General, Organic, and Biological Chemistry Fourth Edition Karen Timberlake 5.8 Shapes and Polarity of Molecules Chapter 5 Compounds and Their Bonds © 2013 Pearson Education, Inc. Lectures

© 2013 Pearson Education, Inc. Chapter 5, Section 8 2 Valence-Shell Electron-Pair Repulsion Theory (VSEPR) In the valence-shell electron-pair repulsion (VSEPR) theory, the electron groups around a central atom  are arranged as far apart from each other as possible.  have the least amount of electron-electron repulsion.  are used to predict the molecular shape.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 3 Shapes of Molecules The three-dimensional shape of a molecule can be predicted by  the number of bonding groups and lone-pair electrons around the central atom  VSEPR theory (valence-shell-electron-pair repulsion).

© 2013 Pearson Education, Inc. Chapter 5, Section 8 4 Two Electron Groups In a molecule of BeCl 2,  there are two electron groups bonded to the central atom, Be (Be is an exception to the octet rule).  to minimize repulsion, the arrangement of two electron groups is 180 , or opposite each other.  the shape of the molecule is linear.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 5 Three Electron Groups In a molecule of BF 3,  three electron groups are bonded to the central atom B (B is an exception to the octet rule).  repulsion is minimized with 3 electron groups at angles of 120 .  the shape is trigonal planar.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 6 Three Electron Groups with a Lone Pair In a molecule of SO 2,  S has 3 electron groups; 2 electron groups bonded to O atoms and one lone pair.  repulsion is minimized with the electron groups at angles of 120 , a trigonal planar arrangement.  the shape is determined by the two O atoms bonded to S, giving SO 2 a bent (~120  ) shape.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 7 Four Electron Groups In a molecule of CH 4,  there are four electron groups around C.  repulsion is minimized by placing four electron groups at angles of 109 , which is a tetrahedral arrangement.  the four bonded atoms form a tetrahedral shape.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 8 Four Electron Groups with a Lone Pair In a molecule of NH 3,  three electron groups bond to H atoms, and the fourth one is a lone (nonbonding) pair.  repulsion is minimized with 4 electron groups in a tetrahedral arrangement.  the three bonded atoms form a pyramidal (~109  ) shape.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 9 Four Electron Groups with Two Lone Pairs In a molecule of H 2 O,  two electron groups are bonded to H atoms and two are lone pairs (4 electron groups).  four electron groups minimize repulsion in a tetrahedral arrangement.  the shape with two bonded atoms is bent (~109  ).

© 2013 Pearson Education, Inc. Chapter 5, Section 8 10 Molecular Shapes for 2 and 3 Bonded Atoms

© 2013 Pearson Education, Inc. Chapter 5, Section 8 11 Molecular Shapes for 4 Bonded Atoms

© 2013 Pearson Education, Inc. Chapter 5, Section 8 12 Guide to Predicting Molecular Shape (VSEPR Theory)

© 2013 Pearson Education, Inc. Chapter 5, Section 8 13 Predicting the Molecular Shape of H 2 Se Step 1 Draw the electron-dot formula. In the electron-dot formula for H 2 Se, there are four electron groups, including two lone pairs of electrons around Se.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 14 Predicting the Molecular Shape of H 2 Se Step 2 Arrange the electron groups around the central atom to minimize repulsion. The four electron groups around Se would have a tetrahedral arrangement. Step 3 Use the atoms bonded to the central atom to determine the molecular shape. Two bonded atoms give H 2 Se a bent shape with a bond angle of ~109 .

© 2013 Pearson Education, Inc. Chapter 5, Section 8 15 Learning Check What is the shape of a molecule of NF 3 ? A. linear B. trigonal pyramidal C. bent (120  )

© 2013 Pearson Education, Inc. Chapter 5, Section 8 16 Solution What is the shape of a molecule of NF 3 ? B. trigonal pyramidal Step 1 Draw the electron-dot formula. In the electron-dot structure for NF 3, there are three electron groups and a lone pair.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 17 Solution What is the shape of a molecule of NF 3 ? B. trigonal pyramidal Step 2 Arrange the electron groups around the central atom to minimize repulsion. The four electron groups around N would have a tetrahedral arrangement. Step 3 Use the atoms bonded to the central atom to determine the molecular shape. Three bonding groups and a lone pair give NF 3 a trigonal pyramidal shape with a bond angle of 109 .

© 2013 Pearson Education, Inc. Chapter 5, Section 8 18 Polar Molecules A polar molecule  contains polar bonds.  has a separation of positive and negative charge. called a dipole, indicated with  + and  –.  has dipoles that do not cancel.  +  – H–Cl NH 3 dipole Dipoles do not cancel.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 19 Nonpolar Molecules A nonpolar molecule  contains nonpolar bonds Cl–Cl H–H  or has a symmetrical arrangement of polar bonds.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 20 Guide to Determination of Polarity

© 2013 Pearson Education, Inc. Chapter 5, Section 8 21 Molecular Polarity, H 2 O Determine the polarity of the H 2 O molecule. Step 1 Determine if the bonds are polar covalent or nonpolar covalent. From the electronegativity table, O 3.5 and H 2.1 gives a difference of 1.4, which makes the O — H bonds, polar covalent.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 22 Molecular Polarity, H 2 O Determine the polarity of the H 2 O molecule. Step 2 If the bonds are polar covalent, draw the electron-dot formula and determine if the dipoles cancel or not. The four electron groups of oxygen are bonded to two H atoms. Thus, the H 2 O molecule has a net dipole, which makes it a polar molecule.

© 2013 Pearson Education, Inc. Chapter 5, Section 8 23 Learning Check Determine the shape of each of the following molecules and whether they are polar or nonpolar. Explain. 1. PBr 3 2. HBr 3. Br 2 4. SiBr 4

© 2013 Pearson Education, Inc. Chapter 5, Section 8 24 Solution Determine the shape of each of the following molecules and whether they are polar or nonpolar. Explain. 1. PBr 3 pyramidal; polar; dipoles don’t cancel 2. HBr linear; polar; one polar bond (dipole) 3. Br 2 linear; nonpolar; nonpolar bond 4. SiBr 4 tetrahedral; nonpolar; dipoles cancel