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NEET CHEMISTRYStructure of AtomMedium

Question

If n = 6, the correct sequence for the filling of electrons will be:

A

ns → (n-1)d → (n-2)f → np

B

ns → (n-2)f → np → (n-1)d

C

ns → np → (n-1)d → (n-2)f

D

ns → (n-2)f → (n-1)d → np

Step-by-Step Solution

According to the Aufbau Principle, orbitals are filled in the order of increasing energy, which is determined by the (n+l)(n+l) rule. For the 6th period (n=6n=6), the orbitals involved are 6s6s, 4f4f, 5d5d, and 6p6p.

  1. Calculate (n+l)(n+l) values: ns→6s:6+0=6ns \rightarrow 6s: 6 + 0 = 6 (n−2)f→4f:4+3=7(n-2)f \rightarrow 4f: 4 + 3 = 7 (n−1)d→5d:5+2=7(n-1)d \rightarrow 5d: 5 + 2 = 7 np→6p:6+1=7np \rightarrow 6p: 6 + 1 = 7

  2. Determine Order: 6s6s has the lowest energy (lowest sum). For 4f,5d,4f, 5d, and 6p6p, the sums are equal (7). In such cases, the orbital with the lower nn value fills first.

  • Order: 4f(n=4)→5d(n=5)→6p(n=6)4f (n=4) \rightarrow 5d (n=5) \rightarrow 6p (n=6).
  1. Final Sequence: 6s→4f→5d→6p6s \rightarrow 4f \rightarrow 5d \rightarrow 6p Substituting n=6n=6: ns→(n−2)f→(n−1)d→npns \rightarrow (n-2)f \rightarrow (n-1)d \rightarrow np.

Exam Context & Concepts Covered

This question aligns with the NEET CHEMISTRY syllabus, specifically targeting concepts from Structure of Atom. Mastering this topic is crucial for scoring well in the upcoming medical entrance examinations. Solving conceptually related problems will help you understand the nuances of these concepts and improve your problem-solving speed.

CHEMISTRYStructure of Atomcorrectsequencefillingelectrons

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