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The Periodic Table Dmitri Mendeleev –designed periodic table in which the elements were arranged in order of increasing atomic mass Henry Moseley –designed.

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Presentation on theme: "The Periodic Table Dmitri Mendeleev –designed periodic table in which the elements were arranged in order of increasing atomic mass Henry Moseley –designed."— Presentation transcript:

1 The Periodic Table Dmitri Mendeleev –designed periodic table in which the elements were arranged in order of increasing atomic mass Henry Moseley –designed periodic table in which the elements were arranged in order of increasing atomic number

2 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.

3 The properties of the elements exhibit trends. These trends can be predicted using the periodic table and can be explained and understood by analyzing the electron configurations of the elements.

4 There are two important trends. First, electrons are added one at a time moving from left to right across a period. As this happens, the electrons of the outermost shell experience increasingly strong nuclear attraction, so the electrons become closer to the nucleus and more tightly bound to it.

5 Second, moving down a column in the periodic table, the outermost electrons become less tightly bound to the nucleus. This happens because the number of filled principal energy levels (which shield the outermost electrons from attraction to the nucleus) increases downward within each group.

6 These 2 trends explain the periodicity observed in the elemental properties of atomic radius, ionization energy, electron affinity, and electronegativity.

7 Periodicity of atomic radii Atomic radii increase down a group Li  Cs; 2s  7s Atomic radii decrease going across a period – the effective nuclear charge, Z eff increases - a proton is added to the nucleus and shielding remains constant Z eff = Z actual – electron shielding


9 The nuclear charge felt by an electron in an outer shell is called the effective nuclear charge

10 Atomic radii of main-group and transition elements Opposing forces: Changes in n and changes in Z eff Overall Trends (A) n dominates within a group; atomic radius generally increases in a group from top to bottom (B) Z eff dominates within a period; atomic radius generally decreases in a period from left to right

11 Periodicity of atomic radius Large size shifts when moving from one period to the next


13 Ranking Elements by Atomic Size PLAN: SOLUTION: PROBLEM:Using only the periodic table, rank each set of main group elements in order of decreasing atomic size. (a) Ca, Mg, Sr(b) K, Ga, Ca(c) Br, Rb, Kr(d) Sr, Ca, Rb Size increases down a group; size decreases across a period. (a) Sr > Ca > MgThese elements are in Group 2A. (b) K > Ca > GaThese elements are in Period 4. (c) Rb > Br > KrRb has a higher energy level and is far to the left. Br is to the left of Kr. (d) Rb > Sr > CaCa is one energy level smaller than Rb and Sr. Rb is to the left of Sr.

14 IONIZATION ENERGY Cations are formed when an atom loses one or more electrons Na  Na + + e - Mg  Mg 2+ + 2e - Energy input is required for this process The first ionization energy, E i1, is the minimum amount of energy required to remove the outermost electron from an isolated gaseous atom H + 1312 kJ  H + + e -

15 In some cases a second and even a third electron may be removed Ca + 590 kJ  Ca + + e - E i1 Ca + + 1145 kJ  Ca 2+ + e - E i2 Al  Al 3+ E i1, E i2, E i3 For a given element E i1 < E i2 < E i3 - because it is much harder to remove an electron from a positively charged ion than from the corresponding neutral atom

16 Compare E i2 vs. E i3 for Mg Mg +  Mg 2+ + e - 1451 kJ Mg 2+  Mg 3+ + e - 7733 kJ


18 Ionization energy shows a clear periodic trend E i decreases as a group is descended e.g. E i for Li > Na > K > Rb > Cs - the electron is lost from successively higher energy levels which are further away from the nucleus

19 There is a gradual increase in E i as a period is traversed Na < Si < Cl Part of the reason: atomic radii decrease making the outermost electrons closer to the nucleus and thus harder to remove The increase across the period is not smooth – breaks occur at Be/B and N/O E i for B < Be Be: 1s 2 2s 2  Be + : 1s 2 2s 1

20 B: 1s 2 2s 2 2p 1  B + : 1s 2 2s 2 In Be, to form Be + a filled shell is being broken – this is very energy expensive On the other hand, B + has a filled shell; in addition it is easy to remove the single 2p electron In the next case: N: 1s 2 2s 2 2p 3  N + : 1s 2 2s 2 2p 2 O: 1s 2 2s 2 2p 4  O + : 1s 2 2s 2 2p 3 The first three ionization energies of beryllium (in MJ/mol)



23 First ionization energies of the main-group elements Increase within a period and decrease within a group


25 Ranking Elements by First Ionization Energy PLAN: SOLUTION: PROBLEM:Using the periodic table, rank the elements in each of the following sets in order of decreasing IE 1 : (a) Kr, He, Ar(b) Sb, Te, Sn(c) K, Ca, Rb(d) I, Xe, Cs IE decreases down in a group; IE increases across a period. (a) He > Ar > Kr (b) Te > Sb > Sn (c) Ca > K > Rb (d) Xe > I > Cs Group 8A elements- IE decreases down a group. Period 5 elements - IE increases across a period. Ca is to the right of K; Rb is below K. I is to the left of Xe; Cs is further to the left and down one period.

26 Identifying an Element from Successive Ionization Energies PLAN: SOLUTION: PROBLEM:Name the Period 3 element with the following ionization energies (in kJ/mol) and write its electron configuration: IE 1 IE 2 IE 3 IE 4 IE 5 IE 6 1012190329104956627822,230 Look for a large increase in energy that indicates that all of the valence electrons have been removed. The largest increase occurs at IE 6, that is, after the 5th valence electron has been removed. The element must have five valence electrons with a valence configuration of 3s 2 3p 3, The element must be phosphorus. P (Z = 15). The complete electronic configuration is: 1s 2 2s 2 2p 6 3s 2 3p 3.

27 ELECTRON AFFINITY Anions are formed by an atom accepting electron(s) This process is also accompanied by an energy change The electron affinity, E ea, is the energy change that occurs when an electron is added to an isolated gaseous atom The energy change is usually negative – the more negative the E ea, the greater the tendency to form anions

28 Be + e -  Be - E ea = 241 kJ mol -1 Cl + e -  Cl - E ea = -348 kJ mol -  Cl form anions easier than Be The periodic trend is E ea becomes more negative across a period – trend is not regular Again there are breaks at Groups 2A and 5A

29 It is very difficult to add an electron to a 2A metal because its outer 2s orbital is filled Values for 5A elements are less negative than expected because they apply to addition of an electron to a relatively stable half- filled

30 Electron affinities of the main-group elements Negative values = energy is released when the ion forms Positive values = energy is absorbed to form the anion


32 Lets tie it all together:

33 Depicting ionic radii

34 Periodicity of ionic radii For cations: ionic radii is always less than atomic radii - so Li + < Li 1s 2 1s 2 2s 1 For Li + ; 3 p  2 e -  greater attraction here Li; 3 p  3 e - For anions: anions are bigger than their parent atoms Cl: 1s 2 2s 2 2p 6 3s 2 3p 5 Cl - : 1s 2 2s 2 2p 6 3s 2 3p 6

35 Ionic vs atomic radius Ionic size increases down a group Trends in periods are complex For atoms that form more than one cation: the greater the ionic charge, the smaller the ionic radius


37 Ranking Ions by Size PLAN: SOLUTION: PROBLEM:Rank each set of ions in order of decreasing size, and explain your ranking: (a) Ca 2+, Sr 2+, Mg 2+ (b) K +, S 2 -, Cl - (c) Au +, Au 3+ Compare positions in the periodic table, formation of positive and negative ions and changes in size due to gain or loss of electrons. (a) Sr 2+ > Ca 2+ > Mg 2+ (b) S 2 - > Cl - > K + These are members of the same Group (2A) and therefore decrease in size going up the group. These ions are isoelectronic; S 2 - has the smallest Z eff and therefore is the largest while K + is a cation with a large Z eff and is the smallest. (c) Au + > Au 3+ The higher the positive charge, the smaller the ion.



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