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Transcript
Chemistry 11 (HL) Unit 4 / IB Topic 14.2 Covalent Bonding 5 Practice Problems Hybridisation and Sigma & Pi Bonds ANSWERS 1. hybrid orbitals form when atoms form bonds a pure s orbital “blends” with 1, 2 or 3 pure p orbitals to make new orbitals with shapes and energy values that are a mix of the pure orbitals 2. hybrid orbital shape around atoms with these hybrid orbitals description of how it forms angle between hybrid orbitals 1 pure s orbital mixes with 3 pure p orbitals to make 4 3 identical “sp ” orbitals sp 3 pure atomic orbitals p p p tetrahedral hybrid orbitals 109.5º (4 NCCs) blending sp3 sp3 sp3 sp3 s 1 pure s orbital mixes with 2 (of the three) pure p 2 orbitals to make 3 identical “sp ” orbitals (and leaving one pure p orbital) sp 2 pure atomic orbitals p p p trigonal planar 120º hybrid orbitals blending p sp2 sp2 sp2 s (3 NCCs)* 1 pure s orbital mixes with 1 (of the three) pure p 2 orbitals to make 2 identical “sp ” orbitals (and leaving two pure p orbitals) linear sp pure atomic orbitals p s p p hybrid orbitals blending sp sp p 180º (2 NCCs)* p * the pure p orbitals are used to make double/triple bonds 3. What type of hybrid orbitals does the central atom have in these molecules/ions? a) CF4 C = 4 NCCs = sp b) 3 BH3 B = 3 NCCs = sp 2 H B H H p. 1 Chemistry 11 (HL) Unit 4 / IB Topic 14.2 c) CH2O d) NH3 C = 3 NCCs = sp 2 N = 4 NCCs = sp e) CO2 f) C = 2 NCCs = sp g) C2H2 BH4 3 H 1- B = 4 NCCs = sp 3 H B H 1- H H C C H C = 2 NCCs = sp 4. Compare the formation of sigma bonds and pi bonds. • similarities: involve sharing of an electron pair involve the overlap of orbitals from two atoms • differences: sigma bond is end on end (or head to head) orbital overlap; pi bond is the side by side overlap of orbitals sigma bond electrons lie on the axis between the 2 nuclei; pi bond electrons are in a plane above and below the nuclei sigma bonds are stronger than pi bonds 5. How many sigma and pi bonds are in these molecules? a) CO2 b) NH3 sigma = 2, pi = 2 d) C 2H 4 sigma = 3, pi = 0 e) C2H2 sigma = 5, pi = 1 6. sigma = 3, pi = 2 c) H2CO sigma = 3, pi = 1 f) O3 sigma = 2, pi = 1 Draw a Lewis structure for ethanoic acid, CH3COOH. sp sp2 3 sp2 sp3 sigma = 7, pi = 1 p. 2 Chemistry 11 (HL) 7. Unit 4 / IB Topic 14.2 Urea, H2NCONH2, is present in animal urine. O H N C H C atom = sp N H H 2 N atoms = sp 3 shape around C atom = trigonal planar shape around N atom = tetrahedral bond angle around C atom = 109.5º bond angle around N atom = 107º sigma bonds = 7; pi bonds = 1 8. The structure of ascorbic acid (vitamin C) is shown below. a) sigma bonds = 20; pi bonds = 2 (there is a sigma bond between the O and H atoms in the OH groups on the molecule) b) 12 (each O atom has 2 lone electron pairs) c) C atoms with only single bonds = sp C atoms with double bond = sp 3 2 O atoms with only single bonds = sp O atoms with double bond = sp d) 2 Circle the carbon atoms having a trigonal planar shape. p. 3 3 Chemistry 11 (HL) 9. Unit 4 / IB Topic 14.2 Caffeine has the formula C8H10N4O2. A model of caffeine is shown below. Atoms without symbols are hydrogen atoms. a) All N atoms have 1 lone electron pair. All O atoms have 2 lone electron pairs. b) N* sp 3 107º tetrahedral C* sp 2 120º trigonal planar O sp 2 (omit angles and shape around this atom) p. 4