1. The Elements
Chapter 5

2. Electronic Structure of Atoms
Nucleus = protons + neutrons
(~ all mass of the atoms)
p + n (variable)  atomic weight (isotopes)
Atomic weight number is the average atomic weight and includes the number of isotopes of an element.
An isotope of an element has the same number of protons (positive charged) but a different number of neutrons (neutral).
Number of protons and electrons (negative charged) define the atomic number of an element Z.

3. Electronic Structure of Atoms
Electron cloud around the nucleus
gives its statistical size
Atomic radii are in the range Å
1 Å = m K L M
Bohr atomic model: electrons travel along specific orbits
with fixed energy levels: K,L,M,N shells
(These are not the actual positions of the electrons.)

4. Electronic Structure of Atoms
Quantum mechanics, Erwin Schrödinger
Occurrence of electrons are expressed in wave equations ().
The position of electron in space can be specified by four quantum numbers:
Principal quantum number: n
= reflects effective volumes of electron orbits and their energy levels (in Bohr model = shells)
Azimuthal quantum number l (orbital shapes); Bohr model = subshells; for each shell there are l = n - 1 subshells
Magnetic quantum number, m
Electron spin s, spin of electrons only in two directions, +1/2, -1/2

5. Quantum Numbers principal quantum number n = 1, 2, 3, . . .
orbital quantum number l = n-1, n-2, n-3,
magnetic quantum number m = 0, ±1, ±2, . . ., ±(l-1), ±l
spin quantum number m s = ±½
s (n = 1) (l = 0) (m = 0) (m s = ±½) → 2
p (n = 2) (l = 1) (m = 0, ±1) (m s = ±½) → 6
d (n = 3) (l = 2) (m = 0, ±1, ±2) (m s = ±½) → 10
f (n = 4) (l = 3) (m = 0, ±1, ±2, ±3) (m s = ±½) → 14
This yields a capacity of 2n2 for each n.
n = 1 → (1s2)
n = 2 → 8 (2s2, 2p6)
n = 3 → 18 (3s2, 3p6, 3d10)

6. Electronic Structure of Atoms
Quantum numbers / shells
Principal quantum No. n Azimuthal quantum No. l
shells subshells or orbitals
innermost K (n = 1) 0 = s
lower energy L (n = 2) 0,1 = s, p
M (n = 3) 0,1,2 = s, p, d
outermost N (n = 4) 0,1,2,3 = s, p, d, f
higher energy

7. s-orbitals, max. 2 electrons
p orbitals, each dumbbell
= 2 electrons, max. 6 electrons

8. d orbitals, each dumbbell 1 electron = max. 10 electronsz x y d xz z y x d xy x y z d yz z y x d z2 y x z d
x2-y2

9. Electronic Structure of Atoms
f orbitals =
max. 14 electrons
Quantized energy levels s p d f
Relative Energy
Note that the energy does not necessarily increase K  L  M  N etc.
4s < 3d
n = 1 K L 3 M N O P Q

10. Note that with increasing atomic number the s and p subshells, with a maximum capacity of eight electrons, are invariably filled first. All the noble gases except for helium have filled s (2 electrons) and p (6 electrons) orbitals. This is called the noble gas configuration, and for reasons within the realm of quantum mechanics, this is the most stable configuration for all of the other atoms. The result is called the octet rule, in which the atom by various means fills the octet in order to achieve the noble gas configuration.

11. Progressive filling of orbitals as energy increases
Again, energy does not increase regularly K  L  M  N etc.
Some complications with spin V  Cr, etc.

12. A handy trick for determining
the order of filling of electron
shells is to draw the slanting
lines as illustrated to the left.
There are some exceptions
that arise e.g. with the transition
metals.

13. Pauli exclusion principle:
There are no two electrons in any one atom,
which have all four quantum numbers the same.
Hund’s rule:
The number of unpaired electrons
in a given energy level is a maximum
Electron configuration of N     
1s2, 2s2, 2p3

14. Schematic diagram showing the relative average energy of atomic orbitals

15. First ionization potential is the
energy required to remove one
electron from a neutral atom in
a vacuum and to place it at rest
an infinite distance away. It
increases as the orbitals are filling up and it has values in the case of filled or half-filled orbitals

16. Electronegativity is the ability of an atom in a crystal structure to attract electrons into its outer shell
In general, electronegativity increases in a period from left to right and in a group from bottom to top.
(except for inert gases which are very low)

17. Elements are classified as:
Metals w/ e-neg < 1.9 thus lose e- and  cations
Nonmetals > 2.1 thus gain e- and  anions
Metalloids intermediate (B, Si, Ge, As, Sb, Te, Po..)

18. Valence states of the elements

19. RULES TO DETERMINE VALENCE STATES IN COMPOSITIONS
1) The oxidation state of all pure elements is zero.
2) The oxidation state of H is +1, except in hydrides (e.g., LiH, PdH2), where it is -1.
3) The oxidation state of O is -2, except in peroxides (e.g., H2O2), where it is -1.
4) The algebraic sum of oxidation state must equal zero for a neutral molecule or the charge on a complex ion.
The oxidation state may be thought of as the charge an atom would acquire if it were dissolved in water as a simple ion. It is important to be able to determine the oxidation state of atoms in compounds for a number of reasons. First and foremost, we must make this determination when dealing with oxidation-reduction reactions. Another reason is that the chemical properties of elements vary according to their oxidation state.
Many elements can exist in variety of oxidation states. However, the oxidation states of oxygen and hydrogen exhibit relatively little variation. As pure elements, O2 and H2, their oxidation state is 0. In almost all other compounds, O exists as -2 and H as +1. The main exceptions for oxygen are the relatively rare peroxides, such as hydrogen peroxide (H2O2) and sodium peroxide (Na2O2). In the case of hydrogen, the exception is the class of compounds known as hydrides, in which the oxidation state of H is -1. Such compounds generally do not occur in nature.
Knowing the possible oxidation states of O and H, and using rule number 4, we can determine the oxidation states of many other elements.
Example 1: H2S(g). We know that each H atom has an oxidation state of +1 and there are 2 of them. We also know that the overall H2O molecule is neutral. Thus, if we let x = the oxidation state of S, we can write: 2(+1) + x = 0. Solving this, we get x = -2 as the oxidation state of S in H2S(g).
Example 2: CrO42-. We know each O atom has an oxidation state of -2. The overall charge on CrO42- is -2. If we let y = the oxidation state of Cr, we can write: 4(-2) + y = -2, or solving we get y = +6 as the oxidation state of Cr in CrO42-.

20. The oxidation state may be thought of as the charge an atom would acquire if it were dissolved in water as a simple ion. It is important to be able to determine the oxidation state of atoms in compounds for a number of reasons. First and foremost, we must make this determination when dealing with oxidation-reduction reactions. Another reason is that the chemical properties of elements vary according to their oxidation state.
Many elements can exist in variety of oxidation states. However, the oxidation states of oxygen and hydrogen exhibit relatively little variation. As pure elements, O2 and H2, their oxidation state is 0. In almost all other compounds, O exists as -2 and H as +1. The main exceptions for oxygen are the relatively rare peroxides, such as hydrogen peroxide (H2O2) and sodium peroxide (Na2O2). In the case of hydrogen, the exception is the class of compounds known as hydrides, in which the oxidation state of H is -1. Such compounds generally do not occur in nature.
Knowing the possible oxidation states of O and H, and using rule number 4, we can determine the oxidation states of many other elements.
Example 1: H2S(g). We know that each H atom has an oxidation state of +1 and there are 2 of them. We also know that the overall H2O molecule is neutral. Thus, if we let x = the oxidation state of S, we can write: 2(+1) + x = 0. Solving this, we get x = -2 as the oxidation state of S in H2S(g).
Example 2: CrO42-. We know each O atom has an oxidation state of -2. The overall charge on CrO42- is -2. If we let y = the oxidation state of Cr, we can write: 4(-2) + y = -2, or solving we get y = +6 as the oxidation state of Cr in CrO42-.