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2 Periodic table Key to understand trends in the periodic table is the concept of effective nuclear charge Z eff. Z eff is the nuclear charge of an atom minus the screening constant, which takes into account the effect of shielding electrons on the charge experienced by the outermost electrons. Connection between the effective nuclear charge Z eff, atomic number Z and screening constant σ is the following: Z eff = Z σ There are several ways to determine the screening constant. The most simple one is to take σ to represent core electrons. For example, Neon has electron configuration of 1s 2 2s 2 2p 6 Z = 10 σ = 2 Z eff = 10 2 = 8 It can be seen that the effective nuclear chargez eff = 8 corresponds to the group number of the element (= 8 for Ne)
3 Write down the electronic configuration of the element as shown below. (1s) (2s, 2p) (3s, 3p) (3d) (4s, 4p) (4d) (4f) (5s, 5p) (5d) Fill the electrons according to Aufbau principle. 1. Any electrons to the right of the electron of interest do not contribute to shielding constant. 2. The shielding constant for each group is formed as the sum of the following contributions: a) All other electrons in the same group as the electron of interest shield to an extent of 0.35 nuclear charge units except 1s group, where the other electron contributes only b) If the group is of the [s, p] type, an amount of 0.85 from each electron (n-1) shell and an amount of 1.00 for each electron from (n-2) and lower shell. c) If the group is of the [d] or [f] type then an amount of 1.00 for each electron from all lying left to that orbital.
4 For example: (a) Calculate effective nuclear charge in Nitrogen for 2p electron. Electronic configuration- (1s 2 ) (2s 2, 2p 3 ). Screening constant, σ = (0.35 4) + (0.85 2) = 3.10 Effective nuclear charge, Z* = Z σ = = 3.90 Calculate effective nuclear charge in Zinc for 4s electron & for 3d electron. Electronic configuration- (1s 2 ) (2s 2, 2p 6 )(3s 2, 3p 6 )(3d 10 )(4s 2 ). For 4s electron, σ = (0.35 1) + ( ) + (1 10) = Z* = Z σ = = 4.35 For 3d electron, σ = (0.35 9) + (1 18) = Z* = Z σ = = 8.85
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6 Periodic Table Atomic radius Z eff increases Principal quantum number increases
7 Periodic Table Ionization energy is the energy required to remove an electron from a neutral gaseous atom Few exceptions!
8 Periodic Table When moving from 2A to 3A group the removable electron in 3A is on np orbital. Thus, the energy of the electron to be removed is higher in the 3A group than in the 2A group. Difference between 5A and 6A (see below) [jalokaasu] Noble gas ns np x np y np z Within the group as we move down the principal quantum number increases and thus ionization energy decreases
9 Periodic Table Electron affinity is the change in the energy (kj/mol) of a neutral gaseous atom when one electron is added Obs! Different from electronegativity! Z eff n
10 Periodic Table 1A [jalokaasu] Noble gas ns np x np y np z 2A [jalokaasu] Noble gas ns np x np y np z
11 Periodic Table 4A Noble gas [jalokaasu] ns np x np y np z 5A [jalokaasu] Noble gas ns np x np y np z Third exception occurs inside the groups, because the second period elements are much smaller than the following ones. Thus smaller than expected EA.
12 Periodic Table Electronegativity describes the tendency of an atom to withdraw electrons to it. This property increases straightforwardly from left to right in a given period. It also decreases as we move down in a group. There are no exceptions in these trends (within the main group elements)
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14 Periodic Table
15 Periodic Table Uniqueness principle states that the chemistry of the second-period elements (Li, Be, B, C, N, O, F, ja Ne) is quite often significantly different from their congeners Three main reasons: (1) small size, (2) enhanced ability to form π- bonds, and (3) unavailability of d-orbitals.
16 Small size Bonds formed by 1A and 2A elements have more covalent character than one would expect
17 Enhanced ability to form π-bonds Small size of the second period elements enhances their ability to form π-bonds among themselves as with other elements. π-bonds form when two parallel p-orbitals overlap. The larger the atom the smaller is the strenght of the π-bond Thus rich chemistry by using sigma and π-bonds (C=C,C C,O=O,C=O,C O,N N jne.) ethylene
18 Unavailability of d-orbitals Unavailability of d-orbitals influences the way the elements bond (no extended octet). Thus, carbon can form only CF 4 compound whereas silicon can form SiF 6 2- compound.
19 Uniqueness principle (a)small size of the atoms -> high covalent character in bonding (b)enhanced probability to form π-bonds (c)lack of d-orbitals
20 Diagonal effect This effect is seen with first three main group elements. It consists of relations between litium and magnesium, beryllium and aluminium as well as boron and silicon. Three main reasons behind similar behavior are: (1) ionic radius, (2) charge density and (3) electronegativity.
21 Diagonal effect Similar ionic radius means that elements can substitute each other in different compounds easily. Electronegativity is of similar magnitude -> similar bonding, for example Be-X and Al-X (X is a typical non-metallic element). Charge density also influences the nature of the chemical bond -> again similar bonding behaviour.
22 The inert pair effect The effect is strongest with the second and third row transition elements. The effect is manifested in the fact that their 5s 2 and 6s 2 valence electrons are less reactive than one would expect based on effective nuclear charge, atomic size and ionization energy. This is shown with following elements: In, Tl, Sn, Pb, Sb, Bi and also Po to some degree) in that they do not exist with their maximum oxidation numbers (+4), but with the value of 2 (thus two less than their maximum).
23 Lanthanide contraction This contraction can be taken to indicate that d- and f-orbital electrons not only do not shield each other from the nucleus, but they do not shield succeeding electrons from the nucleus either very well. Thus, the inert pair effect can be explained by the fact that 4s, 5s, and 6s electrons feel higher effective nuclear charge and have higher ionization energies.
24 Different regions in the periodic table
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26 Summary
27 Transition metals Small atomic volumes (electons are added to d-orbitals) -> high densities-> close packed structures = high number of bonds-> high mp and bp Similar atomic radiuses Multiple oxidation states (4s and 3d orbital energies are close to each other) Magnetic properties (Co, Fe and Ni) Many transition metals form colored compounds (d-d- transitions) no d 0 or d 10 (incomplete d-orbitals)
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30 Formation of intermediate compounds Intermediate compounds that form between dissimilar materials can be: - electrochemical compounds ( normal valence compounds) - size factor compounds (often solid solutions) - electron compounds ( metallic bonding ) Rule of thumb:stoichiometry of an intermediate phase increases as the electronegativity difference (X A -X B ) between A and B increases. When (X A -X B ) > 1.7, the bonding is predominantly ionic and the structure of the compound is simple (AB, AB 2, A 2 B 3 ) because of the electroneutrality requirement.
31 Formation of intermediate compounds In ionic crystals positive and negative ions are packed to maximize the electrostatic attractive forces and to minimize the electrostatic repulsion. The most stable configuration is thus the one that: (i) Fulfills electroneutrality, (ii) Maximizes electrostatic attraction, (iii) Minimizes electrostatic repulsion and (iv) Satisfies the possible covalent directionality requirement Based on these assumptions Pauling has given four rules to explain the formation of different types of ionic compounds.
32 Formation of intermediate compounds Ionic crystals are thought to consist of co-ordination polyhedra which are formed by planes going through the centers of ANIONs, which contain one CATION and are tightly connected to each other in crystals The co-ordination number of the central cation can be 2, 3, 4, 6 or 8
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34 Formation of intermediate compounds Polyhedra can be linked to each other from: (a) corners, (b) edges or (c) faces.
35 4 rules In ionic crystal where distance between cation and anion is r c + r a the coordination number of the cation is determined by the ratio (r c /r a ). According to the rule cation has to be always in contact with the surrounding anions. KL (rk/ra) (rk/ra) kr Polyedri = kolmio = tetraedri = oktaedri = kuutio
36 4 rules The stability of the polyhedra connections decreases in the order: corners, edges and faces, as cations are moved closer to each other. The effect is stronger when the charge of the cation is high and the radius small and vice versa Thus, by maximizing the co-ordination number (within the limits imposed by the radius of cation) the electrostatic energy is minimized
37 4 rules In a stable structure the total strength of the bonds reaching an anion from surrounding cations (= formal charge of the cation divided by its co-ordination number) should be equal to the charge of the anion. For example: Si has valency of 4, co-ordination is tetrahedral (4) bond strength is 1. Structure of Si 2 O 7 is as follows: two Si bonds are shared by one oxygen anion Si bond stregth is 1 2x1 = 2 = charge of the oxygen anion) Polyhedra formed about cations of low co-ordination number and high charge tend especially to be linked by corners
38 NaCl-structure (B1) Two fcc lattices with origins at (0,0,0) and (½,0,0) So: Anions form a fcc lattice and cations are in octahedral positions. CN = 6, polyhedra are octahedrons, joined by edges and the radius ratio is Following compounds have NaCl structure: TiC, NbC, VC, ZrC, V4C3, TiN, VN, CrN, UN, TiB, MgO, TiO, ZrO, FeO, PbS, MgSe,PbSe, etc.
39 Other common crystal structures CsCl type ( B2 ): Two primitive cubic lattices. Anions form cubic lattice and cations are in octahedral positions. CN = 8. Stable in the range of radius ratios of (CN = 8, polyhedra is a cube) Wurtzite structure (B4): In beryllium oxide the radius ratio is 0.25 requiring tetrahedral coordination of four oxygen about each beryllium ion. The bond strenght is then equal to one-half so that each oxygen must be coordinated with four cations. These requirements can be achieved with hexagonal packing of the large oxygen ions, with half of the tetrahedral interstices filled with beryllium cations so as to achieve maximum cation separation. (CN = 4, polyhedra is a tetrahedron)
40 Substitutional and interstitial solid solutions Especially metals can form solid solutions (subs or inter.) In substitutional solid solutions solute atoms replace solvent atoms in their lattice. Solubility can range from fractions of atomic percentage up to 100 at-% sc. complete solid solubility Complete solid solubility is possible only if the following 4 Hume-Rotheryn conditions are fulfilled: 1. Difference in the atomic radii cannot be more than 15 % 2. Elements have to have identical crystal structures, 3. Electronegativity difference cannot be too large 4. Elements have to have the same valence
41 Substitutional solubility Alkuaine Atomisäde Kiderakenne Elektronegatiivi Valenssi (nm) -suus Kupari FCC Sinkki HCP Lyijy FCC , +4 Pii DC Nikkeli FCC Alumiini FCC Beryllium HCP Systeemi Atomisäteiden ero % Elektronegatiivi -suusero Ennakoitu liukoisuusaste Havaittu max. liukoisuus (at- %) Cu-Zn Korkea 38.3 Cu-Pb Erittäin pieni 0.1 Cu-Si Vaatimaton 11.6 Cu-Ni Erittäin korkea 100 Cu-Al Vaatimaton 19.6 Cu-Be Vaatimaton 16.4
42 Interstitial solutions 1) Elements with same crystal structure=> interstitial solubility decreases in a period from left to right 2) When the radius of interstital atoms decrease their solubility to a given crystal structure increases (many exceptions) 3) Interstitial solubility changes non-continuously in allotropic transitions (typically the phase stable at higher temperatures has higher solubility) 4) Solubility of carbon and nitrogen is higher to fcc- and hcp-metals than to bcc metals
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44 Summary Effective nuclear charge Z eff Uniqueness principle Diagonal effect Different regions in the periodic table Formation of intermediate phases Substitutional and interstitial solutions
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