Periodic Properties of Elements

X

HF

V (1) = dx (3)

2

a r

12

b6 =

a

6

b = a

that is, by summing over all the one-electron potential obtained by averaging the interaction of electron 1

2

|χ

dx (2)| dx x

and electron 2, weighted by the probability that electron 2 occupies the volume element at . For

2 2 2

b

further details of this topic see "Modern Quantum Chemistry", by A.Szabo, N.Ostlund.

3

Figure 2: Plots of radial distribution functions for some hydrogenoid orbitals

r = 0, P (0) = 0, s-orbital

As we can see, even if at in general to is associated a greater radial

p-orbitals. s-electrons

probability near the nucleus than We can conclude that penetrates much

p-electrons d f

more than do. Same considerations hold for and electrons. Then, the effect of

s > p > d > f l.

penetration decreases as , that is it decreases with

σ

Slater’s constant can be calculated according to these four following rules:

n

i) All the electrons in orbitals whose principal quantum number is higher than the one for

σ

which we are calculating do not contribute; σ

ii) Each electron with the same principal quantum number contributes to with a factor 0.35.

d f s p

However, if electron in question occupies a or -orbital, electrons in or orbitals with

the same quantum number contributes with 1.00 each one; −

n 1

iii) Each electron in orbitals with principal quantum number contribute each one with a

d f

factor 0.85, except ones in and orbitals, which contributes with 1.00;

n

iv) All the electrons with lower contributes with 1.00 each one.

These rules are also known as Slater’s rules. To show their application we consider the case of Ca.

Its complete ground state electron configuration is,

2 2 6 2 6 2

1s 2s 2p 3s 3p 4s (11)

4s

Then, for an electron in orbital, we get,

· · ·

σ = 1 0.35 + 8 0.85 + 10 1 = 17.15 (12)

Ca,4s

Z = 20

Since, we find the following effective nucleare charge,

? −

Z = Z σ = 2.85 (13)

Ca,4s

?

Z Z

As expected, increases along a group and along a period (since itself increases much more

?

Z increases because the single electron added step by

than Slater’s constant). Along a period,

step is not completely screened by the nucleare charge "produced" by the other core electrons.

Of course noble gases have the higher effective nuclear charge on a period, which corresponds

f

to their low/absent reactivity. For lanthanides, electrons are added in orbitals with principal

−

n 2, s-electrons.

quantum number since valence electrons are Then, for lanthanides we always

? −

Z = 2.85. (n 2)f,

get This value does not depend on electrons in since these penetrate and

completely screen the nuclear charge. 4

We remind that electron configuration of a polyelectronic atoms is determined by filling orbitals

according to Pauli’s exclusion principle, from the lower to the higher in energy. In doing this we

cannot neglect screening and penetration effects. We know from quantum mechanics that energy

n l, ns < np < nd < nf

of orbitals depends on and in the order . Stability order till Ar is

1s > 2s > 2p > 3s > 3p. Ar has the following electron configuration,

2 2 6 2 6

[Ar] : 1s 2s 2p 3s 3p (14)

4s 3d, s-orbitals

Now, if we add an electron in orbital and not in since are more penetrating and

1

4)

since in the higher quantum number (namely we would have a single electron (4s ), effective

d

nuclear charge will increase more than in the case of filling (due to Slater’s rules). This produces

2

4s 3d

an higher stability for K (and also for Ca with completely filled). However., electrons

4s

penetrates a lot the electron density of (since they are "core" electrons respect to these), thus

?

Z 3d 4s 3d

for brutally increase and their energy decrease. Then, after orbital, are filled. This

holds from Sc to Zn, whose electron configuration is,

2 10

[Zn] : [Ar]4s 3d (15)

3d 4p 4p, 5s

After orbitals, are filled, till Kr. For the same reasons expressed above, after (and not

4d) 4d Z

is filled, yielding Rb and Sr. Now, "feels" the increasing of by 2 units, such that they

5p. 4d 5p) 5p

become less energetic than Thus, from Y to Cd (and not are filled. From In to Xe

6s, 6s, 5d 4f

are filled and from Cs to Ba, as expected. and have almost the same energy, then

5d

in La (the next element after Ba) the additional electron fills a orbital. However this electron,

6s, 4f

together with the ones in do not efficiently screen electrons. The direct consequence is

4f

that orbitals decrease in energy much more rapidly than the other. We will have the valence

2 2

6s 4f

electron configuration , and so on till Yb. With similar considerations, we can explain the

progressive filling of all the others elements.

We have two important exceptions to the above explanation of electron configuration. The expected

4 2

[Ar]3d 4s

electron configuration of Cr is , however it actually is,

5 1

Cr : [Ar]3d 4s (16)

10 1 9 2

[Ar]3d 4s [Ar]3d 4s

The same holds for Cu, whose actual electron configuration is , in stead of .

The reason lies in a quantum mechanical effect known as exchange interaction, which acts in order

2

to stabilize these electron configurations.

3 Ionization Energy

The ionization energy (E ) is the work done to remove an electron from the external shell of an

I

atom in its ground state. The first ionization energy correspond to the reaction,

−

+

→

A A + e (17)

2 Exchange interaction is an effect, which has no counterpart in classical mechanics (even if it is a sort of interaction

between charge distributions), directly correlated to indistinguishable principle, that is to the fact that in a quantum

system, different particles with same spin are not distinguishable. This produces some symmetry properties on the

wavefunctions. To understand qualitatively the exchange effect on electron configuration it is sufficient to consider

X 3d

it in terms of units of a factor . For two indistinguishable electrons in a orbital we could make only one

e 5 1

×

1 X 3d 4s

exchange, thus we set the exchange energy equal to . For Cr in electron configuration we can make

e ×

10 X

ten exchanges (we do not count redundant exchanges), that is an exchange energy equal to . In stead, in

e

4 2 ×

3d 4s 6 X

the case of we would have , that is a lower stabilization!

e 5 ns

There is an important exception to take in account: in transition metals orbitals are filled before

−

(n 1)d ns. E

orbitals, however when they are ionized, electron is removed from As expected, I

increases along a period and decreases along a group. In fact, along a period valence electrons are

n-shell

put in the same and they experiences an increasing effective nuclear charge. In stead, along

n

a group increases for valence shells, that is they are farther and farther away from the nucleus.

?

2s 1s. Z

Passing from He to Li, the third electron fills a orbital, external to The decrease of

E

produces in this case a brutally decrease of . In stead, from Be to B the decrease is less evident,

I

2p

since the fifth electron goes to a orbital, which is less penetrating (it screens a lot the nuclear

charge). From B to N the increase is regular. There’s a new decrease for O since the electron

p E

couples with an electron already filling a orbital. The highest values of are found for gas

I

nobles, and the lowest for alkali metals, as expected. ?

E Z

We have already said that along a group decreases, even if increases. The reason lies in the

I

? 2

E Z n n

fact that is directly proportional to but inversely proportional to . Hence, dominates

I

? → ∞.

Z n

on in the limit The explicit relation is, ? 2 2

(Z ) e

1

−

E = (18)

I 2

n 2a

0

The complete derivation from classical electrostatics of the above relation is given in Appendix B.

Ionization energy of transition metals in sixth period are higher then the corresponding values for

fifth period. This is due to the previous entrance of lanthanides, which do not screen very well,

?

Z d

yielding an increasing of experienced by electrons of third period’s transition metals. This

3

phenomenon is also called lanthanide contraction.

4 Atomic and ionic radii

Atomic radii decrease along a period due to an incomplete screening of nuclear charge, that is an

?

Z n

increase in . On the contrary, they increase along a group since the increase of is related to

?

Z

an higher distance of electrons from the nucleus. This effect wins on the increase of . Variations

d

on the atomic radius are more evident for first elements of a group, because, after the filling of ?

f Z

and orbitals and the weak screening effect of these electrons, there’s a progressive increase of

and thus the "electron cloud" expands with more difficulties.

Elements after the lanthanides’ serie are smaller than expected, due to the already cited lanthanide

contraction. ?

Z

Positive charged ions are smaller then the respective neutral atom, due to the higher ; negative

?

Z

charged ions are bigger than the respective neutral atom, due to the lower . Of course, ionic

radii decrease after an increase of the positive charge, in stead they decrease after an increase of

negative charge.

3 Within the Hartree-Fock approximation, it is possible to have a slightly accurate evaluation of ionization energy.

|Ψ i

N

Given an Hartree-Fock wavefunction for electrons , the total energy is,

HF N

N N N

1

X X X

ha| −

E = ĥ|ai + [aa|bb] [ab|ba] (19)

HF 2

a a b

a, b χ

where are two different spin-orbitals. If we remove an electron from an occupied spin-orbital , we can roughly

a

−1

N

|Ψ i E

approximate the new wavefunction as , and evaluate without any relaxation as,

−1

HF N HF

−1 −1 −1

N N N

1

−1 X X X

N ha| −

E = ĥ|ai + [aa|bb] [ab|ba] (20)

HF 2

a a b

With some calculations on the above expression, it can be proved that,

−1 −1

N N

N N

− ' − −ε

E = E ε =⇒ E E E = (21)

a a

I

HF HF

HF HF

ε

where is the energy of the occupied spin-orbital. For a complete proof of this see "Modern Quantum Chemistry",

a

by A.Szabo, N.Ostlund. 6

5 Electron affinity

1

The electron affinity (E ) can be defined as the energy in gas-phase required to produce a negati