Since L(lower case) can be any integer from 0 to n-1, then if L is 4 n must be 5.
mL can be -L to + L which means -4, -3, -2, -1, 0, +1.......+4.
Stick two electrons in each mL and count or get sophisticated and think it through. For s electrons, L(lower case) = 0, for p electrons L is 1 etc,
# orbitals for s = (2L) + 1 = 1
...................(2*0) + 1 = 1
# orbitals for p = (2*1)+1 = 3
# orbitals for d = (2*2)+1 = 5
# orbitals for f = )2*3)+1 = 7
and for each orbital you can have two electrons so that 2,6,10,14 for s,p,d,f.
So g is what? (2*4)+1= ? and that x 2. See if that's the same number you obtained by counting.
Although no currently known elements contain electron in g orbitals in the ground state, it is possible that these elements will be found or that electrons in excited states or known elements could be in g orbitals. For g orbitals, the value of L(lowercase) is 4.
What is the lowest value of n for which g orbitals could exist?
What are the possible values for mL(subscore lowecase)?
How many electrons could a set of g orbitals hold?
1 answer
2 answers