4 Oct 2019 Agenda Per 2 - Colloquia Tues
● Check the PPC is there for you to practice. ● Topic 2.7 VSEPR and molecular shape - make some models? ● VSEPR electron domains and shape, predicting bond angles, influence of nonbonding pairs ● FRQ Practice ● Hybridization ● Exam 2 next Thursday 40 questions - Unit 1 and 2 Qu. 4 Oct 2019 Agenda Per 4 Colloquia Tues
● Check the PPC is there for you to practice. ● VSEPR electron domains and shape, predicting bond angles, influence of nonbonding pairs ● FRQ Practice ● Hybridization ● Exam 2 next Thursday 40 questions - Unit 1 and 2 Qu. Molecular shape categories - based on the number of electron domains around a central atom in the LE model Linear 2 Trigonal planar 3 What Tetrahedral 4 constitutes an electron Trigonal bipyramidal 5 “domain” ? Octahedral 6 Valence Shell Electron Pair Repulsion (VSEPR) Model - useful for predicting geometries of molecules formed from nonmetals The valence electron pairs around a central atom can be
● Lone pairs ● Bonded pairs (or groups in the case of bonds of higher order)
Each group or pair of electrons can be considered an electron domain. It is the number of electron domains around a central atom that dictates the shape of the molecule. Valence Shell Electron Pair Repulsion (VSEPR) Model - useful for predicting geometries of molecules formed from nonmetals Main postulate of the model is that the structure around a given atom is determined principally by minimizing electron-pair repulsions. The electrons around a central atom in a molecule are attracted to the nuclei of the central and surrounding atoms, but repelled by each other. The molecule naturally finds a shape that brings the valence electron pairs as close to the nuclei as possible, while keeping them as far away as possible from other valence electrons. Molecular Geometry (handout) Look at the bond angles in molecules containing lone pairs and compare them to molecules with the same number of electron domains that are all bonds. Propose an explanation for the trends in bond angles. Molecular Geometry - actual experimental data Molecular Geometry (more information from textbook)
Propose an explanation for the trends in bond angles. One interpretation of the trend Lone pairs require more space than bonding pairs (as the number of lone pairs increase, the bonding pairs are increasingly squeezed together.) Makes sense - bonding pair shared between 2 nuclei, and electrons can be close to either nucleus. A lone pair is localized to only one nucleus, and both electrons will be close only to that nucleus, so takes up more space. One interpretation of the trend Lone pairs require more space than bonding pairs (as the number of lone pairs increase, the bonding pairs are increasingly squeezed together.) Makes sense - bonding pair shared between 2 nuclei, and electrons can be close to either nucleus. Valence Shell Electron Pair Repulsion (VSEPR) Model - useful for predicting geometries of molecules formed from nonmetals Main postulate of the model is that the structure around a given atom is determined principally by minimizing electron-pair repulsions. Add: Lone pairs require more room than bonding pairs and tend to compress the angles between the bonding pairs. Valence Shell Electron Pair Repulsion (VSEPR) Model - useful for predicting geometries of molecules formed from nonmetals
Main postulate of the model is that the structure around a given atom is determined principally by minimizing electron-pair repulsions.
Add: Lone pairs require more room than bonding pairs and tend to compress the angles between the bonding pairs.
Also - multiple bonds (double, triple) count as one effective electron pair (electron domain). One more thing:
When a molecule exhibits resonance, any one of the resonance structures can be used to predict the molecular structure using the VSEPR model. Molecular Shapes - sheet (part of VSEPR lab) Complete as much as you can on your own (ok to work with a friend) - THEN after you have made your predictions use the simulation to see how close you got. (Per 4) https://phet.colorado.edu/sims/html/molecule-shapes/latest/molecule-shapes_en.ht ml Unit 2
FRQ Practice 1. White gold is a common alloy of gold and palladium that is often used in jewelry. The atomic radii of the metals are: Au 135 pm Pd 140 pm a) A particular ring is made from an alloy that is 75 mole percent gold and 25 mole percent palladium. Using the box below, draw a particle-level diagram of the solid alloy consisting of 12 atoms with a representative proportion of atom types. Your diagram should clearly indicate whether the alloy is interstitial or substitutional. Use empty circles for gold and and shaded circles for palladium. 1. 12 circles about the same size in a regular array, fill the box as best you can - 9 empty and 3 shaded. 3:1; substitutional; radii approx. same size
2 2. Propanoic acid, C2H5COOH, is an organic acid that is liquid at room temperature.
a. An incomplete Lewis diagram for the propanoic acid molecule is provided in the box below. Complete the diagram, showing how the remaining atoms in the molecule are arranged around the carbon atom marked with an asterisk (*). Your structure should minimize formal charge and include any lone pairs of electrons. 2.a C2H5COOH Ve = 3 C(4) + 6(H) + 2 O(6) =30 Used so far 14 so Remaining at 30 - 14 = 16 2.a C2H5COOH Ve = 3 C(4) + 6(H) + 2 O(6) =30 Used so far 14 so Remaining at 30 - 14 = 16 Need to add OOH - H has to be terminal atom 2.a C2H5COOH Ve = 3 C(4) + 6(H) + 2 O(6) =30 So now used 20 electrons Re = 30 - 20 =10 Satisfy C* octet first leaving 8 electrons 2.a C2H5COOH Satisfy C* octet first leaving 8 electrons Then 2 lone pairs on each O atom. 2.a C2H5COOH Check number of electrons used and that each atom has an octet - duet for H.
1 Drat - need to do more reading and notes. b) hybridization of C with asterix.
1 2. Initial thoughts - pencil in. c) intermolecular forces - dispersion (everything has) also dipole-dipole forces (since its a polar molecule) and hydrogen bonds. 2. Answer c) i) The intermolecular forces in propanoic acid will be hydrogen bonds, dipole-dipole forces and London dispersion forces (LDF). 2. Answer - compared to butanoic acid 2 c) ii) The hydrogen-bonding and the dipole-dipole forces will be similar in both propanoic and butanoic acid. The forces responsible for the lower boiling point of propanoic acid compared to butanoic acid are the London dispersion forces since the propanoic acid has a smaller carbon chain and hence a smaller electron cloud to disperse. 3. Answer the following questions about N2 and N2H4. a. In the box below, draw the complete Lewis electron-dot
diagram of N2.
Ve = 10 Re = 10 - 2 = 8 if put 2 lone pairs on each N the neither will have an octet. What else can I do? Try a double bond
N - N 3. Answer the following questions about N2 and N2H4. a. In the box below, draw the complete Lewis electron-dot
diagram of N2.
Ve = 10 Re = 10 - 4 = 6 3 more pairs to use...
N = N 3. Answer the following questions about N2 and N2H4. a. In the box below, draw the complete Lewis electron-dot
diagram of N2.
Ve = 10 Re = 10 - 4 = 6 one lone pair each and one more bond - bingo!
All octets done. 2 3. a) Based on the Lewis-electron-dot diagram that you drew, is the N2 molecule polar? Explain.
The N2 molecule consists of 2 nitrogen atoms which have identical electronegativities, so the electron bonding pairs in the molecule will be evenly shared between the 2 nitrogen atoms and each N has one nonbonding electron pair so the molecule has symmetrical electron density and hence the molecule will be non-polar. 1 The following graph shows the potential energy of two nitrogen atoms versus the distance between their nuclei. On the graph, indicate the distance that corresponds to the bond length of the N2 molecule by placing an X on the horizontal axis. The response shows an X drawn on the axis and located directly above the lowest part (minimum) of the potential energy curve. X The lowest potential energy on the curve corresponds to the most stable intermolecular distance (i.e. bond length). 1 d) On the graph, which shows the potential energy curve of two N atoms, carefully sketch a curve that corresponds to the potential energy of two O atoms versus the distance between their nuclei. Need a curve similar shape but with:
2 d) On the graph, which shows the potential energy curve of two N atoms, carefully sketch a curve that corresponds to the potential energy of two O atoms versus the distance between their nuclei. Need a curve similar shape but with: a minimum to the right of where the given curve has its minimum (because the length of the O-O double bond is greater than that of the N-N triple bond.
Minimum value of curve needs to be higher than the minimum of the given curve (O=O not as strong as the N-N triple bond). 2 e) N2 can react with H2 to form the compound, N2H4. A sample of N2H4 has a mass of 25g. Identify the numerical quantity that is needed to convert the number of grams of
N2H4 to the number of moles of N2H4. (You do not need to do the calculation.) e) N2 can react with H2 to form the compound, N2H4. A sample of N2H4 has a mass of 25g. Identify the numerical quantity that is needed to convert the number of grams of
N2H4 to the number of moles of N2H4. (You do not need to do the calculation.)
We need the molar mass of N2H4 (32 g/mol) to convert the number of grams of N2H4 to the number of moles of N2H4 . 1 f) i) Based on the Lewis electron-dot diagrams of N2 and
N2H4, compare the length of the nitrogen-to-nitrogen bond in
N2 with the length of the nitrogen-to-nitrogen bond in N2H4.
1 f) i) Based on the Lewis electron-dot diagrams of N2 and
N2H4, compare the length of the nitrogen-to-nitrogen bond in
N2 with the length of the nitrogen-to-nitrogen bond in N2H4.
N to N bond in N2 is shorter than the nitrogen-to-nitrogen bond in N2H4, because the triple bond will be shorter than the single bond. 1 f) ii) Based on the Lewis electron-dot diagrams of N2 and
N2H4, compare the strength of the nitrogen-to-nitrogen bond in N2 with the strength of the nitrogen-to-nitrogen bond in
N2H4.
1 f) ii) Based on the Lewis electron-dot diagrams of N2 and
N2H4, compare the strength of the nitrogen-to-nitrogen bond in N2 with the strength of the nitrogen-to-nitrogen bond in
N2H4.
The nitrogen-to-nitrogen bond in N2 is stronger than the nitrogen-to-nitrogen bond in N2H4 (triple bond stronger than a single bond). 1 f) iii) Identify the hybridization of the N atoms in N2H4.
Rats, we better find out what hybridization is and how we can do this sort of identification. 1 4a) i) Write on the lines below the box, not next to the question. Angle x is approximately 120°. There are three electron domains around the carbon atom and a bond angle of 120o which will maximally separate the electrons and minimize the energy (trigonal planar geometry for electron domains). 1 4a) ii) 1 Angle y will be less than 109.5°. Four electron domains around the nitrogen atom will maximally separate the electrons and minimize the energy when the bond angles are 109.5°, but here one of the electron domains is a nonbonding or lone pair or electrons. According to the VSEPR model, nonbonding (lone) pairs require more room than bonding pairs and tend to compress the angle between bonding pairs (reduce the size of the angle). 4b) But always use the box when they give you one.
Ve 2 + 4+ 2(6) = 18
18 - 10 =8
Put lone pairs on most electronegative 2 atoms - the Os.