1. Periodic Relationships Among the Elements
Chapter 8
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2. 1s < 2s < 2p < 3s < 3p < 4s
What is the electron configuration of Mg?
Mg 12 electrons
1s < 2s < 2p < 3s < 3p < 4s
1s22s22p63s2
= 12 electrons
[Ne]: 1s22s22p6
Abbreviated as [Ne]3s2
core electrons
valence electrons
electrons in the outermost (highest) principal energy level of an atom
inner electrons
7.8

3. Practice problem of chap8:Q2
Which of the following electron configurations do represent similar chemical properties of their atoms?
1s22s22p63s2
(ii) 1s22s22p3
(iii) 1s22s22p63s23p64s23d104p5
(iv) 1s22s1
(v) 1s22s22p6
(vi) 1s22s22p63s23p3
(ii), (vi)
Practice problem of chap8:Q2

4. Electron Configurations of the Elements
Valence Electrons
• the outer electrons of an atom, the ones involved in
bonding
• for the representative elements (main group elements), the number of valence electrons is equal to the A Group Number

5. Ground State Electron Configurations of the Elements
ns2np6
Ground State Electron Configurations of the Elements
ns1
ns2np1
ns2np2
ns2np3
ns2np4
ns2np5
ns2
d10 d1 d5 4f 5f
8.2

6. Electron Configurations of Cations and Anions
Of Representative Elements
Atoms lose electrons so that cation has a noble-gas outer electron configuration.
Na [Ne]3s1
Na+ [Ne]
Ca [Ar]4s2
Ca2+ [Ar]
Remember: electrons are first removed from orbitals with the highest principal quantum number.
Al [Ne]3s23p1
Al3+ [Ne]
H 1s1
H- 1s2 or [He]
Atoms gain electrons so that anion has a noble-gas outer electron configuration.
F 1s22s22p5
F- 1s22s22p6 or [Ne]
O 1s22s22p4
O2- 1s22s22p6 or [Ne]
N 1s22s22p3
N3- 1s22s22p6 or [Ne]
8.2

7. Cations and Anions Of Representative Elements+1 +2 +3 -3 -2 -1
8.2

8. Practice problem of chap8:Q5
Na+: [Ne]
Al3+: [Ne]
F-: 1s22s22p6 or [Ne]
O2-: 1s22s22p6 or [Ne]
N3-: 1s22s22p6 or [Ne]
Na+, Al3+, F-, O2-, and N3- are all isoelectronic with Ne
Isoelectronic: have the same number of electrons
What neutral atom is isoelectronic with H- ?
H-: 1s2
same electron configuration as He
Which of the following pairs are not isoelectronic with Ar?
K+ and Cl- (b) Ca2+ and S2-
Al3+and N3- (d) P3- and Sc3+
Practice problem of chap8:Q5
8.2

9. Electron Configurations of Cations of Transition Metals
Suppose you were asked to write the orbital diagram for manganese…
The electron configuration is given by:
Mn (25e-): 1s2 2s2 2p6 3s2 3p6 4s2 3d5
The orbital diagram would be written… 1s 2s 2p 3s 3p 4s 3d

10. Electron Configurations of Cations of Transition Metals
When a cation is formed from an atom of a transition metal, electrons are always removed first from the ns orbital and then from the (n – 1)d orbitals.
Fe: [Ar]4s23d6
Mn: [Ar]4s23d5
Fe2+: [Ar]4s03d6 or [Ar]3d6
Mn2+: [Ar]4s03d5 or [Ar]3d5
Fe3+: [Ar]4s03d5 or [Ar]3d5
8.2

11. Practice problem of chap8:Q3
A metal ion with a net +3 charge has five electrons in the 3d subshell. What is this metal?
(a) Cr (b) Mn (c) Fe (d) Co (e) Ni
This species has +3 charges, which indicates that it has three more protons than the electrons. According to the question that it has five electrons in the 3d subshell, and thus the total electrons in valence shells for its atomic type will be 8. Note that the transition metals (with d electrons) losing its s electrons prior to its d electrons and thus the valence electron configuration will be 4s23d6, which is Fe.
Practice problem of chap8:Q3

12. Periodic Variation in Physical Properties

13. 8.3

14. Sizes of atoms: Trends
• size increases while going down a Group
• Why?
• Because orbital size increases with increasing n number, the size of atom increases. The highest energy electrons can be farther away from the nucleus.
• size decreases going across Period
• effective nuclear charge increases. The larger the effective nuclear charge, the stronger hold of the nucleus on electrons. The electron clouds shrink.
• Remember n does NOT increases--e- cannot be farther from nucleus.

15. Ionic Radius
Cation is always smaller than atom from which it is formed. This is because the nuclear charge remains the same but the reduced electron repulsion resulting from removal of electrons make the electron clouds shrink.
Anion is always larger than atom from which it is formed. This is because the nuclear charge remains the same but electron repulsion resulting from the additional electron enlarges the electron clouds.
8.3

16. From top to bottom, both the atomic and ionic radius increase within group.
Across a period the anions are usually larger than cations.
For ions derived from different groups, size comparison is meaningful only if the ions are isoelectronic.
E.g. Na+ (Z=11) is smaller than F- (Z=9).
The larger effective charge results in a smaller radius.
Radius of tripositive ions < dipositive ions<unipositive ions
Al3+ <Mg2+ <Na+
Radius of uninegative ions < dinegative ions
N3- > O2->F-

18. Practice problem of chap8:Q6,7
Which of the following is a correct order of atomic radii?
(a) Ar < P < Na < Ca < Cs (b) Cs < Rb < K < Na < Li
(c) Na < Mg < Al < Si < P (d) Rb < Mg < C < F < He
(e) Li < H < Al < K < Ar
Which of the following is a correct order of ionic radii?
Na+ < Mg2+ < Al3+ < O2- < F- < N3-
Al3+ < Mg2+ < Na+ < F- < O2- < N3-
(c) F- < O2- < N3- < Na+ < Mg2+ < Al3+
(d) Al3+ < Mg2+ < Na+ < N3- < O2- < F-
(e) Na+ < Mg2+ < Al3+ < O2- < N3- < F-
Practice problem of chap8:Q6,7

19. Ionization energy (I.E.) is the minimum energy (kJ/mol) required to remove an electron from a gaseous atom in its ground state.
I1 + X (g) X+(g) + e-
I1 first ionization energy, remove 1st electron
I2 + X+(g) X2+(g) + e-
I2 second ionization energy, remove 2nd electron
I3 + X2+(g) X3+(g) + e-
I3 third ionization energy, remove 3rd electron
I1 < I2 < I3
8.4

20. General Trend in First Ionization Energies
Increasing First Ionization Energy
Increasing First Ionization Energy
8.4

21. First IE:
Trends down a Group
• remember that Cs is more reactive than Na
• Easier to remove an e- from Cs than from Na
• As size increases, I.E. decreases
Trends across a Period
• remember that Cl gains rather than loses an e-
• Easier to remove an e- from Al than from Cl
• As size decreases, I.E. increases

22. Cs < Na < Al < S < Cl
Arrange the following in order of the increasing first ionization energy: Na, Cl, Al, S, Cs
Hint: Ionization energy increases across a row of the periodic table and decreases down a column or group.
Cs < Na < Al < S < Cl
Group practice problem:Q10

23. Electron affinity (EA) is the negative of the energy change that occurs when an electron is accepted by an atom in the gaseous state to form an anion.
X (g) + e X-(g)
F (g) + e X-(g)
DH = -328 kJ/mol
EA = +328 kJ/mol
O (g) + e O-(g)
DH = -141 kJ/mol
EA = +141 kJ/mol
8.5

24. Variation of Electron Affinity With Atomic Number (H – Ba)
Overall trend: increases from left to right across the period. The values varied little in the group. The EA of metals are generally lower than those of nonmetals.
8.5

25. Properties of Oxides Across a Period
basic
acidic
Across a period, oxides change from basic to amphoteric to acidic. Going down a group, the oxides become more basic.
8.6

27. Exceptions: Ionization Energy
Occur between Group 2A and 3A
Group 3A: ns2np1. one single electron in p orbital.
• removing the first p electron is less than expected
•Why? This p electron is shielded by inner ns2 electrons. Less energy is needed to remove a single p electron than to remove a pair of s electron of the same n level.
2. Occur Group 5A and 6A
5A: ns2np3 6A: ns2np4
• removing the fourth p electron is less than expected
•Why? 2nd electron in a p orbital increases the
electron-electron repulsion, which makes it easier to ionize an atom of Group 6A.