Chemistry · Periodic Classification Of Properties · NEET
In the first period, the only orbital available is 1s. An s-orbital holds a maximum of 2 electrons, so only 2 elements (hydrogen and helium) fit. There is no 1p orbital in nature, so the period ends after 2 elements.
In period 2, the orbitals filled are 2s and 2p. That is 1 s-orbital + 3 p-orbitals = 4 orbitals. Each orbital holds 2 electrons, so 4 x 2 = 8 elements. Period 3 fills 3s and 3p in the same way, so it also has 8 elements. The 3d orbital is higher in energy and does not fill in period 3, so no extra elements are added there.
Period 4 fills 4s, then 3d, then 4p. That is 1 (s) + 5 (d) + 3 (p) = 9 orbitals. 9 x 2 = 18 elements. The 3d orbitals (10 elements, the transition metals) appear here for the first time, which is why period 4 jumps from 8 to 18.
Period 6 fills 6s, then 4f, then 5d, then 6p. That is 1 (s) + 7 (f) + 5 (d) + 3 (p) = 16 orbitals. 16 x 2 = 32 elements. The 4f orbitals (the lanthanoids, 14 elements) appear for the first time, adding 14 more elements on top of period 4's pattern.
It comes from which subshells are being filled at each level, in order of increasing energy (Aufbau order). Period 1: s. Periods 2 and 3: s+p. Periods 4 and 5: s+d+p. Period 6 and 7: s+f+d+p. Count the orbitals and double them. It is NOT simply 2n squared, because subshells fill by energy, not strictly by shell number.
No, that is a common trap. 2n squared gives the maximum electrons a SHELL can hold (K=2, L=8, M=18, N=32), but a PERIOD does not equal a full shell. For example, the 3rd shell holds 18 electrons, but period 3 has only 8 elements because 3d fills later, in period 4. Use the orbitals-being-filled method instead.
Because a new type of orbital (d or f) enters only every second period. Period 2 opens the p-block, and period 3 just repeats s+p, so both give 8. Period 4 opens the d-block, and period 5 repeats s+d+p, so both give 18. The 'new orbital' shows up in the earlier of each pair.
The element Z = 114 has been discovered recently. It will belong to which of the following family/group and electronic configuration?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
The 6th period, with 32 elements, is the longest completed period. A theoretical 8th period (with g-orbitals) could hold up to 50 elements, but no such period is complete yet, so 32 is the largest real period length.
Period 1 has 2, period 2 has 8, period 3 has 8, period 4 has 18, period 5 has 18, period 6 has 32, and period 7 (still being completed) would also be 32 when full.
Period 7 fills 7s, 5f, 6d, and 7p, the same 16-orbital pattern as period 6, so its full length is 32. It is written as incomplete in older books, but with synthetic (man-made) elements it now reaches its theoretical length.
Yes: count the subshells being filled in that period and multiply by 2 (electrons per orbital, summed over the subshells). s adds 2, p adds 6, d adds 10, f adds 14. Period 4 = s(2)+d(10)+p(6) = 18.
NEET often asks you to place a heavy element (like Z=114 or Z=119) in the correct period and group. Knowing the 2,8,8,18,18,32 pattern and which orbital fills where lets you count to the element quickly and read off its group.
