2006-09-25 03:56:59 · 4 answers · asked by viren_pndy 1 in Science & Mathematics ➔ Chemistry
Has to do with the solutions to the Schrodinger equation governing quantum mechanics and the possible combinations of the four principal quantum numbers: n, l, m(l) and m(s).
n must be a whole number 1 or greater.
l must be a whole number from 0 up to n-1.
Thus, for n=2, l can only be 0 or 1. For d orbitals, l = 2, so no 2d orbital exists, just as there are no 1p, 1d, 1f, 2f, or 3f orbitals.
2006-09-25 07:59:32 · answer #1 · answered by TheOnlyBeldin 7 · 0⤊ 0⤋
Simply because the second shell can only hold up to 8 electrons, and since that's the case, only s and p orbitals would be involved (thus no need for 2d orbitals).
2006-09-25 04:04:41 · answer #2 · answered by chemistry_freako 3 · 1⤊ 0⤋
That is acocording to the rule, i.e. the no. of subshells is equal to (n-1) , where n = Principal quantum number.
So fro n = 2, L shell, subshell = 2-1=1, no of sub shells 0 and 1 which denotes s and p subshells.
2006-09-25 20:30:30 · answer #3 · answered by dinu 3 · 1⤊ 0⤋
the 2 energy shell can have only 2 subshells i.e. s, p, i.e. the total no. of subshells should be n square.
Hence 3 energy shell ( and not 2) has s, p, d subshells.
total no.= 1+3+5 = 9
2006-09-25 04:03:51 · answer #4 · answered by GodLuvsU:)) 4 · 0⤊ 0⤋