AP Chemistry Compounds 2015-09-14 www.njctl.org Slide 3 / 126 - PDF document

Slide 1 / 126 Slide 2 / 126 AP Chemistry Compounds 2015-09-14 www.njctl.org Slide 3 / 126 Table of Contents: Compounds Pt. B Click on the topic to go to that section · Molecular Geometry · Intermolecular Forces · States of Matter · Introduction to Solubility

Slide 4 / 126 Molecular Geometry Return to Table of Contents Slide 5 / 126 Molecular Shapes Molecules are three dimensional objects. Their shapes affect their properties. In order to determine the shape of a molecule we use Valence-Shell Electron-Pair Repulsion theory (VSEPR). According to VSEPR theory, the molecules will adopt a shape/geometry that will reduce the repulsion between the bonded electrons; the electrons want to be as far apart as possible. Slide 6 / 126 VSEPR Numbers VSEPR theory assigns a 3-digit code to each molecular shape. To determine this code we must first sketch the Lewis structure. 1. The first digit of the VSEPR number is the total number of electron-domains around the central atom. That means the total number of bonds and lone pairs. Note: Double and triple bonds only count as 1 electron domain. 3 6 3 O O O

Slide 7 / 126 Molecular Shape and VSEPR Theory Electron Domain Geometries # of unbonded "AB" pairs of Bond Designation Shape Example electrons on "A" Angles (# of bonds) atom 0 AB 2 linear 180 trigonal AB 3 0 120 planar AB 4 0 tetrahedral 109.5 trigonal 90, 120, AB 5 0 bypyramidal 180 AB 6 0 octahedral 90, 180 **Note: Pi bonds act with the sigma bonds to contribute to the repulsions that result in the molecular shape, however they do not act as a separate constituent around the "A" atom. Slide 8 / 126 1 What is the electron domain geometry of CH 4 ? linear A trigonal planar B tetrahedral C octahedral D Slide 8 (Answer) / 126 1 What is the electron domain geometry of CH 4 ? linear A trigonal planar B tetrahedral C octahedral D Answer C [This object is a pull tab]

Slide 9 / 126 2 What is the EDG of H 2 O? O linear A trigonal planar B H H tetrahedral C octahedral D Slide 9 (Answer) / 126 2 What is the EDG of H 2 O? O linear A trigonal planar B H H tetrahedral C octahedral D Answer C [This object is a pull tab] Slide 10 / 126 3 What is the EDG of CO 2 ? O C O linear A trigonal planar B trigonal bipyramidal C octahedral D

Slide 10 (Answer) / 126 3 What is the EDG of CO 2 ? O C O linear A trigonal planar B trigonal bipyramidal C octahedral D Answer B [This object is a pull tab] Slide 11 / 126 VSEPR Numbers 2. The second digit of the VSEPR number is the total number of bonding-domains around the central atom. That means the number of single, double or triple bonds. Remember, double and triple bonds only count as 1 domain. 3 3 6 4 3 2 O O O Slide 12 / 126 4 VSEPR Numbers Students type their answers here 3. The third digit of the VSEPR number is the total number of lone pairs around the central atom. You can check your work - the first digit is always equal to the sum of the second and third. What are the shapes of BH 3 , XeF 4 , and O 3 ? 3 3 0 6 4 2 3 2 1 O O O

Slide 12 (Answer) / 126 4 VSEPR Numbers Students type their answers here 3. The third digit of the VSEPR number is the total number of lone pairs around the central atom. BH 3 - Trigonal Planar You can check your work - the first digit is always equal to the Answer sum of the second and third. XeF 4 - Square Planar What are the shapes of BH 3 , XeF 4 , and O 3 ? O 3 - Bent 3 3 0 6 4 2 3 2 1 117 O [This object is a pull tab] O O Slide 13 / 126 Molecular Geometry The molecular geometry of a molecule is the shape formed by the bonded atoms. Lone pairs may play a role by decreasing the angle between bonded elements. This occurs because lone pairs generate a greater force of repulsion. Click here to view a PhET simulation Slide 14 / 126 Practice NF 3 Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). Elements Xe B Bonds & Electrons F For Ions + - I Cl H C O N S

Slide 14 (Answer) / 126 Practice NF 3 Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). Elements Xe B Bonds & Electrons F For Ions + - I Cl H C O N S F N F Answer F 4 3 1 EDG: tetrahedral MG: trigonal pyramidal [This object is a pull tab] Slide 15 / 126 SiF 4 Practice Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). F Si Elements Xe B For Ions Bonds & Electrons + - I Cl H C O N S Slide 15 (Answer) / 126 Practice SiF 4 Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). F Si Elements Xe B Bonds & Electrons For Ions + - I Cl H C O N S F Si F F Answer F 4 4 0 EDG: tetrahedral MG: tetrahedral [This object is a pull tab]

Slide 16 / 126 Practice IF 5 Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). F Si Elements Xe B Bonds & Electrons For Ions + - I Cl H C O N S Slide 16 (Answer) / 126 IF 5 Practice Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). F Si Elements Xe B For Ions Bonds & Electrons + - I Cl H C O N S F F F Answer I F F 6 5 1 EDG: octahedral MG: square pyramidal [This object is a pull tab] Slide 17 / 126 Practice NO 2- Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). F Si Elements Xe B For Ions Bonds & Electrons + - I Cl H C O N S

Slide 17 (Answer) / 126 Practice NO 2- Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). F Si Elements Xe B For Ions Bonds & Electrons + - I Cl H C O N S - N O O Answer 3 2 1 EDG: trigonal planar MG: bent [This object is a pull tab] Slide 18 / 126 5 Which of the following would have a see-saw shape? A I only I. XeO 2 F 2 II. IBr 3 B II only III. SeH 2 C III only D I and II only Slide 18 (Answer) / 126 5 Which of the following would have a see-saw shape? A I only I. XeO 2 F 2 II. IBr 3 B II only III. SeH 2 Answer C III only A D I and II only [This object is a pull tab]

Slide 19 / 126 6 Which of the following is ranked properly from largest to smallest bond angles within the molecule? I. CH 4 , PCl 3 , SF 5 A II only II. XeF 2 , H 2 O, XeF 4 B III only III. NO 3- , NO 2- , CH 4 C II and III only D I, II, and III Slide 19 (Answer) / 126 6 Which of the following is ranked properly from largest to smallest bond angles within the molecule? I. CH 4 , PCl 3 , SF 5 A II only II. XeF 2 , H 2 O, XeF 4 B III only III. NO 3- , NO 2- , CH 4 C II and III only Answer D D I, II, and III [This object is a pull tab] Slide 20 / 126 7 Which of the following does NOT have a bent shape? A BeCl 2 B SeH 2 C SCl 2 D OH 2

Slide 20 (Answer) / 126 7 Which of the following does NOT have a bent shape? A BeCl 2 B SeH 2 Answer A C SCl 2 D OH 2 [This object is a pull tab] Slide 21 / 126 8 Which of the following has a planar shape? A C 2 H 4 B PH 3 C SiH 4 D PF 5 Slide 21 (Answer) / 126 8 Which of the following has a planar shape? A C 2 H 4 B PH 3 C SiH 4 Answer A D PF 5 [This object is a pull tab]

Slide 22 / 126 9 Which of the following contribute to the shape of the molecule? A Only the number of bonded e- pairs around atom B Only the number of un-bonded and bonded e- pairs around the atom C Only the atomic radii of atoms D Only the atomic radii and bonded e- pairs around the atom Slide 22 (Answer) / 126 9 Which of the following contribute to the shape of the molecule? A Only the number of bonded e- pairs around atom B Only the number of un-bonded and bonded e- pairs around the atom Answer B C Only the atomic radii of atoms D Only the atomic radii and bonded e- pairs around the atom [This object is a pull tab] Slide 23 / 126 10 Which of the following is TRUE regarding the effect of substitution of un-bonded pairs of electrons in place of bonded pairs of electrons on the molecular shape? A The bond angle increases due to the decreased repulsions B The bond angle decreases due to the decreased repulsions C The bond angle decreases due to the increased repulsions D The bond angle increased due to the increased repulsions

Slide 23 (Answer) / 126 10 Which of the following is TRUE regarding the effect of substitution of un-bonded pairs of electrons in place of bonded pairs of electrons on the molecular shape? A The bond angle increases due to the decreased repulsions Answer B The bond angle decreases due to the decreased C repulsions C The bond angle decreases due to the increased repulsions [This object is a pull tab] D The bond angle increased due to the increased repulsions Slide 24 / 126 11 Which of the following would be the correct shape of the BF 3 molecule? A bent B trigonal planar C trigonal pyramidal D see-saw Slide 24 (Answer) / 126 11 Which of the following would be the correct shape of the BF 3 molecule? A bent B trigonal planar Answer B C trigonal pyramidal D see-saw [This object is a pull tab]