Exceptional Configurations — Cr, Cu
Atomic Structure
9
JEE Qs
8%
Hard
20
min
Thoroughly understand the stability rationale behind these exceptions to confidently apply them and avoid simple memorization errors, especially when dealing with ions.
🧮 Key Formulas
✅ Key Points for JEE
- 1The Aufbau principle, while generally accurate, has exceptions due to the extra stability associated with half-filled and completely filled degenerate orbitals (like d and f orbitals).
- 2For Chromium (Cr, Z=24), instead of the expected [Ar] 3d⁴ 4s², one electron from the 4s orbital shifts to the 3d orbital, resulting in [Ar] 3d⁵ 4s¹. This provides a stable half-filled 3d orbital and a half-filled 4s orbital.
- 3For Copper (Cu, Z=29), instead of the expected [Ar] 3d⁹ 4s², one electron from the 4s orbital shifts to the 3d orbital, resulting in [Ar] 3d¹⁰ 4s¹. This provides a highly stable completely filled 3d orbital.
- 4The energy difference between 3d and 4s orbitals is very small for elements in the first transition series, making these electron shifts energetically favorable.
- 5These exceptional configurations are crucial for predicting the chemical properties, oxidation states, and magnetic behavior of these elements.
⚠️ Common Mistakes
- ✕Incorrectly writing the electron configurations of Cr and Cu based solely on the Aufbau principle without considering the stability of half-filled/fully-filled orbitals.
- ✕Applying the 'exception' rule indiscriminately to other elements where it doesn't apply or where other factors dominate.
- ✕Forgetting that even though 4s is filled before 3d according to Aufbau, electrons are *removed* from the 4s orbital first when forming ions of transition metals.
📝 Practice Questions
See allQ57.Given below are two statements : Statement (I) : For a given shell, the total number of allowed orbitals is given by n2 . Statement (II) : For any subshell, the spatial orientation of the orbitals is given by −l to +l values including zero. In the light of the above statements, choose the correct answer from the options given below : (1) Both Statement I and Statement II are false (2) Statement I is true but Statement II is false (3) Both Statement I and Statement II are true (4) Statement I is false but Statement II is true
Q57.In a multielectron atom, which of the following orbitals described by three quantum numbers will have same energy in absence of electric and magnetic fields? A. n = 1, l = 0, m1 = 0 B. n = 2, l = 0, m1 = 0 C. n = 2, l = 1, m1 = 1 D. n = 3, l = 2, m1 = 1 E. n = 3, l = 2, m1 = 0 Choose the correct answer from the options given below: (1) B and C Only (2) A and B Only (3) C and D Only (4) D and E Only
Q51.Radius of the first excited state of Helium ion is given as : a0 → radius of first stationary state of hydrogen atom. (1) r = 4a0 (2) r = 2a0 (3) r = a02 (4) r = a04
Q62.Given below are two statements : Statement (I) : A spectral line will be observed for a 2px →2py transition. Statement (II) : 2Px and 2py are degenerate orbitals. In the light of the above statements, choose the correct answer from the options given below : (1) Both Statement I and Statement II are true (2) Statement I is false but Statement II is true (3) Both Statement I and Statement II are false (4) Statement I is true but Statement II is false
Q51.For hydrogen atom, the orbital/s with lowest energy is/are : (A) 4 s (B) 3px (C) 3 dx2−y2 (D) 3 dz2 (E) 4pz Choose the correct answer from the options given below : (1) (B), (C) and (D) only (2) (A) and (E) only (3) (A) only (4) (B) only
Q44.The frequency of revolution of the electron in Bohr's orbit varies with n, the principal quantum number as (1) 1 (2) 1 n4 n2 (3) 1 (4) 1 n n3
NCERT Chapters
- Class 11 Chemistry Ch 2: Structure of Atom