Chemistry, Paper II

Fluorine atoms contribute 4 electrons to form 4 $S-F$ sigma bonds. Lone Pairs: Calculated as $\frac{1}{2}(V - X)$ , where $V=6$ (valence electrons) and $X=4$ (bonding electrons), resulting in $(6-4)/2 = 1$ lone pair. Steric Number: 4 $4 \text{ (bonding)} + 1 \text{ (lone pair)} = 5$ . Hybridization: $sp^3d$ . According to VSEPR theory, a steric number of 5 corresponds to a trigonal bipyramidal electron geometry. To minimize electron-pair repulsion, the lone pair occupies an equatorial position rather than an axial position, as the equatorial site provides larger bond angles (120° vs 90°), reducing the repulsion between the lone pair and bonding pairs. Resulting Geometry: The displacement of axial fluorine atoms by the lone pair results in a see-saw (or distorted tetrahedral) molecular shape. The bond angles are compressed slightly from ideal values, typically observed at approximately < $101.6^\circ$ (axial) and < $173^\circ$ (equatorial) due to the greater spatial requirement of the lone pair. (a) vi) IF5 Molecular Geometry of $IF_5$ (Iodine Pentafluoride) According to the VSEPR theory , the shape of $IF_5$ is determined by the total number of electron pairs around the central iodine atom. 1. Electronic Configuration and Hybridization Valence Electrons: Iodine ( $I$ ) contributes 7 electrons, and five fluorine atoms ( $F$ ) each contribute 1, totaling 12 electrons (6 electron pairs). Bond Pairs: 5 (associated with five $I-F$ single bonds). Lone Pairs: 1 ( 6 − 5 = 1 6 - 5 = 1 6 − 5 = 1 ). Hybridization: The central iodine atom undergoes $sp^3d^2$ hybridization to accommodate 6 electron domains. 2. Geometry and Shape Electron Geometry: Octahedral, as there are 6 electron pairs arranged to minimize repulsion. Molecular Shape: Square Pyramidal . Bond Angles: Due to the greater repulsive force exerted by the single lone pair, the equatorial $F-I-F$ bond angles are compressed slightly below $90^\circ$ , causing the axial fluorine to tilt away from the lone pair. The lone pair occupies one of the positions of the octahedron, distorting the symmetry and resulting in a square pyramidal structure where the iodine atom sits slightly above the plane of the four equatorial fluorine atoms. 6 a) Write IUPAC names of the following complexes:

Cr-OH-Cr(NH 3 ) 5 ]Cl 5 is a binuclear coordination compound featuring a bridging hydroxo ligand. Following IUPAC guidelines for bridging ligands and homonuclear centers, the name is derived as follows: Bridging Ligand: The $\mu$ -hydroxido group connects the two chromium centers. Coordination Sphere: Each chromium (Cr) is bonded to five ammine (NH 3 ) ligands. These are named as pentaammine. Oxidation State Calculation: Let the oxidation state of Cr be $x$ . 2 x + 1 ( − 1 ) + 10 ( 0 ) + 5 ( − 1 ) = 0 2x + 1(-1) + 10(0) + 5(-1) = 0 2 x + 1 ( − 1 ) + 10 ( 0 ) + 5 ( − 1 ) = 0 2 x − 6 = 0 $2x - 6 = 0 \implies x = +3$ Each Chromium center is in the +3 oxidation state. Naming Sequence: The prefix $\mu$ -hydroxido is placed before the ligand/metal component. Since both centers are identical, the prefix "di-" is used for the complex unit. Final IUPAC Name $\mu$ -hydroxido-bis[pentaamminechromium(III)] chloride Note: The five chloride ions outside the coordination sphere are named as "chloride" at the end of the nomenclature sequence, with the stoichiometric prefix omitted as per standard IUPAC coordination naming conventions. 7 a) Explain in detail the Contact process for the manufacture of Sulphuric acid along with flow sheet diagram. b) Define and give general formula of glass. What is meant by annealing of glass.