Question

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Exercise 4 1. Use the molecular orbital model to predict the bond order and magnetism of each of the following molecules. a- co b- CO C- CO2- 2- Compare and contrast bonding molecular orbitals with antibonding molecular orbitals. 3- Consider the following electron configuration: (0:) (03.*) (03p) (T3p)* (13p*)* Give four species that, in theory, would have this electron configuration. 4- Predict which substance in each of the following pairs would have the greater intermolecular forces. 1. CO2 or OCS II. SeO2 or SO2 III. CH CH CH_NH, or H2NCHCH2NH2 IV. CH CH; or H.CO V. CH3OH or H2CO 5. Explain why water forms into beads on a waxed car finish. 6- At which temperature does water have its greatest density? 7- At room temperature, F2 and Cl2 are gases, Br2 is a liquid, and 12 is a solid. This is because:

Answer 1

1. @ co the 15%* d 5² 25 & 25 do 282 ² & Tapu Tapy? Bond order= Number of Bondining e - Namber of - antibonding e 10-4 = 3/4 rden - cot 615%* 152 253 25² 52p ² Tapu rapy - Bé Bond order = "9-9 = 2.5 © co2+ lae 815 %* 15%25²2502p2²rapu rapy Bond order = 8-4-2 do 4. Coa or oes & ocs molecule polar , thus capable of forming t . Seoa or S0₂ bond more stronger. Sell, polar, thus capable of forming I strong intermolecular forces polac band intermolecula

3 CH ₂ CH ₂ CH₂NH₂ or NCH, CH₂ NH₂. - H₂ N CH₂ CH₂ NH₂ → two active groups, this makes Strong intermolemlar force CH₂ CH3 or H₂CO - co polae molecule, thus makes strong intermolembar bond. 6. enzy & HCO strong interm- 2.0 more polar, thus makes olenular bond.

Bonding molecular Orbitals. Antibonding molecular orbitals. Involved in the formation at these are orbitals conteuna of a chemical bond i ng electrons outside the region between two atomic nuclei Elee bon density is higher Electron density is lower. O Energy is comparatively ® Energy is comparatively lower. © Represented without an © Represented with an asterisk mark (*) higher asterisk mark (*) ③ Geomeby of a molecule Geometry of a molecule is represented by the does not depend on the spatial arrangement. spatial arrangement. © Elechons contribute to tleehons do not conbibute the formation of a bond to the formation of a bond. Stability is comparahvely a stability is comparabvely w lower 2.0 higher boods 3. Number of electrons = 34 ! Because & 152 * 150 250 25² ra pa nape napy? t ap 30 ap? od 282 ² 6352 835283020 3p 47384 . Molecules with 34 number of e = (₂, 5 , 12 )