1	AHL Atomic Structure III: Evidence for subshells
2	Learning Objectives Concepts: Ionization, ionization energy, shielding (screening) effect Skills: Account for the existence of sub-levels (sub shells) using evidence from variation of first ionization energy across a period
3	Electrostatic Attraction within an atom Remember, the presence of both protons and electrons in an atom means that there is both attractive (between the nucleus and the electrons) and repulsive forces (between the electrons). These forces are however balanced; hence the stability of atoms over time, and the need for energy to ionize them. And further, that the force of attraction between the nucleus and the valence electrons in an atom depends on 1. The nuclear charge 2. The number of shells
4	1. Ionization energy Once again, X_(g) ® X⁺_(g) + 1e^-_(g) E_IE (DH^q_IE) = positive Where E_IE is the ionization energy, the energy required to ionize a mole of the substance in the gaseous state to produce a mole of positively charged gaseous ions. (See Core Periodicity: Ionization Energy for more.)
5	Periodicity of first ionization energy
6	Exceptions to the General Trend: Shielding (screening) effect Two exceptions: first decrease going from the second to the third group and second decrease going from fifth to sixth group. That is removing the first p electron and removing the fourth p electron. Let’s compare Be and B in period 2: They both have the same number of shells B has a bigger nuclear charge than Be. 2. Ionization of Be involves the removal of an electron from the sub-shell s whereas that of B from subshell p. Based on what we know about nuclear charge, shells and ionization energy, we would expect B to have a higher ionization energy…unless… Within a shell there are different energy levels (thus providing evidence for the existence of subshells). The different energy levels within a shell are subshells, in this case s and p subshells.
7	Shielding Effect Within the same shell (n), electrons in p orbitals are, on average, at a greater distance from the nucleus than those in s orbital. The boundary of p orbital extends farther out than that of s obital, and as such electrons in p orbitals experience significant repulsion from those in the s orbital, significant enough to exceed the increase in attractive force accompanying the increase in the nuclear charge. p orbitals electrons are said to be shielded, or screened (from the nuclear attraction) by s orbitals reducing the energy required to remove them. Another way of putting this is to say that, within a shell, p orbitals are at a higher energy level than s. The s electrons are more effective at shielding than p electrons. Therefore, forming the s²p⁰ (B⁺) becomes more favorable.
8	Shielding Effect The same can be said of the difference in ionization energies between group 2 and 3 elements in the other periods. Similarly, d orbitals are shielded by p and s orbitals etc. In general, within a shell, f orbitals are shielded the most (by d, p and s), then d (by p and s) then p (by s) and s the least. That is, the trend in the energetic stability of different subshells is f < d < p < s. The relative energy of subshells increases going from s to f.
9	Second Exception: Going From Fifth to Sixth Group The second anomaly is the decrease in going from group V to VI. The electron in box configurations look like so for the two elements:
10	Most stable configuration for a subshell and Hund’s rule Look at the configurations of oxygen and nitrogen again and notice that in nitrogen subshell p is half filled (all p orbitals contain one electron each), while in oxygen one p orbital is completely filled. When a second electron is placed in a p orbital (as in oxygen), the electron-electron repulsion increases. This increase in repulsion is significant enough to counter the increase in attractive force accompanying the increase in nuclear charge. When this electron is removed, the resulting s²p³ (O⁺) is much more stable than the starting s²p⁴ (O) configuration. Therefore, there is a decrease in ionization energy going from N to O. As a footnote to the points in the previous slide, it is worth noting that a subshell is most stable when it is completely filled.
11	Hund’s Rule and Orbital Stabilities Now you know why in the Aufbau principle each orbital is first half filled until all orbitals in that sub-shell are half filled, and then electrons are paired. This is a statement of Hund’s rule. So as far as stabilities of orbitals and subshells are concerned, in general, 1. An orbital is more stable when occupied by a single electron 2. (i) A half-filled subshell (containing only unpaired) electrons is more stable than either one that is less than half-filled or one that contains a combination of unpaired and paired electrons (ii) A completely filled subshell (containing only paired) electrons is more stable than either one that is less than half-filled or one that contains a combination of unpaired and paired electron
12	Summary of factors that affect Ionization energy 1. Nuclear charge Everything else (i.e. number of shells) remaining the same, the bigger the nuclear charge, the higher the ionization energy (explains the general trend across a period) 2. Shielding or screening effect Everything else remaining the same, it has the effect of lowering the ionization energy (explains variations within a period) 3. Distance between the nucleus and the valence shell—the number of shells This has the effect of reducing the electrostatic force of attraction which in turn results in a lowered ionization energy (explains trend down a group)
13	Practice Questions 1. Use only the position in the periodic table (not the atomic number) to write the outermost electron configuration (beyond the rare gas core) of (a) Cs, (b) Pb, (c) I, (d) Au 2. State and where possible explain which of the pair of p subshell configurations is more stable
14	Practice Questions
15	Practice Questions 3. Consider the graph to the right. i) Why is there a general increase in ionization energy on passing from Li to Ne?
16	Practice Questions iii) Explain why He has the largest ionization energy of all the noble gases. 4. Write equations corresponding to (a) the first ionization energy of sodium (b) the second ionization energy of lead.
17	Practice Questions 5. List the elements Be, N and Mg in order of increasing ionization energy and explain the order. 6. Choose the species with the larger ionization energy: (a) Li or Be, (b) C or N, (c) Ne or Na, (d) Na⁺ or Mg⁺
18	Practice Questions: Structured 1. M05/8. (a) For the elements of period 3 (Na to Ar), state and explain (i) the general trend in ionization energy [3] (ii) any exceptions to the general trend. [2]
19	Practice Questions: Structured 2. N03/6b. Outline the reasoning for the following in terms of electronic configuration: (i) The first ionization energy of Al is lower than that of Mg. [2] 3. N01/5d. (i) Explain why lithium, sodium, and potassium are placed in the same group of the periodic table. Your answer should refer to their melting points, ionization energies and electronic arrangements. [4]
20	Practice Questions: Structured (ii) State and explain the trend in the chemical reactivity down group 1. Describe, with the aid of balanced equations, the chemical reactions of sodium metal with water and with chlorine gas. [8]