Unit 2: Topic 7 - VSEPR and Bond Hybridization Flashcards
State the fundamental idea of the VSEPR theory.
VSEPR, or valence-shell electron-pair repulsion theory, is a model assuming that atoms in a molecule assume a shape that minimizes the repulsion between electrons. Any atoms with lone pair electrons would like to be as far away from each other as possible, as elecrons that are too close will repel each other.
For example, if there are two atoms connecting to a central atom (BeF2), then repulsion is minimized by pointing the fluorine atoms completely opposite, giving a linear shape. If there are three atoms (BF3), the repulsion is minimized by utilizing an equilateral triangle. 4 atoms (CCl4) is minimized with a tetrahedron, 5 (PF5) with trigonal bipyramidal, and 6 (SF6) with octahedral.
Based on molecular geometry, consider the possible bond angles when a central atom is bonded to two atoms, and explain how VSEPR plays a role. Also, give an example and explain why the example assumes a specific shape.
2 atoms: Linear (180 degrees, CO2, BeF2)
The electrons will experience the least repulsion if they are as far away from each other as possible. Since in a linear model only the two terminal atoms have lone pairs (the central atom cannot), the two atoms should point in opposite directions, as sexplained by VSEPR. Therefore, this shape would be linear, and the bond angles would be 180 degrees.
CO2 is an example of a linear atom. When drawing Lewis structures, carbon, the central atom, bonds to each terminal oxygen in a double bond. Therefore, carbon has no lone pairs and is bonded to two oxygen atoms, and is linear.
Based on molecular geometry, consider the possible bond angles when a central atom is bonded to three atoms, and explain how VSEPR plays a role.
A central atom with bonding to three atoms either has three bonds and no lone pairs, or two bonds and one lone pair.
Trigonal Planar: if a central atom has three bonds, the lone pairs on the electrons want to be as far away from each other as possible. The terminal atoms would form the vertices of an equilateral triangle, the bond angles would be 120 degrees, and the shape is “trigonal planar.”
Bent: if a central atom has two bonds and a lone pair of electrons, the lone pair electrons on the terminal atoms want to be further away from the lone pair on the central atom by VSEPR. Therefore, the bond angle slightly decreases to about 117 degrees.
Based on molecular geometry, consider the possible bond angles when a central atom is bonded to four atoms.
Tetrahedral: if a central atom is bonded to four elements with no lone pairs on the center, the terminal atoms will take the shape of a tetrahedron. These bond angles will then be 109.5 degrees. (109.5 degrees, CH4)
Trigonal Pyramidal: if a central atom is bonded to three elements and has a single lone pair, the three terminal atoms will be slightly closer together than tetrahedral to reduce repulsion from the central lone pairs. (107 degrees, NH3)
Bent: if a central atom is bonded to two elements and has two lone pairs, the bond angles will again decrease slightly to mminimize repulsion from the central lone pairs. (104 degrees, H2O)
Linear: if a central atom is only bonded to one element with three lone pairs, only one element has lone pairs, so thiss molecule or ion is linear. (180 degrees, HCl)
Based on molecular geometry, consider the possible bond angles when a central atom is bonded to five atoms, and explain how VSEPR plays a role.
Notice that any atom with 5 bonds will violate an octet rule. However, the octet rule does have exceptions, such as listed in this example and the next.
Trigonal Bipyramidal: if a central atom bonds to 5 terminal atoms and itself has no lone pairs, the terminal atoms take up a shape known as “trigonal bipyramidal,” as shown here. There are axial and equatorial atoms: the axial atoms form a 90 degree bond angle, the equatorial atoms form a 120 degree bond angle.
Seesaw: a Seesaw shape is achieved with atoms occupying two axial positions and two equatorial positions: the remaining equatorial position is occupied by a lone pair. The bond angles are again 90 and 120 (SF4).
T-shaped: A T-shape is achieved with atoms occupying two axial and one equatorial position, with two lone pairs. Here, all bond angles are 90 degrees (ClF3)
Linear: if atoms only occupy the axial positions and equatorial positions are all lone pairs, the shape is linear. (ICl2)
Based on molecular geometry, consider the possible bond angles when a central atom is bonded to six atoms, and explain how VSEPR plays a role.
Octahedral: This shape occurs when a central atom forms 6 different bonds to terminal atoms. All bond angles are 90 degrees.
Square Pyramidal: This shape occurs when there are 5 bonds and one lone pair. Again, bond angles are 90 degrees.
Square Planar: This shape occurs when there are 4 bonds and two lone pairs. Again, bond angles are 90 degrees.
Consider bond types (and bond lengths) with molecular geometries.
By the octet rule (which shows that VSEPR and Lewis structures have to work together), any tetrahedral compound consists of single bonds. For trigonal bipyramidal and octahedral, these would be exceptions to the octet rule. However, lone pairs or not, they would also have single bonds.
Exceptions to the octet rule for linear and trigonal planar compounds involve Beryllium and Boron (BeF2, BF3). However, if a compound does not have either of these exceptions, a reasonable conclusion is that there would be at least one bond that is not a single bond (either double or triple) in order to satisfy the octet rule. For example, ozone (O3) consists of a double bond in both of its resonance structures.
Consider bond energies with molecular geometries.
Recall that the shorter a bond, the higher the bond energy. Triple bonds have the highest bond energy, followed by double bonds and single bonds. For example, if VSEPR predicts a model with tetrahedral geometry, then all of the bonds will be single bonds, so the bond energies will be comparatively low. For example, VSEPR predicts that CO2 will have double bonds, and CO3 2- will have resonance structure (a mix of single and double bonds). From here, we can conclude that CO2 will have carbon-oxygen bonds with higher bond energies.
How could you use a combination of Lewis structures and VSEPR to predict dipole moments?
Lewis structures and VSEPR together will help determine the exact structure of a molecule. If the structure is symmetric, the compound is nonpolar: if the structure is nonsymmetric, the compound is polar: this is because in a symmetric molecule, every individual dipole moment cancels out.
Consider CH4: every H-C bond has a small but existent dipole moment because carbon is more electronegative than hydrogen, but when all four bonds are arranged in a tetrahedral shape, the dipole moments in each direction will cancel out with each other, so CH4 is nonpolar.
How could you use a combination of Lewis structures and VSEPR to predict hybridization?
VSEPR alone cannot predict hybridization on a compound (again, H2O seems like there are only two
bonds, and VSEPR without Lewis structures cannot determine that there are two lone pairs). However,
using a combination of the octet rule and the VSEPR, we would be able to figure out how many lone
electron pairs there are and to determine the hybridization.
For example, with NH3, there are three single bonds, as each nitrogen is single bonded to hydrogen
(as hydrogen can only contribute one electron). However, this does not satisfy the octet rule: in this
case, nitrogen only has 6 electrons, and must include a lone pair to reach 8. From here, there are three
bonds, and a lone pair, which allows for sp3 hybridization.
What is the hybridization of carbon in methane, CH4? What about on the nitrogen in ammonia?
Since the carbon in methane is bonded to 4 atoms by single bonds there are no other electron pairs, the hybridization on the carbon in methane would be sp3. For NH3, the nitrogen is bonded to 3 atoms and there is one electron pair. However, the lone pair constitutes a bond when considering hybridization, so the nitrogen is also sp3 hybridization.
Explain why when the central atom is sp hybridized, the ideal bond angle is 180, and for sp2 is 120, and for sp3 is 109.5
When a central atom is sp hybridized, the central atom would expect to bond to two other atoms. Repulsion is minimized in a linear formation and a bond angle of 180.
Similarly, sp2 hybridization is bonding with three atoms, and repulsion is minimized in a triangular (trigonal planar) shape with a 120 bond angle.
Finally, sp3 is bonding with four atoms, where repulsion is minimized by a tetrahedral shape, with bond angles 109.5.
Define a sigma bond and pi bond and list some of their properties
Sigma bonds are the result of head-to-head overlapping atomic orbitals. They are the strongest covalent bonds due to direct overlap of orbitals. Pi bonds formed by lateral overlap of two atomic orbitals. Since there is a lesser degree of overlap, they are weaker than sigma bonds.
In a carbon-carbon triple bond, how many sigma bonds and pi bonds are there?
In any bond, there will exist the strong sigma bond. The remaining bonds are all weaker pi bonds. Since this is a triple bond, the first bond will be a sigma bond, and the remaining two are pi bonds.
(EXTRA) What is the number of unpaired electrons in a ground-state gaseous NF molecule? (Source: USNCO National Exam 2022).
