Energy levelSublevel# of orbitals/sublevel n = 11s (l = 0)1 (m l has one value) n = 2 2s (l = 0) 1 (m l has one value) 2p (l = 1) 3 (m l has three values) n = 3 3s (l = 0) 1 (m l has one value) 3p (l = 1) 3 (m l has three values) 3d (l = 2) 5 (m l has five values) n = principal quantum number (energy) l = azimuthal quantum number (shape) m l = magnetic quantum number (orientation) Quantum numbers and orbital energies Each atom’s electron has a unique set of quantum numbers to define it { n, l, m l, m s }

2 Concept: Each electron in an atom has a unique set of quantum numbers to define it { n, l, m l, m s }

Fig. 7.14

What is the reason that the periodic table organizes elements according to similarities in chemical properties?

There is a relationship between the quantum number (n) and its the number of subshells. Principal quantum number (n) = number of subshells Arrangement of electrons in atoms

6 H atom All other atoms As n increases, the difference in energy level _______. multi-electron atoms

Energy levels for multi-electron atoms 2n 2 electrons energy level # of electrons

Orbital energy ladder n = 1  2 e’s n = 2  8 e’s n = 3  18 e’s

Basic Principle: electrons occupy lowest energy levels available

Aufbau Principle -- “Bottom Up Rule”

Electron spin How could an orbital hold two electrons without electrostatic repulsion?  Stern- Gerlach Experiment

1 1 s value of energy level sublevel no. of electrons spdf NOTATION for H, atomic number = 1 spdf Notation Orbital Box Notation Arrows show electron spin (+½ or -½) ORBITAL BOX NOTATION for He, atomic number = 2 1s1s 2 1s1s  2 ways to write electron configurations

Example: Determine the electron configuration and orbital notation for the ground state neon atom. An orbital can contain a maximum of 2 electrons, and they must have the opposite “spin.” Pauli exclusion principle

Hund’s Rule - Write the ground state configuration and the orbital diagram for oxygen in its ground state

Rules for Filling Orbitals Bottom-up (Aufbau’s principle) Fill orbitals singly before doubling up (Hund’s Rule) Paired electrons have opposite spin (Pauli exclusion principle) Basic Principle: electrons occupy lowest energy levels available

Orbital energy ladder s p n = 2 s d p n = 3 f s d p n = 4 s n = 1 Energy

Phosphorus Symbol: P Atomic Number: 15 Full Configuration: 1s 2 2s 2 2p 6 3s 2 3p 3 Valence Configuration: 3s 2 3p 3 Shorthand Configuration: [Ne]3s 2 3p 3    1s 2s 2p 3s 3p Box Notation

Paramagnetic : atoms with unpaired electrons that are weakly attracted to a magnet. Diamagnetic : atoms with paired electrons that are not attracted to a magnet. Paramagnetic : atoms with unpaired electrons that are weakly attracted to a magnet. Diamagnetic : atoms with paired electrons that are not attracted to a magnet. Electron spin & magnetism For the ground state oxygen atom: spdf configuration: orbital box notation:

Apparatus for measuring magnetic properties

Identify examples of the following principles: 1) Aufbau 2) Hund’s rule 3) Pauli exclusion

Note: Not written according to Aufbau, but grouping according to n

Electron distribution for the argon atom Never zero electron distribution

Electron configuration for As

Silicon’s valence electrons Shorthand notation for silicon 1s 2 2s 2 2p 6 3s 2 3p 2 [Ne] 3s 2 3p 2 [Ne] 

Shorthand notation practice Examples ●Aluminum: 1s 2 2s 2 2p 6 3s 2 3p 1 [Ne]3s 2 3p 1 ●Calcium: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 [Ar]4s 2 ●Nickel: 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 8 [Ar]4s 2 3d 8 {or [Ar]3d 8 4s 2 } ●Iodine: [Kr]5s 2 4d 10 5p 5 {or [Kr]4d 10 5s 2 5p 5 } ●Astatine (At): [Xe]6s 2 4f 14 5d 10 6p 5 {or [Xe]4f 14 5d 10 6s 2 6p 5 } [ Noble Gas Core ] + higher energy electrons

Outer electron configuration for the elements

Electronic configuration of Br 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 4s 2 4p 5 [Ar] 3d 10 4s 2 4p 5 [Ar] = “noble gas core” [Ar]3d 10 = “pseudo noble gas core” (electrons that tend not to react) Atom’s reactivity is determined by valence electrons valence e’s in Br: 4s 2 4p 5 highest n electrons

Valence e ’ s for “main group” elements

Valence e - shells for transition metalsmain group elements transition metals v. main group elements d orbitals sometimes included in valence shell d orbitals not included in valence shell (pseudo noble gas cores)

Rule-of-thumb for valence electrons Examples ●Sulfur: 1s 2 2s 2 2p 6 3s 2 3p 4 or [Ne]3s 2 3p 4 valence electrons: 3s 2 3p 4 ●Strontium: [Kr]5s 2 valence electrons: 5s 2 ●Gallium: [Ar]4s 2 3d 10 4p 1 valence electrons: 4s 2 4p 1 ●Vanadium: [Ar]4s 2 3d 3 valence electrons: 4s 2 or 3d 3 4s 2 Identify all electrons at the highest principal quantum number (n)

Selenium’s valence electrons Pseudo noble gas core includes:  noble gas electron core  d electrons (not very reactive) Written for increasing energy:

Core and valence electrons in Germanium Pseudo noble gas core includes:  noble gas core  d electrons Written for increasing energy:

d-block: some exceptions to the Aufbau principle

Mendeleev’s periodic table generally organized elements by increasing atomic mass and with similar properties in columns. In some places, there were missing elements whose properties he predicted. When gallium, scandium, and germanium were isolated and characterized, their properties were almost identical to those predicted by Mendeleev for eka-aluminum, eka-boron, and eka-silicon, respectively.

Figure 8.14: Mendeleev’s periodic table.

Periodic law states that when the elements are arranged by atomic number, their physical and chemical properties vary periodically. We will look in more detail at three periodic properties: atomic radius, ionization energy, and electron affinity.

Effective Nuclear Charge Effective nuclear charge is the positive charge that an electron experiences from the nucleus. It is equal to the nuclear charge, but is reduced by shielding or screening from any intervening electron distribution (inner shell electrons).

Effective nuclear charge increases across a period. Because the shell number (n) is the same across a period, each successive atom experiences a stronger nuclear charge. As a result, the atomic size decreases across a period.

Atomic Radius While an atom does not have a definite size, we can define it in terms of covalent radii (the radius in covalent compounds).

Figure 8.17: Representation of atomic radii (covalent radii) of the main- group elements.

Atomic radius is plotted against atomic number in the graph below. Note the regular (periodic) variation.

A representation of atomic radii is shown below.

Refer to a periodic table and arrange the following elements in order of increasing atomic radius: Br, Se, Te. 35 Br 34 Se 52 Te Te is larger than Se. Se is larger than Br. Br < Se < Te

First Ionization Energy (first ionization potential) The minimum energy needed to remove the highest-energy (outermost) electron from a neutral atom in the gaseous state, thereby forming a positive ion

Periodicity of First Ionization Energy (IE 1 ) Like Figure 8-18

Fig. 8.15

Left of the line, valence shell electrons are being removed. Right of the line, noble- gas core electrons are being removed.

Identifying Elements by Its Successive Ionization Energies Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE 1 IE 2 IE 3 IE ,815 2,740 11,600 Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table! Solution:

Identifying Elements by Its Successive Ionization Energies Problem: Given the following series of ionization energies (in kJ/mol) for an element in period 3, name the element and write its electron configuration: IE 1 IE 2 IE 3 IE ,815 2,740 11,600 Plan: Examine the values to find the largest jump in ionization energy, which occurs after all valence electrons have been removed. Use the periodic table! Solution: The largest jump in IE occurs after IE 3 so the element has 3 valence electrons thus it is Aluminum ( Al, Z=13), its electron configuration is : 1s 2 2s 2 2p 6 3s 2 3p 1

Fig. 8.16

Ranking Elements by First Ionization Energy Problem: Using the Periodic table only, rank the following elements in each of the following sets in order of increasing IE! a) Ar, Ne, Rn b) At, Bi, Po c) Be, Na, Mg d) Cl, K, Ar Plan: Find their relative positions in the periodic table and apply trends! Solution:

Trends Going down a group, first ionization energy decreases. This trend is explained by understanding that the smaller an atom, the harder it is to remove an electron, so the larger the ionization energy.

Generally, ionization energy increases with atomic number. Ionization energy is proportional to the effective nuclear charge divided by the average distance between the electron and the nucleus. Because the distance between the electron and the nucleus is inversely proportional to the effective nuclear charge, ionization energy is inversely proportional to the square of the effective nuclear charge.

Small deviations occur between Groups IIA and IIIA and between Groups VA and VIA. Examining the valence configurations for these groups helps us to understand these deviations: IIAns 2 IIIA ns 2 np 1 VA ns 2 np 3 VIA ns 2 np 4 It takes less energy to remove the np 1 electron than the ns 2 electron. It takes less energy to remove the np 4 electron than the np 3 electron.

Electrons can be successively removed from an atom. Each successive ionization energy increases, because the electron is removed from a positive ion of increasing charge. A dramatic increase occurs when the first electron from the noble-gas core is removed.

Refer to a periodic table and arrange the following elements in order of increasing ionization energy: As, Br, Sb. Sb is larger than As. As is larger than Br. Ionization energies: Sb < As < Br 35 Br 33 As 51 Sb

Overall periodic trends Note: Electronegativity has similar trend as electron affinity

61 Reactivity of the Alkali Metals Potassium video Sodium video Lithium video 2Na (s) + 2H 2 O (l) → 2NaOH (aq) + H 2 (g) 2K (s) + 2H 2 O (l) → 2KOH (aq) + H 2 (g) 2Li (s) + 2H 2 O (l) → 2LiOH (aq) + H 2 (g) Trend?

62 More Sodium Reaction Videos Preppin g Na 150 g Na in small pieces 2Na (s) + 2H 2 O (l) → 2NaOH (aq) + H 2 (g) g Na in one piece

Electronic Configuration Ions Na 1s 2 2s 2 2p 6 3s 1 Na + 1s 2 2s 2 2p 6 Mg 1s 2 2s 2 2p 6 3s 2 Mg +2 1s 2 2s 2 2p 6 Al 1s 2 2s 2 2p 6 3s 2 3p 1 Al +3 1s 2 2s 2 2p 6 O 1s 2 2s 2 2p 4 O - 2 1s 2 2s 2 2p 6 F 1s 2 2s 2 2p 5 F - 1 1s 2 2s 2 2p 6 N 1s 2 2s 2 2p 3 N - 3 1s 2 2s 2 2p 6

Isoelectronic Atoms and Ions H - 1 { He } Li + Be +2 N - 3 O - 2 F - { Ne } Na + Mg +2 Al +3 P - 3 S - 2 Cl - { Ar } K + Ca +2 Sc +3 Ti +4 As - 3 Se - 2 Br - { Kr } Rb + Sr +2 Y +3 Zr +4 Sb - 3 Te - 2 I - { Xe } Cs + Ba +2 La +3 Hf +4

Trends when atoms form chemical bonds Empirical Observation “when forming ionic compounds, elements tend to lose or gain electrons to be more like the nearest noble gas” Metals tend to lose e - ’s Nonmetals tend to gain e - ’s

Are ions bigger or smaller than atoms? Representative cation Na → Na + + e  Representative anion F + e  → F 