Sigma-Pi Bonding Answer Key

(a) In the acetylene molecule there are: 1 π bond. 2 VSEPR electron pairs around C1 and C2. 3 σ and 1 π bond. 2 σ bonds. (b) In ethylene C1 and C2 are joined: to each other by 1 π and 2 σ bonds to each other by 2 VSEPR pairs by a π bond formed from the overlap of their 2px orbitals. to the hydrogens by 4 σ bonds (c) Consider the molecules as shown at the beginning of the animation. In this series which of the following does not decrease from left to right? total number of σ bonds total number of bonds of all types total number of π bonds total number of VSEPR pairs (d) Rotation can occur around: the C to C bond in ethylene. the C to C bond in acetylene. the C to C bond in ethane. all the C to C bonds shown in the animation. (e) In ethylene, VSEPR predicts: unhybridized d orbitals. formation of no σ and π bonds. trigonal planar geometry. sp3 hybridization occurs. 3. A "p-p" overlap of atomic orbitals would occur in which of the following bonds? H-F H-H Li-H S-S Solution or Explanation Two (empty) p orbitals are available for overlap. 4. A pi orbital is formed from ______ atomic orbitals aligned _______ . p, end-to-end hybrid, parallel to one another p, parallel to one another hybrid, end-to-end Solution or Explanation Parallel overlap of "p" orbitals form pi bonds. 5. Which of the following is a hybrid orbital? 3d pi sigma sp2 Solution or Explanation Combination of orbitals written as one. 6. Specify which hybrid orbitals are used by carbon atoms in the following species. (a) NCO- sp 3 of 13 4 of 13 (b) CHCl3 sp3 (c) CN- sp 7. Question DetailsBurdgeChem2 9.EOCP.040. [1413715] How many sigma bonds and pi bonds are there in each of the following molecules? (b) (a) sigma bonds 4 sigma bonds 5 pi bonds 0 pi bonds 1 (c) 8. sigma bonds 10 pi bonds 3 Question DetailsBurdgeChem2 9.EOCP.042. [1413920] Tryptophan is one of the 20 amino acids in the human body. Describe the hybridization state of the C and N atoms, and determine the number of sigma and pi bonds in the molecule. N-1 C-2 sp3 sp2 5 of 13 C-3 sp2 C-4 sp2 C-5 sp2 C-6 sp2 C-7 sp2 C-8 sp2 C-9 sp2 C-α sp2 C-β sp3 C-γ sp3 N-δ sp3 6 of 13 number of σ bonds 28 number of π bonds 5 9. What is the ideal Cl-C-Cl bond angle in C2Cl4? 60° 109.5° 180° 120° Solution or Explanation AX3 arrangement predicts 120 degree bond angle. 10. Which of the following molecules has both an electron group geometry and a molecular geometry described as trigonal planar? SiH4 BF3 OF2 CHF3 Solution or Explanation AX3 is indicated with no nonbonding pairs present in the structure. 11. Indicate the type of hybrid orbitals used by the central atom in CCl4. sp3 sp3d2 sp sp3d sp2 12. In which one of the following molecules is the central atom sp2 hybridized? SO2 PF5 NF3 N2O BeCl2 13. The number of pi bonds in the molecule below is 7 of 13 3 1 9 5 2 14. 15 4 10 2 6 15. Consider the species Cl2+, Cl2, and Cl2-. Which of these species will be paramagnetic? Cl2 and Cl2All three are paramagnetic Only Cl2 Cl2+ and Cl2Cl2+ and Cl2 16. The electron pairs on the central nitrogen atom can be considered to be in sp2-hybridized orbitals in all of the following species except HNNH NNN NO2NO3- Solution or Explanation Since NNN - is linear, sp hybridization is indicated. 17. For which of the following species can the bonds formed by the central atom be described in terms of sp2 hybrid orbitals? H2O CO32SO42ClF3 8 of 13 Solution or Explanation Triangular arrangement of orbitals indicates sp2. 18. Which statement is incorrect? An sp hybridization produces a tetrahedral molecule. Hybrid orbitals are combination of atomic orbitals. The bond length is the internuclear distance. There are the same number of hybrid orbitals produced as the number of atomic orbitals combined. Solution or Explanation Sp3 is characteristic of tetrahedral geometry. 19. For BeCl2, what is the hybridization on the central atom? sp2 sp 3 sp d sp3 Solution or Explanation Linear shape indicates sp hybridization. 20. All of the following are true concerning the bonding in methane, CH4, except: these sp3 orbitals combine with the s orbitals of the hydrogen to form molecular orbitals. the compound has tetrahedral geometry. some of the C-H bonds are stronger than others. the carbon s and p orbitals combine to form four equivalent sp3 orbitals. Solution or Explanation All the C-H would have the same bond dissociation energy and are equivalent. 21. Which of the pairs of molecules below have the same hybridization on the central atom in each molecule? (The central atom is in bold.) HOCl, ClF2 HCN, CO2 BeH2, NH3 H2O, HNO Solution or Explanation Both compounds are linear and sp hybridized. 9 of 13 22. Consider the following three molecules, and identify the compound(s) that contain(s) both sp and sp3-hybridized carbon atoms: 1. 2. 3. 2 only 1 only 2 and 3 only 3 only Solution or Explanation The carbon-carbon triple shows "sp" while the tetrahedral part or C-H 's indicate sp3. 23. One resonance structure of N2O is shown in the diagram [ N N-O]. The hybridized atomic orbitals of the central nitrogen atom, which are consistent with this structure, are: four sp3 orbitals three sp2 orbitals and a "p" orbital two sp2 orbitals and two sp orbitals two sp orbitals and two "p" orbitals Solution or Explanation Linear shape suggests sp hybridization. 24. The geometry of sp3 hybridized orbitals is tetrahedral linear octahedral triangular Solution or Explanation sp3 is characteristic of tetrahedral geometry. 25. What is the type of hybridization used by carbon in C2H2, (acetylene)? sp2 sp sp3 sp2d 10 of 13 Solution or Explanation Only two orbitals are needed which suggests sp hybridization. 26. According to the valence-bond method, which of the following molecules involves sp2 hybridization of orbitals on the carbon atom? C2H6 CO2 CO C2H4 Solution or Explanation C=C bonds involve sp2 due to the triangular arrangement of orbitals. 27. How many sigma bonds and pi bonds are there in each of the following molecules? (a) (b) (c) (a) sigma bonds 4 pi bonds 0 (b) sigma bonds 5 pi bonds 1 (c) sigma bonds 10 pi bonds 3 28. Hybrid Orbitals (a) The fact that the BCl3 molecule is planar means that the B atom is which of the following? unhybridized sp3 hybridized sp2 hybridized sp hybridized Hint (b) Hybridizing one s atomic orbital and one p atomic orbital would not yield which of the following? a linear geometry 2 equivalent hybrid orbitals sp2 hybridization 11 of 13 a 180° angle Hint (c) Which combination of hybrid orbitals and electron pair geometries is incorrect? sp2 = trigonal planar sp3 = tetrahedral sp3d = square planar sp3d2 = octahedral Hint (d) Which of the following statements is true concerning sp3d hybrid orbitals? One s atomic orbital, three p atomic orbitals and 2 d atomic orbitals combine to make sp3d hybrid orbitals. The resulting angles of sp3d hybrid orbitals are 90° and 109.5° Trigonal bipyramidal geometry is a result of sp3d hybrid orbitals. There are four equivalent sp3d hybrid orbitals produced. Hint (e) What is a common feature of sp, sp2, sp3, sp3d, and sp3d2 hybrid orbitals? They all incorporate at least one s atomic orbital and one p atomic orbital. There are at least 3 equivalent hybrid orbitals of each type. They all only use s atomic orbitals. They all use at least s atomic orbital, one p atomic and one d atomic orbital. Hint 29. Practice Exercise 10.3 Enter the hybridization state of the underlined atoms in the following compounds. (a) CO2 sp (b) CCl4 sp3 30. Practice Exercise 10.5 Describe the bonding in the thiocyanate ion, SCN‾. Assume that N is sp-hybridized. geometry 12 of 13 bent linear trigonal planar tetrahedral hybridization of C atom sp 31. According to the valence-bond method, which of the following molecules involves sp2 hybridization of orbitals on the carbon atom? C2H6 C2H4 CO2 CO Solution or Explanation C=C bonds involve sp2 due to the triangular arrangement of orbitals. 32. The geometry of sp3 hybridized orbitals is tetrahedral linear octahedral triangular Solution or Explanation sp3 is characteristic of tetrahedral geometry. 33. In which of the following molecules would you expect the nitrogen to nitrogen bond to be the shortest? N2H4 N2O4 N2 N2O Solution or Explanation N2 has the highest bond order. Assignment Details Name (AID): Assignment 5 (1524883)