A German scientist named Johann Dobereiner put forward his law of triads in 1817. Each of Dobereiner's triads was a group of three elements. The appearance.

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Presentation transcript:

A German scientist named Johann Dobereiner put forward his law of triads in Each of Dobereiner's triads was a group of three elements. The appearance and reactions of the elements in a triad were similar to each other. At this time, scientists had begun to find out the relative atomic masses of the elements. Dobereiner discovered that the relative atomic mass of the middle element in each triad was close to the average of the relative atomic masses of the other two elements. This suggested that atomic mass might be important to arranging the elements.

With this idea in mind nearly 50 years later, in 1864, an Englishman, John Newlands, arranged the known elements into seven rows. Newlands' Octave Arrangement from Chemical News [1866] No. H 1 Li 2 G 3 Bo 4 C 5 N 6 O 7 F 8 Na 9 Mg 10 Al 11 Si 12 P 13 S 14 Cl 15 K 16 Ca 17 Cr 19 Ti 18 Mn 20 Fe 21 Co & Ni 22 Cu 23 Zn 25 Y 24 In 26 As 27 Se 28 Br 29 Rb 30 Sr 31 Ce & La 33 Zr 32 Di & Mo 34 Ro & Ru 35 Pd 36 Ag 37 Cd 38 U 40 Sn 39 Sb 41 Te 43 I 42 Cs 44 Ba & V 45 Ta 46 W 47 Nb 48 Au 49 Pt & Ir 50 Os 51 Hg 52 Tl 53 Pb 54 Bi 55 Th 56 The elements were placed in order of increasing atomic mass and arranged so elements with similar chemical properties are in the same group.

Since the properties were repeated every eighth element, Newlands referred to his arrangement as the Law of Octaves. Unfortunately for Newland’s, there were a number of problems with his arrangement and it really only worked up through calcium. About five years later, Dmitri Mendeleev a Russian chemist who was unaware of the work of Newlands, designed his own table. After about 18 months of gathering information and arranging element cards, it was finished. Like Newlands, Mendeleev arranged the 63 known elements in order of increasing atomic mass, having elements with similar chemical properties in the same group.

Unlike Newlands his table had groups of varying lengths. He also left gaps in the table for elements he believed had not yet been discovered. He even made predictions about the properties of some of these elements.

Although his successful predictions allowed many to accept his periodic idea, there were still anomalies that Mendeleev could not explain. One of these is the order of iodine and tellurium. The problem was finally solved in 1913 by a 25- year-old English physicist named Henry Moseley. Moseley showed that the ordering of X-ray spectral lines was dependent upon the ordering of nuclear charge, that is, in order of the atomic number. When the elements were placed in order of increasing atomic number, the anomalies in Mendeleev’s table were eliminated. Moseley’s work gave rise to the modern periodic law:

The properties of the elements are a periodic function of their increasing atomic numbers. Tragically for the development of science, Moseley was killed in battle during World War I, only two years later. Moseley’s revised periodic table looked something like this

The Modern Periodic Table seven horizontal rows called periods seven horizontal rows called periods 18 vertical columns called groups or families 18 vertical columns called groups or families groups 1 and 2 and groups are called representative elements groups 1 and 2 and groups are called representative elements groups 3-12 are the transition metals groups 3-12 are the transition metals

Modern Periodic Table cont. elements in any group have similar physical and chemical properties elements in any group have similar physical and chemical properties properties of elements in periods change from group to group properties of elements in periods change from group to group symbol placed in a square symbol placed in a square atomic number above the symbol atomic number above the symbol atomic mass below the symbol atomic mass below the symbol

Metallic Character Metals are found on the left of the table, nonmetals on the right, and metalloids in between Metals are found on the left of the table, nonmetals on the right, and metalloids in between Most metallic element always to the left of the Period, least metallic to the right, and 1 or 2 metalloids are in the middle Most metallic element always to the left of the Period, least metallic to the right, and 1 or 2 metalloids are in the middle Most metallic element always at the bottom of a column, least metallic on the top, and 1 or 2 metalloids are in the middle of columns 4A, 5A, and 6A Most metallic element always at the bottom of a column, least metallic on the top, and 1 or 2 metalloids are in the middle of columns 4A, 5A, and 6A

Metallic Character Metals Metals – malleable & ductile – shiny, lustrous – conduct heat and electricity – lose electrons in reactions Nonmetals brittle in solid state dull electrical and thermal insulators gain electrons in reactions Metalloids Also known as semi- metals Show some metal and some nonmetal properties

Other Important Groups to Know Group IA  alkali metals Group IIA  alkaline earth metals Group VIIIA  noble gases Group VIIA  halogens – “salt formers” Group VIA  chalcogens Group VA  Nitrogen group Group IVA  IVA group Group IIIA  IIIA group

Other Groups s & p block filled  representative elements d block filled  transition metals f block filled  inner transition metals – 4f  lanthanides – 5f  actinides f elements that are naturally occurring  rare earth elements

What are Periodic Trends also called “atomic trends” – take place at the atomic level trends are general patterns or tendencies – they are general not definite – there are exceptions when looking at trends we look for increases & decreases – across  periodic – down  group

Effects on the Trends 1.Nuclear Charge -the “pull” of the nucleus -proportional to the number of protons in an atom -the greater the number of protons, the stronger the nuclear charge (“pull”) -this generally affects periodic trends

Effects on Trends Cont. 2.Shielding - the electron protection from the nuclear “pull” - shield = an energy level of electrons - we are not concerned with single electrons, only energy levels of electrons - these electrons reduce the nuclear pull - affects group trends

Effects on Trends Cont. 3.Stability - where electron arrangement is compared to stable octet (or other special stabilities) - determines if atom gains or loses electrons -can be used to explain anomalies in trends

Trend in Atomic Size Trend in Atomic Size Increases down column Increases down column – valence shell farther from nucleus because of increased shielding Decreases across period Decreases across period – left to right because of the nuclear “pull” – adding electrons to same valence shell – valence shell held closer because more protons in nucleus Illustration Illustration Illustration

Ionic Size cations – lose electrons (positively charged) anions – gain electrons (negatively charged) elements gain or lose e - to become stable – being like noble gases (filled outer sublevel) IA - +1VA - -3 IIA - +2VIA - -2 IIIA - +3VIIA - -1 IVA – shareVIIIA – 0, stable

Ionic Size Cont. GOOD RULE OF THUMB – anions are always larger than their neutral atom – cations are always smaller than neutral atom Across a Period – cations decrease (I-III) because of greater pull on electrons – anions decrease (V-VII) because of less pull on electrons and repulsion of the electrons Down a Group – both cations and anions increase size

Trend in Ionization Energy Trend in Ionization Energy Minimum energy needed to remove a valence electron from an atom Minimum energy needed to remove a valence electron from an atom – 1 mole of electrons in the gaseous state (kJ/mol) The lower the ionization energy, the easier it is to remove the electron The lower the ionization energy, the easier it is to remove the electron – metals have low ionization energies Ionization Energy decreases down the group Ionization Energy decreases down the group – valence electron farther from nucleus Ionization Energy increases across the period Ionization Energy increases across the period – left to right – harder to remove an electron from the atom because of the increased nuclear “pull” Exceptions: Group 3, Group 6 (chalcogens) Exceptions: Group 3, Group 6 (chalcogens)

Harder to Remove More “Stable”

Harder to Remove More “Stable”

Harder to Remove More “Stable”

Ionization Energy Cont. Li + energy  Li + + e - – 1 st ionization = 520 kJ/mol Li + + energy  Li +2 + e - – 2 nd ionization = 7297 kJ/mol Li +2 + energy  Li +3 + e - – 3 rd ionization = 11,810 kJ/mol Notice, each successive ionization energy is greater than the preceding one – there is a greater “pull” between the nucleus and the electron and thus more energy is needed to break the attraction. Examining ionization energies can help you predict what ions the element will form. – easy to remove an electron from Group IA, but difficult to remove a second electron. So group IA metals form ions with a 1+ charge.

Electron Affinity atoms attraction to an electron it is the energy change that accompanies the addition of an electron to a gaseous atom “opposite” of ionization energy (Concept NOT actual trend) Across a Period – electron affinity increases because of increased “pull” Down a Group – electron affinity decreases because the electrons are shielded from the pull of the nucleus Exceptions: Nitrogen Group & Noble Gases

Electronegativity the ability of an atom to attract electrons when the atom is in a compound electron “tug-of-war” very similar to electron affinity Across a Period – increases because of increased pull Down a Group – decreases because of shielding Fluorine – most electronegative element

Increase Decrease

Reactivity Reactivity of metals increases to the left on the Period and down in the column Reactivity of metals increases to the left on the Period and down in the column – follows ease of losing an electron Reactivity of nonmetals (excluding the noble gases) increases to the right on the Period and up in the column Reactivity of nonmetals (excluding the noble gases) increases to the right on the Period and up in the column Reactivity Video Reactivity Video Reactivity Video Reactivity Video

Melting Point Most Melting Points are more complex (Depending mostly on bonding type and structure) Most Melting Points are more complex (Depending mostly on bonding type and structure) However…Melting points of groups 1 and 17 are predictable. However…Melting points of groups 1 and 17 are predictable. Li-454 Li-454 Na-371 Na-371 K-337 K-337 Rb-312 Rb-312 Cs-302 Cs-302 Clearly Group 1 MP’s decrease as you move down….But why? Clearly Group 1 MP’s decrease as you move down….But why? These are metals held together by the attraction between outer delocalized electrons and positively charged ions. This attraction decreases with distance (atomic radius)

Melting Point Group 17? Group 17? F2-54 F2-54 Cl2-172 Cl2-172 Br2-266 Br2-266 I2-387 I2-387 At2-575 At2-575 Clearly Group 17 MP’s INCREASE as you move down….But why? Clearly Group 17 MP’s INCREASE as you move down….But why? These are Not metals….So are held together by the attraction of London Dispersion Forces. This attraction increases with the number of electrons in the molecule.

Practice Which element has a greater ionization energy – Mg or Ba Which element has a greater atomic radius – N or F Which element has a great electron affinity – S or Pb REVIEW

Advanced Periodic Properties

Group Ratios: IA with VIIA = IA with VIA=

Properties: Ag with Halides Halogens with Silver Ion: Silver Ion with Chorine Ion: Silver Ion with Bromine Ion: Silver Ion with Iodine Ion: Ag + (aq) + Cl - (aq) → AgCl(s) Ag + (aq) + Br - (aq) → AgBr(s) Ag + (aq) + I - (aq) → AgI(s) Note: Silver Precipitates also have distinct colors that will be helpful to know: AgI – Pale Yellow AgBr – Cream colored AgCl – White These are good to know because they are commonly used as a test for halide ions. Combine aqueous compound with silver nitrate solution (Often solution is acidified by the addition of nitric acid). Ex: AgNO 3 (aq) + NaCl(aq) → AgCl(s) + NaNO 3 (aq)

Properties: Halogens Melting Point? Increases with size of electron cloud. London Dispersion Forces Halogens with halides – Try Cl 2(aq) with Br - Reactions to KNOW: Halogens with Group 1 metals to produces halides 2Na (s) + Cl 2(g) → 2NaCl (s) Halogens with Halides (Theory only explanation next): Cl 2 (aq) + 2Br - (aq) → Br 2 (aq) + 2Cl - (aq)

Properties: Halogens THE OXIDISING ABILITY OF THE GROUP 7 ELEMENTS (THE HALOGENS) A.K.A. Displacement Reactions Relative Reactivity of elements can be seen by placing them in competition for an extra electron. We are going to look at the reactions between one halogen (chlorine, say) and the ions of another one (iodide ions, perhaps). The iodide ions will be in a solution of a salt like sodium or potassium iodide. The sodium or potassium ions will be spectator ions, and are completely irrelevant to the reaction. Cl 2 (aq) + 2I - (aq) → I 2 (aq) + 2Cl - (aq) The iodide ions have lost electrons to form iodine molecules. They have been oxidized. The chlorine molecules have gained electrons to form chloride ions. They have been reduced. This is obviously a redox reaction in which chlorine is acting as an oxidizing agent. We'll have to exclude fluorine from this descriptive bit, because it is too strong an oxidizing agent. Fluorine oxidizes water to oxygen and so it is impossible to do simple solution reactions with it. Cl 2 (aq) + 2KBr - (aq) → Br 2 (aq) + 2KCl(aq) Br 2 (aq) + 2I - (aq) → I 2 (aq) + 2Br(aq) Colorless (Not Clear) Brown Colorless (Not Clear) Dark Orange/Brown Turns.. well.. darker. Can be tested for Iodine though by adding a non-polar solvent. Chlorine has a greater attraction to the electron due to it’s smaller radius so it will “win”. Taking the electrons, turning into aqueous ions and leaving Bromine What about: 2Br - (aq) + 2I 2 (aq) → ???? Yep… Nothing

Properties: Alkali Metals Melting Point? Decreases with increasing shielding: Less attraction to shared delocalized electrons. Reactivity? Increases with increasing shielding: Less “hold/pull” to retain valence electrons. Alkali with H 2 O - Try the first three (Hint what gas does this reaction produce?) Reactions to KNOW: Alkali Metals with water: 2Li(s) + 2H 2 O(l) → 2LiOH(aq) + H 2 (g) 2Na(s) + 2H 2 O(l) → 2NaOH(aq) + H 2 (g) 2K(s) + 2H 2 O(l) → 2KOH(aq) + H 2 (g) Alkali with halogens - Try the first three combining with Cl 2, Br 2, I 2

Alkali Metals with Halogens: 2Li(s) + Cl 2 (g) → 2LiCl(s) 2Li(s) + Br 2 (l) → 2LiBr(s) 2Li(s) + I 2 (s) → 2LiI(s) 2Na(s) + Cl 2 (g) → 2NaCl(s) 2Na(s) + Br 2 (l) → 2NaBr(s) 2Na(s) + I 2 (s) → 2NaI(s) 2K(s) + Cl 2 (g) → 2KCl(s) 2K(s) + Br 2 (l) → 2KBr(s) 2K(s) + I 2 (s) → 2KI(s)

Properties: Period 3 Oxides Ionic to Molecular: The Oxides we’ll be looking at are: Na 2 O(s), MgO(s), Al 2 O 3 (s), SiO 2 (s), P 4 O 10 (s), SO 3 (g) Acid to Base: Increases with increasing shielding: Less “hold/pull” to retain valence electrons. Reactions with Water: These oxides are known as the highest oxides of the various elements. These are the oxides where the Period 3 elements are in their highest oxidation states. In these oxides, all the outer electrons in the Period 3 element are being involved in the bonding The trend in structure is from the metallic oxides containing giant structures of ions (Ionic Obviously) on the left of the period via a giant covalent oxide (silicon dioxide) in the middle to molecular oxides on the right.

Properties: Period 3 Oxides Acid to Base: The trend is from strongly basic oxides on the left-hand side to strongly acidic ones on the right, via an amphoteric oxide (aluminium oxide) in the middle. WHY??? An amphoteric oxide is one which shows both acidic and basic properties. n.b. For this simple trend, you have to be looking only at the highest oxides of the individual elements. The pattern isn't so simple if you include the other oxides as well. n.b. For the non-metal oxides, their acidity is usually thought of in terms of the acidic solutions formed when they react with water - for example, sulfur trioxide reacting to give sulfuric acid.

Sodium oxide Sodium oxide is a simple strongly basic oxide. It is basic because it contains the oxide ion, O 2-, which is a very strong base with a high tendency to combine with hydrogen ions. Reaction with water Sodium oxide reacts exothermically with cold water to produce sodium hydroxide solution. Depending on its concentration, this will have a pH around 14. Na 2 O(s) + H 2 O(l) → 2Na + (aq) + 2OH - (aq) Reaction with acids As a strong base, sodium oxide also reacts with acids. For example, it would react with dilute hydrochloric acid to produce sodium chloride solution. Na 2 O(s) + 2HCl(l) → 2Na + (aq) + 2Cl - (aq) + H 2 O (l)

Magnesium oxid e Magnesium oxide is again a simple basic oxide, because it also contains oxide ions. However, it isn't as strongly basic as sodium oxide because the oxide ions aren't so free. In the sodium oxide case, the solid is held together by attractions between 1+ and 2- ions. In the magnesium oxide case, the attractions are between 2+ and 2-. It takes more energy to break these. MgO(s) + H 2 O(l) → Mg(OH) 2 (s) Reaction with acids Magnesium oxide reacts with acids as you would expect any simple metal oxide to react. For example, it reacts with warm dilute hydrochloric acid to give magnesium chloride solution. MgO(s) + 2HCl(l) → MgCl 2 (aq) + H 2 O (l) Reaction with water If you shake some white magnesium oxide powder with water, nothing seems to happen - it doesn't look as if it reacts. However, if you test the pH of the liquid, you find that it is somewhere around pH 9 - showing that it is slightly alkaline. There must have been some slight reaction with the water to produce hydroxide ions in solution. Some magnesium hydroxide is formed in the reaction, but this is almost insoluble - and so not many hydroxide ions actually get into solution.

Acid/Base Behavior of Period 3 Oxides Na 2 O(s) + H 2 O(l) → 2Na + (aq) + 2OH - (aq) MgO(s) + H 2 O(l) → Mg(OH) 2 (s) Al 2 O 3 (s) insoluble and no rxn with H 2 O but amphoteric: SiO 2 (s) + 2OH - (aq) → SiO 3 2- (aq) + H 2 O(l) (don’t need to know this equation) P 4 O 6 (l) + 6H 2 O(l) → 4H 3 PO 3 (aq) (Don’t need this one) P 4 O 10 (s) + 6H 2 O(l) → 4H 3 PO 4 (aq) Or P2O5 SO 2 (g) + H 2 O(l) → H 2 SO 3 (aq) (Don’t need this one) SO 3 (g) + H 2 O(l) → H 2 SO 4 (aq)

To Do: Work on Period 3 oxides portion Add Transition metal definition and explanation Color Complex Magnetism