Chemistry · Structure Of Atom · NEET
Fill in this energy order: 1s, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p. This is the Aufbau order. It is NOT the same as counting shells 1, 2, 3, 4 in order, because energies overlap. For NEET, either memorise this list or use the n+l rule to build it yourself.
Because filling order follows ENERGY, not just the shell number n. Use the n+l rule: lower (n+l) means lower energy and fills first. For 4s, n+l = 4+0 = 4. For 3d, n+l = 3+2 = 5. So 4s (sum 4) has lower energy and fills before 3d (sum 5). This is why potassium (Z=19) is [Ar]4s1, not [Ar]3d1.
The n+l rule says: the orbital with the lower value of (n+l) has lower energy and fills first. If two orbitals have the SAME (n+l) value, the one with the smaller n fills first. Example: 4p (n+l = 5) and 3d (n+l = 5) are tied, so 3d (n=3) fills before 4p (n=4). This rule is also called Bohr-Bury or Madelung rule and is heavily tested in NEET.
s holds 2, p holds 6, d holds 10, f holds 14. This comes from orbitals: s has 1 orbital, p has 3, d has 5, f has 7, and each orbital holds 2 electrons. So s = 1x2 = 2, p = 3x2 = 6, d = 5x2 = 10, f = 7x2 = 14.
Find the noble gas that comes just before your element in the periodic table. Write it in a square bracket, then add only the extra electrons. Example: iron (Z=26). The noble gas before it is argon (Z=18). So Fe = [Ar] 3d6 4s2. This saves time and is what NEET options usually use, like [Xe]4f7 6s2 for europium.
For Cr (Z=24), the expected [Ar]3d4 4s2 shifts to [Ar]3d5 4s1, and for Cu (Z=29), expected [Ar]3d9 4s2 shifts to [Ar]3d10 4s1. One 4s electron jumps to 3d because a half-filled (d5) or fully-filled (d10) d-subshell is extra stable. This exception is covered in detail in the next concept.
When you FILL, 4s comes before 3d. But when you WRITE the final answer, many books group by shell, so you may see 3d written before 4s (e.g. [Ar]3d6 4s2). Both orders mean the same electrons. NEET accepts either, but writing 3d before 4s helps when you remove electrons to form ions, because the 4s electrons leave first.
The electronic configurations of Eu (Atomic no. 63), Gd (Atomic no. 64) and Tb (Atomic no. 65), respectively, are:
Magnesium reacts with an element X to form an ionic compound. If the ground-state electronic configuration of X is 1s2 2s2 2p3, the simplest formula for this compound is:
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.
You FILL by energy (Aufbau order, using the n+l rule), but you may WRITE the final answer grouped by shell. For example iron fills 4s before 3d, but is often written [Ar]3d6 4s2. The electrons are the same either way.
First write the neutral atom. For a positive ion (cation), remove electrons from the HIGHEST n shell first (for transition metals, remove 4s before 3d). For a negative ion (anion), add electrons in normal Aufbau order. Example: Fe is [Ar]3d6 4s2, so Fe2+ is [Ar]3d6 (4s electrons leave first).
They mean the same total, but the second form shows Hund's rule: the 3 electrons go into separate p orbitals with parallel spins before any orbital gets a second electron. NEET 2017 tested that writing nitrogen as 2p3 is fine, but the fully expanded form shows one electron in each of px, py, pz.
Yes, memorise up to about 7p, or learn the diagonal n+l diagram so you can rebuild it fast. Most NEET questions on this topic are direct: given Z, write the configuration, or spot which option is wrong. Speed here saves time for harder questions.
