Core Logic From the de-Broglie equation: lambda propto frac1p Rightarrow lambda p = h text (constant) This represents the equation of a rectangular hyperbola (xy = c). ### Step 1: Graph Identification The plot of lambda versus p will be a rectangular hyperbola curve in the first quadrant. Graph 1 correctly depicts this hyperbolic relationship.
Dual Behaviour of Matter diagram for Q66 - JEE Main 2024 Morning
Dual Behaviour of Matter diagram for Q66 - JEE Main 2024 Morning
### Pattern Recognition Inverse proportionality y = frackx always graphs as a rectangular hyperbola in the positive quadrant. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Structure of Atom
Q65 jee_main_2024_27_jan_morning Electronic Configuration and Magnetic Moment
Which of the following electronic configuration would be associated with the highest magnetic moment?
  • A. [Ar] 3d^7
  • B. [Ar] 3d^8
  • C. [Ar] 3d^3
  • D. [Ar] 3d^6

Solution

### Related Formula Spin-only magnetic moment formula: mu = sqrtn(n+2)text BM where n is the number of unpaired electrons. ### Step 1: Audit configurations and count unpaired electrons
ConfigurationUnpaired e^- (n)Magnetic Moment (BM)
[Ar] 3d^73sqrt15
[Ar] 3d^82sqrt8
[Ar] 3d^33sqrt15
[Ar] 3d^64sqrt24
### Step 2: Conclusion Since [Ar] 3d^6 contains the maximal count of 4 unpaired electrons, it yields the highest spin-only magnetic moment value. ### Pattern Recognition Maximal unpaired configuration in high-spin 3d^n series yields the highest mu. Check n systematically. ### Chapter Mix Class 11 Chemistry: Structure of Atom Class 12 Chemistry: d-and f-Block Elements
Q88 jee_main_2024_27_jan_morning Quantum Numbers and Electron Capacity
The number of electrons present in all the completely filled subshells having n=4 and s=+frac12 is textquadquad. (Where n= principal quantum number and s= spin quantum number)
Numerical Answer. Answer: 16 to 16

Solution

### Step 1: Identify all available subshells within the n=4 energy shell For principal quantum level n=4, the allowed values of azimuthal quantum numbers (l) are: - 4texts (l=0) rightarrow 1 text orbital rightarrow 2 text electrons capacity - 4textp (l=1) rightarrow 3 text orbitals rightarrow 6 text electrons capacity - 4textd (l=2) rightarrow 5 text orbitals rightarrow 10 text electrons capacity - 4textf (l=3) rightarrow 7 text orbitals rightarrow 14 text electrons capacity ### Step 2: Filter capacity using spin values Every single spatial orbital holds exactly 2 electrons maximum; one with spin s=+frac12 and one with spin s=-frac12. Total number of orbitals across n=4 is: 1 + 3 + 5 + 7 = 16text orbitals Thus, the total count of electrons featuring spin value s=+frac12 across these completely filled configurations is exactly 16. ### Pattern Recognition Total orbitals in shell n is n^2. Since each orbital contributes exactly 1 electron with s=+frac12, capacity is simply n^2 = 4^2 = 16. ### Chapter Mix Class 11 Chemistry: Structure of Atom
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

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