lecture notes on inorganic chemistry

lecture notes on introduction to inorganic chemistry and Modern Inorganic Chemistry pdf free download
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Modern Inorganic Chemistry Inorganic Materials Metal ions in Biology B Bu uiillt t o on n p pr riin nc ciip plle es s e es st ta ab blliis sh he ed d llo on ng g llo on ng g a ag go o Let us go through a small tour of some examples & current topics which make inorganic chemistry interesting and meaningful The goals are … 1. To give an overview of the basic trends in Inorganic Chemistry 2. Interpret collection of data in terms of common theory involved 3. Rationalize chemical and physical properties in terms of established theories. 4. ApplicationsCourse Coverage 1. Periodic Table (trends, anomalies, application, nomenclature) 2. Extraction of metals from ores, purification, etc. 3. Transition Metal Chemistry (complexes, bonding, magnetism) 4. Metal ions in biology 5. Organometallic Chemistry & Catalysis Recommended Text Books: (1) Concise Inorganic Chemistry - J.D. Lee (2) Inorganic Chemistry-D.F. Shriver, P.W. Atkins, C.H. Langford (3) Chemistry: Principles and Properties, M. J. Sienko, & R.A. Plane (4) Some class notes available at: www.iitb.ac.in/rmv www.chem.iitb.ac.in/rmv/ Topic 1 Periodic Table & Periodic Properties rd Ref. Chapter 1, Inorganic Chemistry, Shriver & Atkins, 3 EditionThe periodic table is the most important tool in the chemist’s toolbox Periodic Table and Periodicity What is so special about it? •Helps us to bring order into inorganic chemistry •Concept of chemical periodicity –central to study of inorganic chemistry •It systematizes and rationalizes –chemical facts –predict new ones –suggest fruitful areas for further research Periodic Law: The properties of chemical elements are not arbitrary, but depend upon the electronic structure of the atom and vary with the atomic number in a systematic way. Therefore: periodic table may be useful for • the interpretation of the periodic law in terms of the electronic structure of atoms • the systematization of trends in physical & chemical properties, and to detect possible errors, anomalies, & inconsistencies • the prediction of new elements & compounds and to suggest new areas of research Dmitri Mendeleev 1869 : Proposed his periodic law that “the properties of the elements are a periodic function of their atomic weights”. He published several forms of periodic table, one containing 63 elements. 1871 : Mendeleev modified and improved his tables and predicted the discovery of 10 elements (now known as Sc, Ga, Ge, Tc, Re, Po, Fr, Ra, Ac and Pa). He described with amazing prescience the properties of four of these (Sc, Ga, Ge, Po). He did not predict the existence of noble gases and the 1834 - 1907 number of lanthanide elementsHistory of the Periodic Table • 1894-8: Lord Rayleigh, W. Ramsay and M. W. Travers detected and isolated the noble gases (He, Ne, Ar, Kr, Xe). • 1913: N. Bohr explained the form of the periodic table on the basis of his theory of atomic structure and showed that there could be only 14 lanthanide elements. • 1913 : H. G. J. Moseley observed regularities in the characteristic X-ray spectra of the elements; he thereby discovered atomic numbers Z and provided justification for the ordinal sequence of the elements. • 1940: E. McMillan and P. Abelson synthesized the first 93 transuranium element Np. Others were synthesized by G. T. Seaborg during the next 15 years. Glenn T. Seaborg After co-discovering 10 new elements, in 1944 he moved 14 elements out of the main body of the periodic table to their current location below the Lanthanide series. These became known as the Actinide series. 1912 - 1999Glenn T. Seaborg He is the only person to have an element named after him while still alive. "This is the greatest honor ever bestowed upon me - even better, I think, than winning the Nobel Prize." 1912 - 1999IUPAC Nomenclature of elements with atomic number above 100 • Digit Name Abbreviation • • 0 nil n • 1 un u • 2 bi b • 3 tri t • 4 quad q • 5 pent p • 6 hex h • 7 sept s • 8 oct o • 9 enn e E. g., 114 Un-un-quad-ium Uuq 118 Un-un-oct-ium UuoBuilding Up the Periodic Table: The Basis 1. Various quantum numbers 1 H 1s 2. Hund's Rule: 2 When more than one orbital has the same He 1s energy (e.g. p , p , p ), electron x y z 2 1 occupy separate orbitals and do so Li 1s 2s with parallel spins. …… 3. Pauli (Exclusion) Principle 2 2 5 F 1s 2s 2p No more than two electrons shall occupy 2 2 6 a single orbital and, if two do occupy Ne 1s 2s 2p a single orbital, then their spins must be paired. or "no two electrons can have the same four quantum numbers" 4. The order of orbitals for a given quantum number depends on Shielding Effects (Z) Penetration of orbitals Shielding 2 2. Energy of an electron in an atom is a function of Z /n Nuclear charge (Z) increases more rapidly than principal quantum no. (n). Therefore continuous increase expected in IE with increase in atomic number. On the other hand -1 IE H 1312 KJ mol -1 Li 520 KJ mol Why? Reasons: Average distance of 2s electron is greater than that of 1s. 2 The 2s electron is repelled by inner core 1s electrons, so that the former is much more easily removed – shielding or screening of the nucleus by inner electrons. Valence electron ‘sees’ only part of the total charge Effective Nuclear charge Z = Z – σ σ σ σ (σ σ σ σ = Screening Constant)How to determine Z? If the electron resides in s or p orbital 1. Electrons in principal shell higher than the e- in question contribute 0 to σ σ σ σ 2. Each electron in the same principal shell contribute 0.35 to σ σ σ σ 3. Electrons in (n-1) shell each contribute 0.85 to σ σ σ σ 4. Eelectrons in deeper shell each contribute 1.00 to σ σ σ σ Example: Calculate the Z for the 2p electron 2 2 5 Fluorine (Z = 9) 1s 2s 2p Screening constant for one of the outer electron (2p): 6 (six) (two 2s e- and four 2p e-) = 6 X 0.35 = 2.10 2 (two)1s e- = 2 X 0.85 = 1.70 σ σ σ σ = 1.70+2.10 = 3.80 Z = 9 - 3.80 = 5.20 What is Z for 1s electron? If the e- resides in a d or f orbital 1. All e-s in higher principal shell contribute 0 2. Each e- in same shell contribute 0.35 3. All inner shells in (n-1) and lower contribute 1.00Effective nuclear charge Z increases very slowly down a group for the “valence” i.e. outermost orbital e.g. H 1.0 Li 1.3 Na 2.2 Valence configuration same K 2.2 Rb 2.2 Cs 2.2 …..but increases rapidly along a period Li Be B C N O F Ne 1.3 1.95 2.6 3.3 3.9 4.6 5.2 5.9 1 2 1 2 3 4 5 6 2s 2s 2p 2p 2p 2p 2p 2pPenetration of Atomic Orbitals • Energy levels for hydrogen show no distinction between energies of different types of orbitals (s, p, d, f) in a given quantum level. • Li nucleus : 3 protons; it has +3 nuclear charge 2+ If Li has only one electron (Li ), this electron would reside 1s orbital; strong attraction to nucleus. Therefore, size of this 1s orbital is smaller than it is for hydrogen (+1 nuclear charge) • Li with 2e-; Some repulsion between electrons, but size of 1s still smaller than it has for hydrogen • Uncharged Li has 3 electrons. Two electrons in 1s; 3rd electron in 2nd quantum shell which is larger than the 1s shell. • Therefore, expect 3rd e- to be attracted by +3 nuclear charge and repelled by two 1s electrons and hence ‘see’ an effective nuclear charge of 1.3. • If 3rd electron could penetrate close to nucleus, effective nuclear charge would be 1.3 Penetration of orbitals The penetration potential of an orbital varies as: ns np nd nf The energy of the orbitals for a given n varies as: ns np nd nfConsiderations of principles such as penetration and shielding have enabled atomic orbitals to be arranged in rough order of increasing energy (order of filling of orbitals). How do you fill electrons ? 1 H 1s 2 He 1s 2 1 Li 1s 2s …… 2 2 5 F 1s 2s 2p 2 2 6 Ne 1s 2s 2p 1 Na Ne3s 23p Al Ne3s 1 …. 2 6 Ar Ne3s 3p Now what next ? 1 19 K Ar4s 2 20 Ca Ar4s 3 0 2 1 Sc (at. No. 21) Ar3d 4s or Ar3d 4s - Is this correct? then? Sc NO; Why?• For most of the d-block, both spectroscopic determination of the ground states and computation show that it is advantageous to occupy higher energy 4s orbitals, even if 3d is lower (Why?) • Two electrons present in the same d-orbital repel each other more strongly than do two electron in a s-orbital . Therefore, occupation of orbitals of higher energy can result in a reduction in the repulsion between electrons that would occur if the lower-energy 3d orbitals were occupied. • It is essential to consider all contributions to the energy of a configuration, and just not one-electron orbital energies • Spectroscopic data show that GS configurations of d- n 2 block elements are of the form 3d 4s , with 4s orbitals full occupied. 1 2 Sc (at. No. 21) is Ar3d 4s This order is followed in most cases - but not always Two atomic configurations do not follow the nuclear sequence of filling of orbitals 5 1 4 2 Z = 24 Cr Ar 3d 4s ; not Ar 3d 4s 10 1 9 2 Z = 29 Cu Ar 3d 4s ; not Ar 3d 4s As atomic number increases, energy of 3d orbitals decrease relative to both 4s and 4p; at z = 29, energy of 3d becomes much lower than 4s, hence order of filling 3d 4s 4pFilling of Orbitals (Aufbau) • Transition series: filling order: 4s, 3d • removal order (cation formation): 4s, 3d (not 3d, 4s) 2 2 e.g. Ti Ar 4s 3d 2+ 2 2 • Ti Ar 3d (not Ar 4s ) Why? • When 2 electrons are removed, regardless of where they come from, all atomic orbitals contract (Z increases because of net ionic charge and reduced shielding) • Contraction has a small effect on 4s orbital which owes its low energy to its deep penetration • Contraction in d orbital causes a considerable decrease in energy – this decrease is evidently enough to lower the energy of 3d well below 4s Periodic Table – Lecture 2 We will look at the following periodic trends in this lecture: •Atomic size (radius), •Ionic size (radius), •Atomic volume •Ionization energy, •Electron affinity •Electronegativity.Atomic Radius The METALLIC RADIUS is half of the experimentally determined distance between the nuclei of nearest neighbors in the solid The COVALENT RADIUS of a non-metallic element is half of the experimentally determined distance between the nuclei of nearest neighbors in the solid The IONIC RADIUS of an element is related to the distance between the nuclei of neighboring cations and anions 2- Ionic radius of O is 1.40 Å; 2+ What is the ionic radius for Mg ? Measure the Mg-O distance in MgO and subtract 1.40 Å Atomic Radius In a period, left to right 1. n (number of shells) remain constant. 2. Z increases (by one unit) 3. Z increases (by 0.65 unit) 4. Electrons are pulled close to the nucleus by the increased Z So atomic radius decreases with increase in atomic number (in a period left to right) In a group, top to bottom 1. n increases 2. Z increases 3. No dramatic increase in Z - almost remains constant So atomic radius increases moving down the groupDecreases with increase in atomic number in a period left to right Increases moving along a group top to bottomMetallic Radius Metallic radii in the third row d-block are similar to the second row d-block, but not larger as one would expect given their larger number of electrons Lanthanide Contraction f-orbitals have poor shielding properties; low penetrating power. All anions are larger than their parent atoms; The cations are smallerMolar Atomic Volume  Volume per mole of atoms of the element Density, melting point, etc. depend on atomic volume; related to compactness or the lack of it Ionisation Energy (IE) The minimum energy needed to remove an electron from a gas phase atom Depends on (a) Size of the atom - IE decreases as the size of the atom increases (b) Nuclear Charge - IE increases with increase in nuclear charge (c) The type of electron - Shielding effect st 1 IE H 1312 KJ mol-1 Li 520 KJ mol -1 Reasons 1. Average distance of 2s electron is greater than that of 1s 2. Penetration effect 3. Electronic configurationOn moving down a group 1. nuclear charge increases 2. Z due to screening is almost constant 3. number of shells increases, hence atomic size increases. 4. there is a increase in the number of inner electrons which shield the valence electrons from the nucleus Thus IE decreases down the group On moving across a period 1. the atomic size decreases 2. nuclear charge increases Thus IE increases along a periodAccount for the decrease st in 1 IE between P and S