Electron Location and Charge of Molecules

Valence Shell Electron Pair Repulsion and Molecular Polarity Simulation Activities

Instructions: Perform the two simulations below and answer all questions. You can either type in your answers in the provided spaces or handwrite your answers and then scan the file to submit in assignments

1.Phet Molecular Shapes VSEPR Simulation Activity

Introduction

Atoms bond to satisfy their need for more electrons. Most atoms will share electrons to satisfy the Octet Rule – every atom wants 8 electrons to fill the s and p orbitals in the outer energy level. But, as you will see, sometimes atoms can deviate from and not follow the Octet Rule.

Because electrons have a negative charge and atoms occupy space, bonds and electrons will spread out as much as possible. Since we write in a two-dimensional plane on paper, it is difficult to visualize the true geometry of these molecules. This activity and the program you are about to use allows us to visualize on a more 3-dimensional scale.

Procedure

Log on to https://phet.colorado.edu/en/simulation/molecule-shapes either by Googling “phet simulations molecule shape.” Click on .

Part 1 – Model Generic Molecules

Click on .

Fill in the chart below by creating the generic molecules below. On your screen in the right side click on “remove all” to be left with just the purple central atom. On your screen in the lower left corner, click on “molecule geometry”, and on “electron geometry”. Add atoms and electron pairs as needed to produce the generic formula. Once the molecule is assembled, click and drag the screen to spin the atom around. Click on the “Show bond angles.” Use the following key:

·

· A – central purple atom – cannot be removed

· B – single bonded white atom

· C – double bonded white atom

· D – triple bonded white atom

· E – Electron pairs not bonded

In each box below:

1. Draw the molecule you create to the best of your ability

2. Write the Electron Geometry (EG) and the Molecule Geometry (MG) name in the box

3. Label the bond angle

4. Look at the central atom, is its octet satisfied?

Before moving on to the next molecule for each of the molecules you create: with your mouse left click/hold and move your mouse around to move the molecule and get the feel for the 3D shape.

Molecule

Molecule

AC

EG ___Linear__________________________

MG _____Linear___________Bond angle: 180 degrees_____________

Octet? ___Yes, for the central atom A, if it is a typical element that follows the octet rule, the triple bond provides it with six electrons, and it\\\\\\\'s presumed that the central atom also has another bond (possibly a hidden lone pair or implied second atom not mentioned in AC) to satisfy its octet.__

ABE3

EG ________Tetrahedral_____________________

MG ________Trigonal Pyramidal Bond Angle: Approximately 107 degrees_____________________ Octet? ____Yes, the central atom\\\\\\\'s octet is satisfied with one bond to atom B and three lone pairs._

AB3E

EG _______tetrahedral ____bond angle: 107______

MG _____trigonal pyramidal________________ Octet? ____Yes_

AB

EG ___________linear__bond angle: 180___________

MG _______linear_____________ Octet? __Yes___

ACE2

EG ______trigonal planar______bond angle: 104.5_

MG _______________bent______________ Octet? _____yes

AB2E2

EG ______ Tetrahedral_____bond angle: 104.5_______

MG _______bent______________________ Octet? _____yes

AB2C

EG __________tetrahedral, bond angle: 109.5___________________

MG ________trigonal planar_____________________ Octet? __yes___

AB3

EG _____Trigonal Planar________bond angle 120________________

MG ____Trigonal Planar_________________________ Octet? _____yes

AB4

EG ___ Tetrahedral____bond angle: 109.5___________

MG _____ Tetrahedral________________________ Octet? _____yes

ADE

EG ______linear, bond angle 180__________

MG ___linear__________________________ Octet? _____depends

Part 2 – Real Molecules

Click on the “Real Molecules” tab at the bottom of the page. Using the pull down menu, select the molecules below and fill in the chart. Match the molecule to the generic structure above in terms of Electron Geometry EG and Molecular Geometry MG. Fill in the generic bond angles as expected according to the generic model shapes from Part 1. Fill in the True bond angles as given by the simulation.

Molecule

Generic Structure

Generic (expected) bond angles (from Part 1)

True Bond Angles

H2O

EG____Tetrahedral___________

MG_________Bent_____

104.5

104.5

CO2

EG____Linear___________

MG______Linear________

180

180

CH4

EG_____Tetrahedral__________

MG________ Tetrahedral ___

104.5

109.5

NH3

EG____ Tetrahedral ____

MG____Trigonal Pyramidal____

4.5

107.8

BF3

EG__Trigonal Planar________

MG__Trigonal Plarnar_____

107.5

120

1. What does VSEPR stand for? Explain its meaning in your own words. VSEPR stands for Valence Shell Electron Pair Repulsion theory. In simple terms, it\\\\\\\'s a way of predicting the shape of molecules based on the idea that electron pairs around a central atom will position themselves as far apart from each other as possible to minimize repulsion. This happens because electrons have the same negative charge, and like charges repel.

2. What molecules in Part 1 consisted of only two atoms? AC, AB

3. The program did not give a bond angle to a molecule consisting of only two atoms. Why? Think geometry class. For a molecule consisting of only two atoms, the concept of a bond angle doesn\\\\\\\'t apply because a bond angle requires three points to define (such as in the case of a triangle). With only two atoms, the structure is linear by necessity, and there\\\\\\\'s no angle to measure between them.

4. Looking at the table in Part 2, some of the angles stayed consistent while others did not. Compare and contrast the two groups of molecules (those with matching angle measurements to those with different measurements). What is causing the angles to skew? Explain why this might be. In Part 2, some bond angles remain consistent with the expected angles from the generic models, while others differ due to real-world deviations from ideal shapes. This discrepancy results from factors like lone pairs of electrons, which exert more repulsion than bonded pairs and can compress bond angles, or due to differences in atomic sizes and electronegativity that can distort angles away from their ideal values. The presence of double or triple bonds can also influence the molecule\\\\\\\'s geometry by affecting electron distribution around the central atom.

5. What angle is needed to spread 4 bonds as far apart as possible? Hint: look at a molecule with four separate bonds.

The ideal angle to spread four bonds as far apart as possible is 109.5 degrees, which is the tetrahedral angle. This angle arises in molecules where the central atom is bonded to four other atoms with no lone pairs affecting the geometry, ensuring maximum spatial separation between bonds.

6. Find the two generic molecules from Part 1 that are made of 3 atoms.

a. Compare and contrast these two molecules by listing two similarities and two differences.

b. Give a real-life example of each.

Molecule

AB3

AB4

Similarities

Both AB3 and AB4 have multiple electron domains around the central atom (A), consisting of bonds with no lone pairs on the central atom in these idealized generic formulas.

The spatial arrangement in both aims to minimize electron pair repulsion, which dictates their molecular geometries.

Differences

AB3 has a trigonal planar molecular geometry if there are no lone pairs on the central atom, characterized by 120° bond angles. In contrast, AB4 adopts a tetrahedral geometry with approximately 109.5° bond angles, assuming no lone pairs.

The AB3 molecule, being trigonal planar, is essentially flat, lying in a single plane. AB4\\\\\\\'s tetrahedral structure gives it a three-dimensional shape, not confined to a single plane.

Real-life Examples

Boron trifluoride (BF3) is an example of a molecule with a trigonal planar geometry where a central boron atom is bonded to three fluorine atoms. BF3 is used in organic chemistry as a catalyst for certain reactions.

Methane (CH4) is a real-life example of the AB4 structure, where a carbon atom is at the center, bonded to four hydrogen atoms. Methane is a fundamental component of natural gas and is crucial for energy production.

2. Phet Molecular Polarity Simulation Activity

Introduction

Molecules polarity ultimately depends on the presence or absence of a permanent dipole within the molecule. A dipole is defined as “separation of charge at a short distance” it occurs when electrons are not symmetrically distributed within the atoms in a molecules, and the molecule has one side with a partial positive charge and a side with a partial negative charge. The factors to evaluate are the of presence (or absence) of polar covalent bonds and the shape of the molecule. A polar covalent bond is a dipole because electrons are shared unequally, the more electronegative element “stealing” them from the less electronegative element and acquiring a partial negative charge. If a molecule contains polar covalent bonds it has the potential of being polar, unless the shape of the molecule is symmetrical and these dipoles “cancel out”, yielding a net zero dipole. In this activity you will be able to experience the effect of electronegativity of atoms and molecular shapes have on the net dipole of different molecules.

Procedure

Log on https://phet.colorado.edu/en/simulation/molecule-polarity. either by Googling “phet simulations molecule polarity.” Click on .

Part 1

1. Choose Two Atoms.

2. Under View, click the box to show Partial Charges.

3. Without changing anything else, answer the following questions:

a. Which atom is more electronegative? B

b. Which atom has the partial negative charge? B

c. Which atom has the partial positive charge? A

d. Which atom does the bond dipole point toward? B