The Angular Momentum Quantum number L, defines how many types of orbitals can exist.
For a particular Quantum Level n, L is defined as any integer from 0 to L = n-1.
For example, for the energy level n = we must have, L = 0 or 1.
L=0 relates to the s-orbital
L=1 relates to the p-orbital
L=2 would relate to the d-orbital, but we can see here that for n=2, L cannot = 2
The s orbital is present in all valid principal quantum number shells.The p orbital is present in n = 2 and higher.The d orbital is present in n = 3 and higher.The f orbital is present in n = 4 and higher.So the invalid ones are b (there are no 2d orbitals) and c (there are no 3f orbitals). 4s and 3p are perfectly legitimate.
3f
2d Only s and p in the 2 level
no because f orbitals are not energetically available until the n=4 quantum state
The 2d sub energy level does not exist. The first shell to contain a d sub-shell is the third shell: the 3d sub-shell contains a maximum of 10 electrons, with two electrons in each of five different d orbitals.
The s orbital is present in all valid principal quantum number shells.The p orbital is present in n = 2 and higher.The d orbital is present in n = 3 and higher.The f orbital is present in n = 4 and higher.So the invalid ones are b (there are no 2d orbitals) and c (there are no 3f orbitals). 4s and 3p are perfectly legitimate.
3f ( 3 only has the d orbital final ) and 2d ( 2p is the final orbital on the 2 level )
3f
3f
2d is incorrect
2d Only s and p in the 2 level
no because f orbitals are not energetically available until the n=4 quantum state
There can be a maximum of 14 electrons in any "f" orbital. However, the 3f orbital does not exist. f orbitals are only found in quantum energy level 4 and above.
2d is invalid. Only s and p orbitals in 2nd energy level
3d orbitals do exist and 2d orbitals dont exist because of the pauli exclusion principle which says only 6 electrons can exist in the 2nd shell, and you need at least 7 to get a d subshell
the 3F then to medium also prelim it goes: 1A 1B 1C 1D 1E 1F Novice: 2A 2B 2C 2D 2E 2F AND ELIMENTRY: 3A 3B 3C 3D 3E 3F
-3f - 14 = 1 -3f = 1+14 -3f = 15 f = -5