Chemistry · Periodic Classification Of Properties · NEET
Because filling order depends on total ENERGY, not just the shell number n. Energy is compared using the (n + l) rule. For 4s: n=4, l=0, so n+l = 4. For 3d: n=3, l=2, so n+l = 5. The orbital with the smaller (n + l) value has lower energy and fills first. So 4s (=4) fills before 3d (=5), even though the shell number 4 looks bigger than 3.
For each orbital, add the principal quantum number n and the azimuthal quantum number l (s=0, p=1, d=2, f=3). Fill the orbital with the LOWER (n + l) first. If two orbitals have the same (n + l), fill the one with the smaller n first. 4s = 4 + 0 = 4. 3d = 3 + 2 = 5. Since 4 is less than 5, 4s fills first. This is why the 3d transition series appears in period 4, after 4s is already filled at K and Ca.
This is the part most students miss. Once 3d starts to fill, the 3d electrons pull closer to the nucleus and 3d actually drops slightly BELOW 4s in energy. So in a filled transition atom, 4s is now the higher, outer orbital. When you form an ion (like Fe to Fe2+), the outermost, highest-energy electrons leave first — that is the 4s electrons. Rule to remember: fill 4s first, but remove 4s first.
Scandium (Z = 21) fills 4s completely first because 4s is lower in energy for the neutral atom. Potassium and calcium already filled 4s to 4s2 (that is why period 4 starts with them). Then the 21st electron goes into 3d, giving [Ar] 3d1 4s2. It is NOT 3d3, because 4s must be filled before 3d gets three electrons. This filling pattern is exactly what NCERT states: 'the fourth period starts at potassium, the added electrons fill up the 4s orbital, and before 4p is filled, filling up of 3d becomes energetically favourable.'
It starts at scandium, Sc (Z = 21), with configuration [Ar] 3d1 4s2, and ends at zinc, Zn (Z = 30), with [Ar] 3d10 4s2. Across these 10 elements the five 3d orbitals fill up. After 3d is full, the 4p orbitals fill (Ga to Kr) and period 4 ends at krypton. Remember: 3d holds a maximum of 10 electrons, which is why the first transition series has exactly 10 elements.
Yes. Chromium is [Ar] 3d5 4s1 (not 3d4 4s2) and copper is [Ar] 3d10 4s1 (not 3d9 4s2). One 4s electron shifts into 3d because a half-filled (d5) or completely filled (d10) 3d set is extra stable (symmetry + exchange energy). The general 4s-before-3d rule still holds; Cr and Cu are just the two famous exceptions NEET loves to ask.
Which among the following electronic configurations belong to main group elements? A. [Ne]3s1 B. [Ar]3d3 4s2 C. [Kr]4d10 5s2 5p5 D. [Ar]3d10 4s1 E. [Rn]5f0 6d2 7s2
Try the real previous-year questions from this chapter — each with the answer and a full solution.
No. For empty (neutral) atoms before 3d starts filling, 4s is lower, so it fills first. But once 3d is occupied, 3d drops below 4s, which is why 4s electrons are removed first in ions like Fe2+ and Ti3+.
Scandium, Sc (Z = 21), is [Ar] 3d1 4s2. It is the first element of the 3d transition series.
Because 3d fills only after 4s, which belongs to period 4 (K and Ca). By the (n + l) rule 4s (=4) fills before 3d (=5), so the 3d block physically appears in the fourth period.
Ten, from scandium (Z = 21) to zinc (Z = 30), because the five 3d orbitals hold a maximum of 10 electrons.
Standard NCERT/NEET practice writes 3d before 4s (for example [Ar] 3d1 4s2), grouping electrons by shell number. But remember 4s was actually filled first and is removed first.
