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

Slide 7 / 126 Slide 8 / 126 Molecular Shape and VSEPR Theory 1 What is the electron domain geometry of CH 4 ? Electron Domain Geometries # of unbonded "AB" pairs of Bond linear Designation A Shape Example electrons on "A" Angles trigonal planar B (# of bonds) atom tetrahedral C AB 2 0 linear 180 octahedral D trigonal 0 AB 3 120 planar 0 AB 4 tetrahedral 109.5 trigonal 90, 120, 0 AB 5 bypyramidal 180 0 AB 6 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 (Answer) / 126 Slide 9 / 126 1 What is the electron domain geometry of CH 2 What is the EDG of H 2 O? 4 ? O linear linear A A trigonal planar trigonal planar B B H H tetrahedral tetrahedral C C octahedral octahedral D Answer D C [This object is a pull tab] Slide 9 (Answer) / 126 Slide 10 / 126 2 What is the EDG of H 2 O? 3 What is the EDG of CO 2 ? O C O O linear linear A A trigonal planar trigonal planar B B H H tetrahedral trigonal bipyramidal C C octahedral octahedral D D Answer C [This object is a pull tab]

Slide 10 (Answer) / 126 Slide 11 / 126 VSEPR Numbers 3 What is the EDG of CO 2 ? O C O 2. The second digit of the VSEPR number is the total number linear A of bonding-domains around the trigonal planar central atom. That means the number B of single, double or triple bonds. trigonal bipyramidal C Remember, double and triple bonds octahedral D only count as 1 domain. Answer B 3 3 6 4 3 2 O O O [This object is a pull tab] Slide 12 / 126 Slide 12 (Answer) / 126 4 VSEPR Numbers 4 VSEPR Numbers Students type their answers here Students type their answers here 3. The third digit of the VSEPR number 3. The third digit of the VSEPR number is the total number of lone pairs is the total number of lone pairs around the central atom. around the central atom. BH 3 - Trigonal Planar You can check your work - the You can check your work - the first digit is always equal to the first digit is always equal to the Answer sum of the second and third. sum of the second and third. XeF 4 - Square Planar What are the shapes of BH 3 , What are the shapes of BH 3 , XeF 4 , and O 3 ? XeF 4 , and O 3 ? O 3 - Bent 3 3 0 6 4 2 3 2 1 3 3 0 6 4 2 3 2 1 117 O O [This object is a pull tab] O O O O Slide 13 / 126 Slide 14 / 126 NF 3 Practice Molecular Geometry Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). The molecular geometry of a molecule is the shape formed by the Elements Xe B For Ions Bonds & Electrons F bonded atoms. Lone pairs may play a role by decreasing the angle + - I Cl H C O N S between bonded elements. This occurs because lone pairs generate a greater force of repulsion. Click here to view a PhET simulation

Slide 14 (Answer) / 126 Slide 15 / 126 NF 3 SiF 4 Practice Practice Draw a Lewis Structure and determine the electron domain geometry Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). (EDG) and the molecular geometry (MG). Elements Xe B For Ions Bonds & Electrons F Si Elements Xe B For Ions Bonds & Electrons F + - + - I Cl H C O N S I Cl H C O N S F F N Answer F 4 3 1 EDG: tetrahedral MG: trigonal pyramidal [This object is a pull tab] Slide 15 (Answer) / 126 Slide 16 / 126 Practice SiF 4 Practice IF 5 Draw a Lewis Structure and determine the electron domain geometry Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). (EDG) and the molecular geometry (MG). F Si Elements Xe B Bonds & Electrons F Si Elements Xe B Bonds & Electrons For Ions For Ions + - + - I Cl H C O N S 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 (Answer) / 126 Slide 17 / 126 IF 5 NO 2- Practice Practice Draw a Lewis Structure and determine the electron domain geometry Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). (EDG) and the molecular geometry (MG). F Si Elements Xe B For Ions Bonds & Electrons F Si Elements Xe B For Ions Bonds & Electrons + - + - I Cl H C O N S 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 (Answer) / 126 Slide 18 / 126 NO 2- Practice 5 Which of the following would have a see-saw shape? Draw a Lewis Structure and determine the electron domain geometry (EDG) and the molecular geometry (MG). A I only F Si Elements Xe B For Ions Bonds & Electrons I. XeO 2 F 2 + - I Cl H C O N S II. IBr 3 B II only - III. SeH 2 C III only N O O D I and II only Answer 3 2 1 EDG: trigonal planar MG: bent [This object is a pull tab] Slide 18 (Answer) / 126 Slide 19 / 126 5 Which of the following would have a see-saw shape? 6 Which of the following is ranked properly from largest to smallest bond angles within the molecule? A I only I. XeO 2 F 2 I. CH 4 , PCl 3 , SF 5 A II only II. IBr 3 B II only II. XeF 2 , H 2 O, XeF 4 B III only III. SeH 2 III. NO 3- , NO 2- , CH 4 Answer C III only A C II and III only D I and II only D I, II, and III [This object is a pull tab] Slide 19 (Answer) / 126 Slide 20 / 126 6 Which of the following is ranked properly from largest to 7 Which of the following does NOT have a bent shape? smallest bond angles within the molecule? A BeCl 2 I. CH 4 , PCl 3 , SF 5 A II only II. XeF 2 , H 2 O, XeF 4 B SeH 2 B III only III. NO 3- , NO 2- , CH 4 C SCl 2 C II and III only Answer D OH 2 D D I, II, and III [This object is a pull tab]

Slide 20 (Answer) / 126 Slide 21 / 126 7 Which of the following does NOT have a bent shape? 8 Which of the following has a planar shape? A BeCl 2 A C 2 H 4 B SeH 2 B PH 3 Answer A C SCl 2 C SiH 4 D OH 2 D PF 5 [This object is a pull tab] Slide 21 (Answer) / 126 Slide 22 / 126 8 Which of the following has a planar shape? 9 Which of the following contribute to the shape of the molecule? A C 2 H 4 A Only the number of bonded e- pairs around atom B PH 3 B Only the number of un-bonded and bonded e- pairs around the atom C SiH 4 Answer C Only the atomic radii of atoms A D PF 5 D Only the atomic radii and bonded e- pairs around the atom [This object is a pull tab] Slide 22 (Answer) / 126 Slide 23 / 126 9 Which of the following contribute to the shape of the 10 Which of the following is TRUE regarding the effect of molecule? substitution of un-bonded pairs of electrons in place of bonded pairs of electrons on the molecular shape? A Only the number of bonded e- pairs around atom A The bond angle increases due to the decreased B Only the number of un-bonded and bonded e- pairs repulsions around the atom Answer B B The bond angle decreases due to the decreased C Only the atomic radii of atoms repulsions D Only the atomic radii and bonded e- pairs around C The bond angle decreases due to the increased the atom repulsions [This object is a pull tab] D The bond angle increased due to the increased repulsions

Slide 23 (Answer) / 126 Slide 24 / 126 10 Which of the following is TRUE regarding the effect of 11 Which of the following would be the correct shape of the substitution of un-bonded pairs of electrons in place of BF 3 molecule? bonded pairs of electrons on the molecular shape? A bent A The bond angle increases due to the decreased repulsions B trigonal planar Answer B The bond angle decreases due to the decreased C trigonal pyramidal C repulsions D see-saw 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 (Answer) / 126 Slide 25 / 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] Slide 25 (Answer) / 126 Slide 26 / 126 Molecular Polarity Molecules in which the electrons are not evenly distributed experience a dipole moment when in an electric field and are said to be polar. Two factors contribute to the polarity of a molecule: Polarity of bonds Polar bonds are necessary for a molecule to be polar but do not guarantee polarity. Symmetry To be polar a molecule must be asymmetrical to ensure an uneven distribution of electrons.