HYBRIDIZATION IN SQUARE PLANER COMPLEXS TOPIC:-HYBRIDIZATION IN SQUARE PLANER COMPLEX. COURES :-M.Sc(Hons)CHEMISTR

ABSTRACT HYBERDIZATION:-The phenomenon of mixing of orbitals of the same atom with slight difference in energies so as to redistribute their energies and give new orbitals of equivalent energy and shape. The new orbitals which get formed are known as hybrid. TYPES OF HYBERDIZATION:-These are:- SP HYBERDIZATION SP2 HYBERDIZATION SP3 HYBERDIZATION (Regular tetrahedron geometry) SP3D HYBERDIZATION (Trigonal bipyramidal geometry ) SP3D2 HYBERDIZATION (Octahedral geometry ) SP3D3 HYBERDIZATION (Pentagonal bipyramedal geometry DSP2 HYBERDIZATION (Square planer geometry )

INTRODUCTION Hybridization and the LE Model of Bonding — Lewis structures of molecules — prediction of geometry of molecules — hybrid orbitals (sp3, sp2, sp, dsp3, d2sp3) — interpretation of structure and bonding Molecular Orbital Model of Bonding in Molecules — molecular orbital diagrams — bond order — magnetism Molecular Spectroscopy — Electronic Spectroscopy — Vibrational/Rotational Spectroscopies — Nuclear Magnetic Resonance (NMR) Spectroscopy

Hybridization and the LE Model of Bonding — Assume bonding involves only valence orbitals — Methane, CH4: Isolated atoms Valence orbitals H 1s1 C 2s22p2 (2p: 2px, 2py, 2pz) H atoms in CH4 will use 1s orbitals Of the two types of orbitals (2s and 2p) Which will C atoms use for bonding in CH4? — If both are used: 2 different types of C-H bonds (Contrary to experimental facts) — Neither of the “native” atomic orbitals of C atoms are used; instead, new hybrid orbitals are used.

Hybridization of atomic orbitals The mixing of the “native” atomic orbitals to form special orbitals for bonding is called hybridization. The 4 new equivalent orbitals formed by mixing the one 2s and three 2p orbitals are called sp3 orbitals. The carbon atom is said to undergo sp3 hybridization, i.e. is sp3 hybridized. Energy-level diagram showing the sp3 hybridization Energy 2p hybridization sp3 2s Orbitals in C in CH4 molecule Orbitals in isolated C atom

Native 2s and three 2p atomic orbitals characteristic of a free carbon atom are combined to form a new set of four sp3 orbitals.

Energy-level diagram showing the formation of four sp3 orbitals

Sp2 Hybridization Consider ethylene C2H4 molecule Lewis structure — 12 valence e-s in the molecule — What orbitals do the carbon atoms use to bond in ethylene? — 3 effective electron pairs around each carbon VSEPR model predicts a trigonal planar geometry sp3 orbitals with tetrahedral geometry and 109.5o angles will not work here. 120o angles

The plastics shown here were manufactured with ethylene.

The hybridization of the s, px, and py atomic orbitals results in the formation of three sp2 orbitals centered in the xy plane.

Energy-level diagram of sp2 hybridization 1 2s orbital + 2 2p (px, py) orbitals 3 sp2 orbitals Energy-level diagram of sp2 hybridization E Energy 2p 2p hybridization sp2 2s C atom orbitals in ethylene Orbitals in isolated C atom Un-hybridized pz orbital Carbon uses the sp2 hybridized orbitals for forming sigmal (σ) bonds within the plane The remaining 2pz orbital is used for forming the pi (π) bond. Note that the double bond consists of one σ and one π bond.

When one s and two p oribitals are mixed to form a set of three sp2 orbitals, one p orbital remains unchanged and is perpendicular to the plane of the hybrid orbitals.

The sigma bonds in ethylene.

A carbon-carbon double bond consists of a sigma bond and a pi bond.

(a) The orbitals used to form the bonds in ethylene (a) The orbitals used to form the bonds in ethylene. (b) The Lewis structure for ethylene.

.. .. Other sp2 hybridized carbon atoms An atom surrounded by 3 effective electron pairs uses sp2 hybridized orbitals for bonding. Example H2CO formaldehyde .. Lewis Structure .. — 12 valence electrons — 3 effective pairs around C Sp2 hybridized orbitals are used to form the C-H bonds and the C-O σ bond, the un-hybridized 2pz orbital is used to form the C=O π bond.

O=C=O sp Hybridization Carbon in carbon dioxide, CO2 uses another type of hybridization (rather than sp2 or sp3) ‧‧ ‧‧ O=C=O ‧‧ 2 hybrid orbitals required to meet the 180° (linear) geometry requirement are sp orbitals. 2 effective pairs around C atom sp hybrid orbitals 3 effective pairs around O atom sp2 orbitals 2p 2p sp Energy Hybridization 1s Orbitals in sp hybridized orbitals in CO2 Orbitals in a free C atom

When one s orbital and one p orbital are hybridized, a set of two sp orbitals oriented at 180 degrees results.

The hybrid orbitals in the CO2 molecule

(a) Orbitals predicted by the LE model to describe (b) The Lewis structure for carbon dioxide

Other Examples of sp Hybridization N2 molecule N atom: 2s22p3 ：N N： Lewis Structure 2 effective pairs around each N atom in the Lewis structure – Linear (180°) geometry – 2 sp orbitals for each N atom: ．1 sp orbital for forming the σ bond ．1 sp orbital for holding the lone pair The remaining un-changed 2p orbitals are used to form the 2 π bonds. Each triple bond consists of one σ and two π bonds.

(a) An sp hybridized nitrogen atom (b) The s bond in the N2 molecule (c) the two p bonds in N2 are formed when

Consider the bonding in phosphorous pentachloride PCl5 dsp3 Hybridization Consider the bonding in phosphorous pentachloride PCl5 ： ： ： ： ： ： ： Lewis structure ： ： (assuming d-orbital participation) ： ： ： ： ： ： – 40 valence electrons in the molecule – 5 electron pairs around the central atom P ．VSEPR predicts trigonal bipyramidal geometry – one 3d orbital one 3s orbital three 3p orbital a set of 5 dsp3 hybrid orbitals oriented in a trigonal bipyramidal configuration – the Cl atoms in PCl5 use sp3 orbitals to form the P-Cl bonds and to hold the lone pairs In general, when there are 5 effective pairs around an atom it uses dsp3 orbitals.

A set of dsp3 hybrid orbitals on a phosphorous atom

The orbitals used to form the bonds in the PCl5 molecule

Arsenic pentafluoride Other Examples of dsp3 Hybridization Triiodide ion I3- ： ： ： Lewis structure [ I – I – I ]- ： ： ： ： ： ： Arsenic pentafluoride AsF5 ： d2sp3 hybridization ： ： ： ： ： ： ： ： Sulfur hexafluoride, SF6 ： ： – 48 valence electrons – 6 effective pairs around S atoms – VSEPR model predicts Octahedral geometry The 6 pairs lead to d2sp3 hybridization of s atom, forming a set of 6 octahedrally oriented d2sp3 orbitals. ： ： ： ： ： ： ：

An octahedral set of d2sp3 orbitals on a sulfur atom

The relationship among the number of effective pairs, their spatial arrangement, and the hybrid orbital set required

The relationship among the number of effective pairs, their spatial arrangement, and the hybrid orbital set required (cont’d)

Turning to Square Planar Complexes Most convenient to use a local coordinate system on each ligand with y pointing in towards the metal. py to be used for s bonding. z being perpendicular to the molecular plane. pz to be used for p bonding perpendicular to the plane, p^. x lying in the molecular plane. px to be used for p bonding in the molecular plane, p|.

ML4 square planar complexes ligand group orbitals and matching metal orbitals

ML4 square planar complexes MO diagram π- bonding s-only bonding

SQUARE PLANER MOLECULE GEOMETRY Idealized structure of a compound with square planar coordination geometry. The square planar molecular geometry in chemistry describes the stereochemistry (spatial arrangement of atoms) that is adopted by certain chemical compounds .As the name suggests, molecules of this geometry have their atoms positioned at the corners of a square on the same plane about a central atom.

STRUCTURE OF SQUARE PLANER MOLECULE

Molecular Geometry bond length, angle determined experimentally Lewis structures bonding VSEPR geometry Valence Shell Electron Pair Repulsion octahedron 90o bond angles small groups big groups trigonal bipyramid equatorial 120o axial 180o

tetrahedron 109.5o trigonal planar 120o linear 180o geometry apply to Chemistry

.. linear 180o BeCl2 valence e- = 2 + (2 x 7) = 16e- fewer than 8e- Cl two valence pairs on Be bonding e- linear molecule

.. .. linear 180o CO2 valence e- = 4 + (2 x 6) = 16e- C O C O two valence pairs on C ignore double bonds single and double bonds same molecular geometry linear molecular shape linear

.. : .. : .. : trigonal planar 120o SO2 valence e- = 6+ (2 x 6) = 18e- three valence pairs on S two bonding pairs one lone pair molecular geometry trigonal molecular shape bent < 120o

tetrahedral 109.5o CH4 valence e- = 4+ (4 x 1) = 8e- C H four valence pairs on C 109.5o molecular geometry tetrahedral molecular shape tetrahedral

: tetrahedral 109.5o NH3 valence e- = 5+ (3 x 1) = 8e- N H four valence pairs on N three bonding pairs one lone pair molecular geometry tetrahedral molecular shape trigonal pyramid < 109.5o

: tetrahedral 109.5o H2O valence e- = 6+ (2 x 1) = 8e- O H four valence pairs on O two bonding pairs two lone pair molecular geometry tetrahedral molecular shape bent < 109.5o

.. bipyramidal 120o and 1800 PCl5 P Cl valence e- = 5+ (5 x 7) = 40e- five valence pairs on P 180o 90o molecular geometry bipyramidal molecular shape bipyramidal 120o

: .. bipyramidal 120o and 1800 SF4 valence e- = 6+ (4 x 7) = 34e- S F five valence pairs on S four bonding pairs one lone pair < 180o molecular geometry bipyramidal molecular shape seesaw

: .. bipyramidal 120o and 1800 ClF3 valence e- = 7+ (3 x 7) = 28e- Cl five valence pairs on Cl three bonding pairs two lone pair molecular geometry bipyramidal molecular shape T 90o 180o

: .. bipyramidal 120o and 1800 ICl2- valence e- = 7+ (2 x 7) + e- five valence pairs on I two bonding pairs three lone pair on I molecular geometry bipyramidal molecular shape linear

: .. octahedral 90o BrF5 valence e- = 7+ (5 x 7) = 42e- Br F six valence pairs on Br five bonding pairs one lone pair molecular geometry octahedral molecular shape square pyramidal

: .. octahedral 90o XeF4 valence e- = 8+ (4 x 7) = 36e- Xe F six valence pairs on Xe four bonding pairs two lone pair molecular geometry octahedral molecular shape square planar

SUBSTITUTION IN SQUARE PLANER Substitution at Square Planar Metal Complexes Examples of Square Planar Transition Metal Complexes: Ni(II) (mainly d8) Rh(I) Pd(II) Ir(I) Pt(II) Au(III) General Rate Law: Factors Which Affect The Rate Of Substitution 1. Role of the Entering Group 2. The Role of The Leaving Group 3. The Nature of the Other Ligands in the Complex 4. Effect of the Metal Centre

SUBSTITUTION IN SQUARE PLANER

GRAPH OF SQUARE PLANER SUBSTITUTION

REFERENCE ^ G. L. Miessler and D. A. Tarr. Inorganic Chemistry, 3rd Ed., Pearson/Prentice Hall.. Miessler and Tarr(inorgnic chemistry)

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