4.2.7 & Shapes, and bond angles for molecules with two, three and four negative charge centers

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Domenic McLaughlin
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1 4.2.7 & Shapes, and bond angles for molecules with two, three and four negative charge centers The shape of a molecule has an important part to play in determining its chemical (e.g. reactivity and ph) and physical properties (e.g. mpt and bpt). A negative charge center or region refers to the number of pairs electrons around the central bonded atom. This includes both the lone (non bonding pairs) pairs and bonded pairs of electrons in single, double or triple bonds. Each double and triple bond counts as one negative charge center. Watch out for non octet examples. Number of negative charge centers Shape with respect to the number of negative charged centers Bond Angle with respect to negative charge centers 2 Linear planar Tetrahedral Valence shell electron pair repulsion theory (VSEPR) GN Lewis proposed that chemical bonds resulted when two atoms shared a pair of electrons. The Lewis concept allowed for "electron dot bookkeeping" in the form of a Lewis dot digram to show how atoms could share electrons to achieve their quota as a noble gas or an octet of electrons. Lewis originally did not set out to describe the shape of the molecule but it soon became apparent to him that the electron pairs around the central atom, being like charged, would repel each other. This lead to his VSEPR theory which is still used today to describe the shape of a molecule. In VSEPR theory lone pairs of electrons are closer to the central atom and therefore closer to one another than the bonding pairs of electrons because they are being attracted by the positive protons in the nucleus of one atom instead of two. The lone pairs and bonding pairs of electrons are arranged around the central atom so as to minimize the repulsion between the lone and bonding pairs of electrons. The relative magnitude of the electron pair repulsions is: Lone pair / lone pair > bonded pair / lone pair > bonded pair / bonded pair repulsion The overall shape according to the VSEPR theory depends on the number of bonding pairs and lone pairs of electrons around the central atom of a molecule. The overall bond angle is the angle between the bonded atoms attached to the central atom. 1

2 To determine the overall shape and bond angle of a molecule by the VSEPR theory the number of bonding and lone pairs of electrons around the central atom need to be identified. Number of negative charge centers Number of bonding pairs of electrons Number of lone pairs of electrons Overall Shape (according to VSEPR) Overall Bond Angle (according to VSEPR) Linear planar Tetrahedral pyramid Bent Bent > (about 117 ) Reason for the shape (according to VSEPR) Examples To minimize the repulsion the two bonding pairs repel each other CO 2, HCN, BeCl 2 equally To minimize the repulsion the three bonding pairs repel each other BF 3, C 2 H 4 equally To minimize the repulsion the four bonding pairs repel each other CH 4, SiCl 4, CCl 4 equally To minimize the repulsion the lone pair-lone pair > bonded pair-lone NH 3, NF 3 pair > bonded pair- bonded pair repulsion To minimize the repulsion Lone pair-lone pair > bonded pairlone pair > bonded pair- bonded pair 2 O, H 2 S H repulsion bonded pair-lone pair > bonded pairbonded pair repulsion O 3, SO 2 2

3 NOTE: The number of negative charge centers may not be reflected in the molecular shape. If the electrons are not being used for bonding then they cannot be seen and the molecule is described as if they were not there. In other words the shape and angle of the molecule is determined with respect to the attached atoms. 3

4 Complete the table below leaving out the last column. Formula and name Structural formula in correct shape (To do this you may need to the Lewis dot diagram first) Number negative charge centers Overall Bond Angle (according to VSEPR) Overall Shape (according to VSEPR theory) Explanation for the overall shape using VSEPR theory or non polar covalent HCN Hydrogen cyanide H-C N linear 2 bonding pairs of electrons & 0 lone pairs around the central atom. In order to minimize the repulsion the bonding pairs of electrons repel each other equally. BeCl 2 Beryllium chloride Cl-Be-Cl linear 2 bonding pairs of electrons & 0 lone pairs around the central atom. In order to minimize the repulsion the bonding pairs of electrons repel each other equally. CO 2 O=C=O linear 2 bonding pairs of electrons & 0 lone pairs of electrons around the central atom. Bonding pairs of electrons repel each other equally in order to minimize the repulsion between them. OR 2 negative charge centers repel each other as much as possible to minimize the repulsion between them. 4

5 H 2 O bent 2 bonding pairs of electrons and 2 lone pairs of electrons. In order to minimize the repulsion between then the 2 lone pairs of electrons exert a greater repulsion than the bonding pair and lone-bonding pair repulsion. CHCl tetrahedral 4 bonding pairs of electrons and 0 lone pairs of electrons around the central atom. To minimize the repulsion between them, the four bonding pairs repel each other equally NH trigonal pyramid 3 bonding pairs of electrons and 1 lone pair of electrons around the central atom. To minimize the repulsion between them the bonded pair / lone pair > bonded pair / bonded pair repulsion H 2 S CH 4 5

6 BF 3 SiCl 4 CClF 3 Cl 2 O C 2 H 2 ethyne 6

7 H 2 CO methanal SO 2 Sulfur dioxide C 2 H 4 ethene POCl 3 O 3 7

8 CH 3 NH 2 Aminomethane (C and N) (C and N) ( with respect to C and N) Extra Problems 1. Predict the bond angle and shape around the carbon atom and the oxygen atom attached to the hydrogen atom in methanoic acid (HCOOH). 2. Compare the bond length and strength between the carbon and oxygen atoms in carbon dioxide and carbon monoxide. 3. State and compare the difference in the bond angle in CH 4 and NF Compare the N-N-H bond angle between the two possible Lewis structures of N 3 H. 8

9 5. State the bond angle around the carbon and nitrogen atoms in aminomethane, CH 3 NH 2 6. Draw the structural formula for ethanoic acid and determine its relative molecular mass. 9

10 ANSWERS Formula Structural formula Number negative charge centers Shape with respect to negative charge centers Bond Angle Overall Shape Explanation for shape using VSEPR theory or non polar HCN Hydrogen cyanide H-C N 2 linear 180º linear 2 bonding pairs of electrons & 0 lone pairs around the central carbon atom. In order to minimize repulsion the bonding pairs of electrons repel each other equally. polar BeCl 2 Beryllium chloride Cl-Be-Cl 2 linear 180º linear 2 bonding pairs of electrons & 0 lone pairs around the central beryllium atom. Bonding pairs of electrons repel each other equally. Non polar CO 2 O=C=O 2 linear 180º linear C 2 H 2 ethyne H-C C-H 2 linear 180º linear BF 3 3 planar 120º planar 2 bonding pairs of electrons & 0 lone pairs of electrons around the central carbon atom. Bonding pairs of electrons repel each other equally. OR 2 negative charge centers arranged as far apart as possible. 2 bonding pairs of electrons & 0 lone pairs of electrons around the central carbon atoms. Bonding pairs of electrons repel each other equally. OR 2 negative charge centers arranged as far apart as possible. 3 bonding pairs of electrons & 0 lone pairs of electrons around the central boron atom. Bonding pairs of electrons repel each other equally. OR 3 negative charge centers arranged as far apart as possible. Non polar Non polar 10

11 H 2 CO methanal 3 SO 2 Sulfur dioxide C 2 H 4 ethene CH 4 methane 3 3 for each C atom planar planar planar for each C atom 120º planar ~117º bent 120º between each H atom planar 4 Tetrahedral 109.5º Tetrahedral SiCl 4 4 Tetrahedral 109.5º Tetrahedral 3 bonding pairs of electrons & 0 lone pairs of electrons around the central carbon atom. Bonding pairs of electrons repel each other equally. OR 3 negative charge centers arranged as far apart as possible. 2 bonding pairs & 1 lone pair of electrons around the central sulfur atom. Lone pair - bonding pair repulsion is greater than bonding pair bonding pair repulsion. 3 bonding pairs & 0 lone pairs of electrons around each carbon atom. 3 bonding pairs of electrons repel each other equally in order to minimize the repulsion. OR 3 negative charge centers on each carbon atom arranged as far apart as possible to minimize repulsion. 4 bonding pairs & 0 lone pairs of electrons around the central carbon atom. In order to minimize the repulsions the 4 bonding pairs of electrons repel each other equally. 4 bonding pairs & 0 lone pairs of electrons around the central silicon atom. 4 bonding pairs of electrons repel each other equally. Non Non Non CClF 3 4 Tetrahedral 109.5º Tetrahedral H 2 O 4 Tetrahedral 104.5º Bent 4 bonding pairs & 0 lone pairs of electrons around the central chlorine atom. 4 bonding pairs of electrons repel each other equally. 2 bonding pairs of electrons and 2 lone pairs of electrons around the central oxygen atom. The 2 lone pairs of electrons exert a greater repulsion than the bonding pair and lonebonding pair repulsion. 11

12 NH 3 ammonia 4 Tetrahedral 107º pyramid POCl 3 4 Tetrahedral Tetrahedral 3 bonding pairs and 1 lone pair of electrons around the central nitrogen atom. Lone pair bonding pair repulsion is greater than bonding pair repulsion. 4 bonding pairs & 0 lone pairs of electrons around the central phosphorus atom. 4 bonding pairs of electrons repel each other equally. CH 3 NH 2 Aminomethane 4 Tetrahedral C (109.5º) N (107º) C Tetrahedral N pyramid 4 bonding pairs & 0 lone pairs of electrons around carbon atom. 4 bonding pairs of electrons repel each other equally. 3 bonding pairs and 1 lone pair of electrons around nitrogen atom. Lone pair bonding pair repulsion is greater than bonding pair repulsion. 12

13 More IB Problems 1. Predict the bond angle around the carbon atom and the oxygen atom attached to the hydrogen atom in methanoic acid (HCOOH). Carbon 3 bonding pairs and 0 lone pairs of electrons. Bond angle 120º (trigonal planar shape) Oxygen 2 bonding pairs of electrons and 2 lone pairs. Bond angle 104.5º (bent shape) 2. Compare the bond length and strength between the carbon and oxygen atoms in carbon dioxide and carbon monoxide. The more bonding electrons between the nuclei of the two bonded atoms, The greater the attraction the protons in the nucleus of each atoms have for the bonding electrons the shorter the bond and the more energy will be required to break it. Therefore the C=O bond in CO 2 is longer and has a lower bond energy than the C O bond in CO. 3. State and compare the difference in the bond angle in CH 4 and NF 3. CH 4 the bond angle between the H atoms is 109.5º due to 4 bonding pairs & 0 lone pairs of electrons around the central carbon atom. 4 bonding pairs of electrons repel each other equally in order to minimize the repulsion between them.. NF 3-107º due to 3 bonding pairs and 1 lone pair of electrons around the central nitrogen atom. Lone pair bonding pair repulsion is greater than bonding pair repulsion, which results in the bonding pairs of electrons being pushed closer together than in CH 4, decreasing the bond angle between the attached atoms. 4. Compare the N-N-H bond angle between the two possible Lewis structures of N 3 H. 5. State the bond angle around the carbon and nitrogen atoms in aminomethane, CH 3 NH 2 C º N - 107º 6. Draw the structural formula for ethanoic acid and determine its relative molecular mass. 13

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VSEPR Theory. Shapes of Molecules. Molecular Structure or Molecular Geometry

VSEPR Theory. Shapes of Molecules. Molecular Structure or Molecular Geometry VSEPR Theory VSEPR Theory Shapes of Molecules Molecular Structure or Molecular Geometry The 3-dimensional arrangement of the atoms that make-up a molecule. Determines several properties of a substance,