Page 115 - Bruno Linder Elementary Physical Chemistry

August 18, 2010 11:36 9in x 6in b985-ch09 Elementary Physical Chemistry

100 Elementary Physical Chemistry

n (shell) l (subshell) m l (orbital)

1(K) 0 (s) 0 (1s)
2(L) 0 (s) 0 (2s)
1(p) −1(2p x), 0 (2p z), 1 (2p y)
3(M) 0 (s) 0 (3s)
1(p) −1(3p x), 0 (3p z), 1 (3p y)
2(d) −2, −1, 0, 1, 2; 5 orbitals

Relations between n, l, m l and shells, subshells, and orbitals.
Fig. 9.2
(b) Orbitals with the same n but different l’s form subshells of the
given shell. Subshells are denoted as s, p, d, f, g, etc. pertaining to
l =0, 1, 2, 3, 4, etc. When n =1,l can have only one value, 0, and
the subshell can have only one orbital 1s.When n =2, l can have the
values 0 and 1, giving rise to the two subshells s and p.The number
of orbitals of a subshell may be calculated from 2l +1. Thus, in the
subshell 1s there is one orbital; in the subshell p there are 3 orbitals, etc.
Figure 9.2 shows relations between shells, subshells, and orbitals. In
many atoms, except hydrogenic atoms, the increase in energy does not
always correspond to an increase in n. For example, the 4s energy is lower
than the 3d energy, etc.

9.6. Shapes of Orbitals
Shapes of orbitals play an important role in determining how atoms bind
to form molecules.
Question: An s orbital has its maximum at the center, yet the probability
∗
of finding the electron there is zero. Why? Reason: The quantity ψ ψ is
maximum at the center, but it has to be multiplied by the volume element,
which is zero at the center, δV =0.

Radial Distribution Function
Most often, one is interested in the distance of an electron from the nucleus
(regardless of angles) rather than in a given volume element. To obtain
2
a working formula, note that the volume element is δV =4πr dr.The
probability of finding the electron in that volume element is
2
Prob. = ψ ψ 4πr dr (9.9)
∗