ADVANCED-ATOMIC-STRUCTURE

Advanced Electronic Structure of Atoms 2.4 Explain how the evidence from first and second ionization energies accounts for the existence of main energy levels and sub levels. Interpretation of graphs of first ionization energies and successive ionization energies vs. atomic number, leading to evidence for the existence of main energy levels and sub-levels. State the numbering of orbitals For n < 5, e.g. n =1 (1s); n = 2 (2s,2p); n = 3 (3s,3p,3d); n = 4 (4s,4p,4d,4f) State the relative energies of the s,p,d and f orbitals. Relative energies s< p < d< f State the number of orbitals in each energy level s=1 p=3 d=5 f=7 Sketch the shapes of the s and p orbitals s is spherical

p is dumbbell shaped

Describe how the Aufbau principle is derived Reference should be made to the Pauli exclusion principle, Hund’s rule and the minimization of the potential energy of an atom. Apply the Aufbau notation for electronic configurations Apply the Aufbau principle for an atom (up to Z = 110). Exceptions to the rule are optional. Relate the electronic configuration of an atom to its position in the Periodic Table Label s, p, d, and f blocks and relate to electronic configuration.

The nucleus of the atom is surrounded by electrons arranged in specific energy levels and sub-levels. The different sub-levels differ in the shape of the electron distribution. Each energy sub-level is divided into orbitals. Each orbital can contain up to two electrons. The two electrons must have opposite spins. The evidence to support this model of electronic structure comes mainly from the study of atomic spectra. The energy level closest to the nucleus only contains one sublevel and one orbital. This orbital has a spherical symmetry and is termed the “s orbital” (as are all other spherical shaped orbitals in an atom). In the case of the first energy level, it is termed to as the 1s orbital. This orbital can hold two electrons of opposite spins. The second energy level can have one “s” orbital with a spherical shape and three “p” orbitals which have a dumbbell shape. This shape gives rise to the characteristic “figure eight” electron distribution. The porbitals differ in orientation. One p-orbital (p1) is orientated along the x-axis, the second (p2), along the yaxis and the third (p3), along the z-axis. Note again, that each orbital can hold only two electrons. The total electrons on the second energy level is , therefore, eight. (2 s electrons and 6 p electrons)

s orbital

px orbital

py orbital

pz orbital

The third energy level can similarly contain two s-orbitals, three p-orbitals. It also has five d-orbitals. (The “d” orbitals have a somewhat complex shape of which I cannot simulate on this computer- duh-h-hh!) Anyway, the third energy level can hold 18 electrons. Can you determine how? There is , however, a complication in that the 3d-orbitals are at a higher energy than the 3p-orbitals. The 3d-orbitals are also, in most atoms, at a slightly higher energy than the 4s-orbital. In the fourth energy level, as well as the s-orbital, the three p-orbitals and the five d-orbitals, there are seven f-orbitals. the different orbitals and their relative energies for a typical atom are given. The Electron Energy Levels in an Atom ______ ______ ______ ______ ______ 4d ______5s ______ ______ ______ 4p ______ ______ ______ ______ ______ 3d ______4s ______ ______ ______ 3p Energy ______3s

______ ______ ______ 2p

Energy levels

______ 2s 1 2 3 4 ______1s

Types of orbitals

orbitals

s s.p s.p.d s.p.d.f

1 1+3 = 4 1+3+5 = 9 1+3+5+7 = 16 n2

n

maximum e2 18 32 2n2

The Electronic Structure of Atoms The electrons in atoms always adopt the lowest energy configuration possible, this is known as the Aufbau principle. In hydrogen therefore, the electron occupies the 1s orbital and in helium this orbital is filled with two electrons. The electronic structures of these atoms can be written as 1s1 and 1s2 respectively. The first energy level is now filled, so in lithium, the third electron must occupy the s-orbital of the second energy level. With beryllium, the 2s orbital is filled with two electrons. The structures of lithium and beryllium are 1s2 2s1 and 1s2 2s2 respectively. The fifth element, boron, now occupies one of the 2p orbitals. Boron has an electronic structure of 1s2 2s2 2p1. Carbon has six electrons so there is the possibility of these electrons occupying separate orbitals, with similar spins, separate orbitals with opposite spins, or the same orbital with opposite spins. a) ______ ______ ______

b) ______ ______ ______

separate orbitals different spins

separate orbitals opposite spins

c) ______ ______ ______ same orbital opposite spin

It turns out that (a) is the most stable configuration (this is known as Hund’s rule) and so in carbon the two outer electrons singly occupy two of the p-orbitals and in nitrogen all three p orbitals are singly occupied. The electronic configurations, however, are 1s2 2s2 2p2 for carbon and 1s2 2s2 2p3 for nitrogen. Going from oxygen, through fluorine to neon, the p-orbitals are each doubly filled. The electronic structures follow the sequence of 1s2 2s2 2p4, 1s2 2s2 2p5, and 1s2 2s2 2p6.

At sodium the outer electrons start to occupy the third energy level in a manner totally analogous to the filling of the second energy level until argon (1s2 2s2 2p6 3s2 3p6) is reached. At this point, the 4s level is at a lower energy than the 3d level so the 4s orbital is to be filled in potassium and calcium. The five 3d orbitals are filled after the two 4s electrons. Starting at scandium, the 3d orbital is filled, each orbital of d is first singly occupied (Hund’s rule) as far as manganese (1s2 2s2 2p6 3s2 3p6 4s2 3d5) and then doubly filled, until zinc (1s2 2s2 2p6 3s2 3p6 4s2 3d10). From gallium to krypton the 4p-orbital is filled in the usual manner ( singly occupied orbital the doubly filled). There are two exceptions to the filling of the 3d orbital. A peculiarity of these elements with both d and s electrons in the valence shell is that when the elements between scandium and zinc form cations, the first electrons that are lost are the 4s electrons, even though this orbital was filled before the 3d orbitals. Therefore, for example, the electronic structure of the iron(II) ion, formed by the loss of two electrons from iron ([Ar] 3d6 4s2) is [Ar]3d6 not [Ar]3d4 4s2. The 3d and 4s sublevels are close in energy, so that once the 4s2 electrons are lost, the 3d electrons also behave as valence electrons, for example iron(III) is [Ar]3d5. The electronic structures of the elements are related to the position of the element in the periodic table. In the elements on the far left of the periodic table, the s-orbitals are being filled up, so this is known as the sblock. Similarly in the middle the d-block of the periodic table the d-orbitals are being filled. In the right hand p-block, the p-orbitals are being filled. The f-block is traditionally separated from the main table, though it should be placed between the s-block and d-block. This is seen in the long form of the table. Long Form Periodic Table

s

f

d

p

Ionization Energy The ionization energy of an atom is the minimum amount of energy required to remove an electron from a mole of gaseous atoms to form one mole of gaseous ions. Using Q as the symbol for the element, it is the energy required for the change: Q+ (g)

Q (g) ⇒

e (g)

+

The second ionization is similarly the energy required to remove the second electron from the ion produced by the loss of one electron, i.e. the energy required for the change. Q+ (g) ⇒

Q2+ (g)

+

e- (g)

The magnitude of the ionization energy will depend on the attraction of the nucleus. This will be counteracted by the repulsion or shielding of electrons in filled inner orbitals. To a first approximation, each electron in a filled inner shell will cancel one unit of nuclear charge and after these have been subtracted, the remaining nuclear charge is referred to as the effective nuclear charge. The third factor that affects the nuclear charge is the repulsion that the electron experiences from other electrons within the valence shell.

Successive Ionization Energies The more electrons that have been removed from an atom, the greater the energy required to remove the next electron. Within an energy level this is because of a reduction in the amount of electron-electron repulsion and a greater amount of nuclear-electron attraction. Consider, for example the successive ionization energies for the magnesium atom, shown in the diagram below. The two outer electrons experience the same effective nuclear charge. The first one to be removed is however repelled by the other valence electron, but this force is absent when the second electron is removed, hence the higher ionization energy. Sometimes the next electron must be removed from a filled inner energy level, so that this electron will experience a much higher effective nuclear charge and there is a sudden large rise in ionization energy. This is the case for the third and eleventh electron removed from magnesium. Note the use of a logarithmic scale in the diagram. The Removal of Successive Electrons from a Magnesium Atom 5.5 5.0 4.5 4.0 3.5 3.0 2.5 1st

2nd

3rd

4th

5th

6th

7th

Electron Removed

8th

9th

10th

11th

12th

Variation of Ionization Energy Within the Group Going down a group of the periodic table, the ionization energy of the elements decreases. This is because while the effective nuclear charge remains approximately constant, the electrons that are being lost are in successively higher energy levels ( further away from the nucleus) An example of this would be the elements in Group 1A, the alkali metals. Element I.E. (kJ mol-1)

Li 526

Na 502

K 425

Rb 409

Cs 382

This trend can be seen for these elements and perhaps even more clearly for the noble gases, the pea ionization energies in the illustration below. Variation of First ionization Energy with Atomic Number

Overall, going across a period (e.g. Li to Ne, or Na to Ar), it can be seen that the ionization energy increases. This is because of the increase in the charge on the nucleus which, as the electrons being removed are all in the same energy level, increases the effective nuclear charge. The increase is not, however, a smooth one. Going from the second to the third element in each period (Be to B and Mg to Al) there is a decrease. This is because the electron removed from the third element is in a p-subshell and these are at a higher energy than those of the s-subshell from which the second element loses its electron. There is also a slight decrease going from the fifth to the sixth element in each period (N to O and P to S). This is because in the fifth element each of the p-orbitals is singly filled. In electron repulsion and hence a lower ionization energy.