Help
Options
Didn't know it?
click below
click below
Knew it?
click below
click below
Don't Know
Remaining cards (0)
Know
retry
shuffle
restart
0:00
Embed Code - If you would like this activity on your web page, copy the script below and paste it into your web page.
Normal Size Small Size show me how
Normal Size Small Size show me how
Chemistry Chapter 9
Molecular Geometry: Shape Determines Function
| Question | Answer |
|---|---|
| Molecular Recognition (9.1) | Enables biomolecular structures (such as nasal membranes) to recognize and react when a molecule with a particular shape fits into a part of the structure known as an active site. |
| Valence-Shell Electron-Pair Repulsion (VSEPR) Theory (9.2) | Electrons have negative charges and repel one another. Valence electrons are situated around an atom in a way that creates the minimum amount of repulsion. |
| Electron-Pair Geometry (9.2) | Defines the relative positions in three-dimensional space of all the bonding and lone pairs of valence electrons around the CENTRAL atom. First step to determining molecular geometry. |
| Molecular Geometry (9.2) | Defines the relative positions in three-dimensional space of the atoms in a molecule. In order to determine molecular geometry, electron-pair geometry must be known. However, mol-geo may or may not be the same as electron-pair geometry (this depends on the addition of lone pairs, which complicates the process). |
| Steric Number (SN) (9.2) | The sum of the atoms bound to the central atom, added to the number of lone pairs in its valence shell. |
| Linear (SN = 2) (9.2) | The steric number of a linear molecule is 2, and the bond angles separating the two non-central atoms are 180° each. The three atoms in the molecule are arranged in a straight line. (360°÷2). |
| Trigonal Planer (SN = 3) (9.2) | The steric number of a trigonal planer molecule is 3, and the bond angles separating the non-central atoms are 120° each. The three non-central atoms are located at the points of an equilateral triangle. (360°÷3). |
| Tetrahedral (SN = 4) (9.2) | The steric number of a tetrahedral molecule is 4. The first case of atoms not being on the same geometric plane. Instead, they are located at the vertices of a four sided pyramid, called a tetrahedron. The four non-central atoms form bond angles of 109.5° with one another. |
| Trigonal Bipyramidal (SN = 5) (9.2) | The steric number of this molecule is 5. The five atoms in this shape exist at the points of a bipyramidal figure; two three-sided pyramids base-to-base. The central atom is in the middle of the plane on which convergence occurs, within the equilateral triangle created by the pyramidal figures. The bond angles between the three equatorial bonds are 120° as in a trigonal planar model. The angle between equatorial and axial (tops of pyramids) is 90° and the two axial bonds are 180° apart. |
| Octahedral (SN = 6) (9.2) | The steric number of a octahedral molecule is 6. This figure is equivalent to two square-basal pyramids linked at the base, similar to the trigonal bipyramidal. Every atom is equidistant, and situated on the three cartesian axes 90° from one another (and 180° from the atom directly across the sphere). |
| Steps for Applying/Drawing VSEPR Theory (9.2) | (1.) Draw a Lewis diagram, including resonance structures. (2.) Determine the steric number of the central atom. (3.) Use the steric number to predict the molecular geometries. To draw three-dimensional VSEPR diagrams on paper, use a solid wedge to indicate a bond that protrudes toward the viewer and a dashed wedge to indicate that the bond is behind the body of the molecule. Solid lines describe bonds that lay on the plane of the page. Orient structures so that the max # of bonds is on this plane. |
| Double Bonds & VSEPR (9.2) | VSEPR theory does not enable us to predict the actual values of the bond angles in molecules containing double bonds to the central atom, but it does allow us to correctly predict how the presence of a double bond causes a bond angle to deviate from the ideal value. Double bonds are made up of TWO pairs of bonding electrons (four electrons) and exert more repulsion than a single bond would. This greater repulsion decreases the angles of its constituents. |
| Lone Pairs & VSEPR (9.2) | Create geometries with names that are not the same as the molecules’ electron pair geometries. (SN ≥ 3) for this to be an issue. Lone pairs cause more repulsion, because typically bonded pairs are closer to the midpoint between nuclei. Unbound electrons therefore are closer to the central atom and have a stronger negative field. For example, trigonal planar model decreases to have an angle smaller than 120° between two bound atoms. Repulsion caused by lone pairs is even stronger than double bond repulsion. |
| Key to VSEPR Predictions (9.2) | Repulsions between pairs of e⁻ ↓ as the angle between them ↑. Two e⁻ pairs at 90° experience greater mutual repulsion than two at 120°, which have greater repulsion than two at 180°. To minimize repulsions w/ lone pairs, VSEPR theory predicts that they preferentially occupy equatorial rather than axial vertices, because an equatorial lone pair has two 90° repulsions with the two axial e⁻ pairs, but an axial lone pair has three 90° repulsions with the three equatorial e⁻ pairs. |
| Disphenoidal (Seesaw) Molecular Geometry (SN = 5) (9.2) | When a single lone pair occupies an equatorial site, we get a molecular geometry called a seesaw, because, when rotated 90° clockwise, it resembles a playground seesaw. (SN = 5) for this structure and bond angles are <90° <120° or <180°. When two lone pairs are added to this trigonal bipyramidal structure, it creates a T-shaped molecular structure. Three lone pairs may even create a linear molecular structure, in which the two remaining bonded atoms are on the axial points. |
| Bent (Angular) Molecular Geometry (SN = 3 or 4) (9.2) | When (SN = 3 or 4), and a lone pair is involved, the molecule will have a bent shape (like a water molecule with a 104.5° bond angle). The lone pair still forces the atoms alongside it into a trigonal planar structure. The two non-central bonded atoms share an angle that is marginally less than 120° because of the extra repulsion of the lone pair. |
| Square Pyramidal (or Square Planar) Molecular Geometry (SN = 6) | When an octahedral molecule has one or two lone pairs, a square pyramidal or square planar structure will be created, respectively. Square pyramidal will have bond angles that are <90° and <180° while planar will be exactly 90° or 180° from the opposite equatorial atoms. The square pyramidal central atom is embedded into the base plane. |
| Bond Dipoles (9.3) | Unequal distribution of bonding electrons produces a partial negative charge in one region of the bond and a partial positive charge in the other. Occurs especially often in Bent Angular molecules with (SN = 4), a tetrahedral electron-pair geometry, and a linear molecular geometry. (ex. H₂O or CS₂ which both have bond angles <109.5°) |
| Dipole Moment (µ) (9.3) | Expressed in units of debyes (D), where 1 D = 3.34 ✕ 10⁻³⁰ coulombs-meters. Measures the uneven dist. of e⁻ density over the entire molecule. The value of μ can be determined by measuring the degree to which molecules align with a strong electric field between parallel plates that have positive (+) and negative (−) charges. Polar molecules align with the field so that their negative regions are oriented toward the positive plate and their positive regions are oriented toward the negative plate. |
| Coulombmeter (9.3) | A tool for measuring the electrostatic charge of a material. A Coulombmeter is used in combination with a Faraday cup or a metal probe for taking charge measures of a material. |
| Permanent Dipole (9.3) | External factors do not affect the permanent dipole moment. Permanent dipole refers to the dipole moment that originally occurs in a compound due to uneven electron distribution. Therefore, a polar compound contains a permanent dipole moment. Polar covalency creates a state in which a more electronegative atom gets a partial negative charge while the less electronegative atom gets a partial positive charge. This establishes a permanent dipole in the molecule. |
| Polarity & Dipole Moment of CO₂ (9.3) | Electron densities around the two ends of the molecule are the same. Oxygen caps both ends of the carbon, and pull electrons away from [C] evenly at both locations. The two bond dipoles, oriented in opposite directions, exactly offset each other. The result is no permanent dipole and a dipole moment of zero (i.e. it is nonpolar). |
| Nonpolar Covalent Molecules (9.3) | Molecular symmetry is an important determining characteristic of what is nonpolar. Water (SN=4, Bent molecular geometry) is polar because the hydrogen dipoles are angled in such a way that they are unable to offset one another properly. However, even CF₄ is nonpolar because of the properly spaced cancellation effect of its Fluorine atoms. |
| Valence Bond Theory (9.4) | A theory for integrating the ideas from chapter seven (electron configuration in specific orbitals) with the molecular and electron-pair geometries described by VSEPR theory. Incorporates the assumption that (a) a chemical bond between two atoms results from overlap of the atoms’ atomic orbitals, and (b) the greater the overlap, the stronger and more stable the bond. |
| σ Bond (sigma) (9.4) | Whenever the region of highest electron density lies along the bond axis, covalent sharing of electrons is called a sigma bond (σ). The overlap of the 1s orbitals on two hydrogen atoms produces a single (σ) bond that holds the two hydrogen atoms together in a H₂ molecule. The covalent bond between H and Cl in HCl results from overlap between the half-filled hydrogen 1s orbital and a half-filled chlorine 3p atomic orbital. |
| Hybridization (Formation of Hybrid Atomic Orbitals) (9.4) | To account for the shapes of double bonds, or the angles separating each bond, atomic orbitals of different shapes and energies are "mixed" together. Their energy is the weighted average of component orbitals. LOOK AT EXERCISE 9.6 on page 469. |
| Possible Hybrid/Non-Hybrid Covalent Bonds (9.4) | Covalent bonds result either from (1) overlap of a hybrid orbital on one atom with an unhybridized orbital on another atom, (2) from overlap of two hybrid orbitals on two atoms, or (3) from overlap of two unhybridized orbitals on two atoms. |
| Tetrahedral Geometry: Hybrid (sp³) Orbitals (9.4) | Methane, CH₄ (SN=4) is a tetrahedral with 109.5° angles. A single [C] atom has an electron configuration of [He]2s²2p² (4 valence e⁻). One electron from the 2s orbital is promoted to 2p. Now, there are four = energy orbitals (each one with only one e⁻). This means the hybrid sp³ orbitals are able to bond with four [H]. They are named this way because they were mixed from (1) S and (3) P orbitals. Any atom with a set of four equivalent sp³ hybrid orbitals has a tetrahedral orientation of valence electrons. |
| Steric Number Indication (9.4) | An atom’s steric number always indicates how many of its orbitals must be mixed to generate the hybrid set, and the number of hybrid orbitals in a set is always equal to the number of atomic orbitals mixed. The carbon atom in methane has SN = 4, so four atomic orbitals on carbon are mixed to form four hybrid orbitals. |
| Trigonal Planar Geometry: Hybrid (sp²) Orbitals (9.4) | The VSEPR model defines (SN=3) for the carbon atom in CH₂O, so three atomic orbitals must be mixed to create 3 (σ) bonds (sp³ creates too many). To produce a trigonal planar geometry, the 2s orbital on carbon is mixed with two of the carbon 2p orbitals, and the third 2p orbital is left unhybridized. (sp²) Orbitals are lower in energy than (sp³) bc of their mixed orbital averages. The two lobes of the unhybridized p orbital lie above and below the plane of the triangle defined by (sp²). |
| Difference Between VSEPR and Valence Bond Theory (9.4) | VSEPR considers only the central atom, while valence bond theory considers the orbitals of every atom in the molecule. |
| π Bond (pi) (9.4) | The overlap of the two unhybridized p-orbitals above and below the plane of the σ-bonding framework. Pi bonds have their greatest electron density above and below the internuclear axis (or in front of and behind the internuclear axis). They can be formed only by the overlap of partially filled p-orbitals that are parallel to each other and perpendicular to the σ bond joining the atoms. Pi bonds are inherently weaker than sigma bonds because the side-on overlap is less effective than head-on overlap. |
| Central Atom (SN=3) Hybridization (9.4) | Central atoms with trigonal-planar electron-pair geometry have a steric number of three and are (sp²) hybridized. |
| Linear Geometry: Hybrid (sp) Orbitals (9.4) | Leaves the other two p-orbitals unhybridized, allowing a triple bond to form (σ/π/π). Pi bonds easily react with other atoms, more often than sigma bonds do. |
| Trigonal Pyramidal (9.2) | In the case of a theoretically tetrahedral structure, like [NH₃] bond angles are less than the 109.5° they'd typically be. A lone pair presses the tetrahedral shape flat, creating a true triagonal pyramid. Three bonding pairs, one lone pair |
| VB Hybrids Summary (9.4) | (sp³ hybridization uses all three p-orbitals and cannot create pi bonds. This is why atoms cannot create more than a certain number of double/triple bonds. Atoms like CH₄ use all of the p-orbitals making four sigma bonds.) (sp² hybridization can create two or three sigma bonds, because it creates 3 hybrid orbitals. One pi bond is able to form because an original p-orbital is unused.) (sp hybridization can form two sigma bonds and two pi bonds). |
| Condensed Structures (9.5) | Uses subscripts to indicate the number of times a subgroup is repeated. The numerical subscript after the groups in parentheses means that ___ of them connect to the terminal groups in a compound. |
| Skeletal Structures (9.5) | Uses no element symbols for carbon or hydrogen atoms; each line segment symbolizes one carbon–carbon bond within the molecule. These structures are assumed to contain the appropriate number of hydrogen atoms so that each carbon atom has a steric number of 4. |
| Resonance Stabilization (9.5) | Delocalization reduces the potential energy of the electrons in π bonds and lowers the energy of the entire molecule. Delocalized pi bonds are found in molecules with at least two pi bonds that alternate with single bonds |
| Aromatic Compounds (9.5) | Contain carbon rings with delocalized π electrons above and below the plane of the ring. An important class of these compounds, known as polycyclic aromatic hydrocarbons (PAHs), consists of molecules containing several benzene rings joined together. [C₆H₆]. |
| Intercalation (9.5) | To insert or position between or among existing elements or layers. The reversible inclusion or insertion of a molecule (or ion) into layered materials with layered structures. Examples are found in graphite and transition metal dichalcogenides, as well as PAHs into DNA. |
| Chirality (9.5) | Optical Isomers. generate mirror images of both molecules, and then we rotate the mirror images 180° to superimpose each mirror image on its original image. If the reflected, rotated image is superimposable on the original image, then the substance is not chiral. Scientists call it achiral. Although several features within molecules can lead to chirality, the most common is the presence of a carbon atom that has four different atoms or groups of atoms attached to it. |
| Molecular Orbital (MO) Theory (9.6) | Hybrid atomic orbitals are associated with a single atom in a molecule, but molecular orbitals are spread out over two or more atoms in a molecule; they represent discrete energy states. Electrons in molecules can move to higher-energy MOs when molecules absorb quanta of electromagnetic radiation. When the electrons return to lower-energy MOs, distinctive wavelengths of UV and visible radiation are emitted. |
| Bonding Orbitals (9.6) | Lobes of high electron density that lie between bonded pairs of atoms. The energies of bonding MOs are lower than the energies of the atomic orbitals that combined to form them, so populating them with electrons (each MO can hold two electrons, just like an atomic orbital) stabilizes the molecule and contributes to the strength of the bonds holding its atoms together. These bonding orbitals are oval and spans the atomic centers. |
| Antibonding Orbitals (9.6) | Lobes of high electron density that are not located between bonded pairs of atoms. When electrons are in antibonding orbitals, they destabilize the molecule. If a molecule were to have the same number of electrons in its bonding and antibonding orbitals, then there would be no net energy holding the molecule together, and it would never have formed. |
| Key Point in MO Theory (9.6) | Number of MOs = Number of orbitals involved in formation. Combining two atomic orbitals, one from each of two atoms, produces one low-energy bonding orbital and one high-energy antibonding orbital. |
| Sigma Molecular Orbital (9.6) | When the region of highest density is located along the bond axis it is labelled a sigma MO and the bond is sigma as well. An example MO is labelled as σ₁ₛ because it involves the 1s orbitals. The less stable antibonding MO is written σ*₁ₛ and describes two separate lobes of electron density and a region of zero electron density (a node) between the two atoms. |
| Molecular Orbital Diagram (9.6) | A molecular orbital diagram shows the relative energies of bonding and antibonding molecular orbitals and of the atomic orbitals that formed them. σ₁ₛ<1s² < σ*₁ₛ by energy level. |
| Bond Order Requirement (9.6) | A bond order of zero indicates that a molecule is unstable, and it doesn't even existed. It is calculated this way: Order = ½(bonding electrons - antibonding electrons). The greater the bond order, the stronger the bond and the more stable the molecule |
| Net Gain in Stability (9.6) | Given that the bonding orbital of a molecule is lower in energy than the original atomic orbitals, and the antibonding orbitals are higher in energy, There must be a net GAIN in stability in order for the molecule to exist. He₂ does not exist, because two full 2s orbitals combine to create a full antibonding orbital and a full bonding orbital. These new higher/lower energies evenly cancel out and create no net gain in stability. |
| General Guidelines for Constructing an MO diagram (9.6) | (1) # of MOs = # of AOs used (2) AOs with similar energy/orientation mix better than those w/ different energies/orientations. An s-orbital mixes more easily with another s-orbital than with a p-orbital. The 1s and 2s orbitals have different sizes/energies, resulting in less effective mixing than two 1s or two 2s orbitals (3) Better mixing = greater energy difference btw. bonding/antibonding = more stable (4) An MO can hold two opp. spin e⁻ (5) Ground state molecules follow Hund's rule + Aufbau's principle. |
| Can p-orbitals form σ bonds? (9.6) | Yes. Each set of degenerate p-orbital subshell is made up of three separate AOs. The orbitals are each oriented on a cartesian axis, and the p(z)-orbitals of two bonding atoms face each other and are level at the bond axis. |
| Verticle Sequences (N₂ vs. O₂) (9.6) | The vertical sequence of orbitals for N₂ applies to MOs of the homonuclear diatomic molecules of elements (Z ≤ 7).The O₂ sequence applies to homonuclear diatomic molecules of all elements beyond oxygen (Z ≥ 8), including the halogens. |
| MO sequence for O₂ (valence of 12) (9.6) | (σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)² |
| MO sequence for N₂ (valence of 10) (9.6) | (σ2s)²(σ*2s)²(π2p)⁴(σ2p)² |
| Predicting Bond Types Using MO Theory (9.6) | Calculate bond order using the number of electrons in bonding orbitals minus the number in antibonding orbitals. Divide this number by two, and the resulting value describes the number of bonds between the two bonded atoms (Bond order 2 = double bond, as in O₂). |
| Diamagnetic (9.6) | Complete electron pairing means these molecules are repelled slightly by a magnetic field. |
| Paramagnetic (9.6) | Incomplete pairing of electrons within a molecule will result in the entity being attracted to magnetic fields. The more unpaired electrons, the greater its attraction. |
| Elements before [N] that display modified MO Energy Levels (9.6) | Li, Be, B, C, and N itself. All of these fill in this order: (σ2s)²(σ*2s)²(π2p)⁴(σ2p)²(π*2p)⁴(σ*2p)² Whereas the typical order is: (σ2s)²(σ*2s)²(σ2p)²(π2p)⁴(π*2p)⁴(σ*2p)² In other words, for diatomic homonuclear bonds after and including O₂ (F₂, Ne₂) the π2p (bonding and antibonding) orbitals are located BETWEEN the energy levels of σ2p bonds, while they are LATERALLY integrated with the ≤ N₂ atoms' σ2p bonds. [NOTE: Be₂ & Ne₂ both have bond orders of zero and do not exist] |
| Number of MOs (2p subshell) (9.6) | First atom contributes: 1 pz 1 px 1 py. Second atom contributes: 1 pz 1 px 1 py. This creates a total of 6 MOs. Since pz orbitals are oriented ON the bond axis, they are sigma bonds that experience direct overlap. There is thus 1 σ2p bonding orbital and 1 σ2p antibonding orbital (1 pz + 1 pz = 2MOs). The remaining two py orbitals and two px orbitals form pi bonds AROUND the bond axis. There are thus 2 π2p bonding orbitals and 2 π2p antibonding orbitals ( 1 px + 1 py + 1 px + 1 py = 4MOs). |
| Drawing MOs for Heteronuclear Molecules (9.6) | Bonding MOs tend to be closer in energy to the atomic orbitals of the more electronegative atom, whereas antibonding MOs tend to be closer in energy to the atomic orbitals of the less electronegative atom. In the case of carbon monoxide (CO) orbitals fill according to O₂'s energy order (π bonds and antibonds sandwiched between 2p sigma bonds). |
| Aurora Phenomenon (9.6) (Related to light spectra, Chapter 7) | Collisions with ions in the solar wind result in the promotion of electrons from the σ2p orbitals in N₂ molecules to π*2p orbitals. Higher-energy collisions create N²⁺* molecular ions, which also have electrons in π*2p orbitals. When these electrons return to the ground state, red and blue-violet light are emitted. |
| Nonbonding Molecular Orbital (n) (9.6) | Mixing three orbitals (such as when hybridizing) creates 3 MOs, 1 antibonding, 1 bonding, and 1 nonbonding (n). These orbitals do not lower/raise a molecule's energy; null when calculating stability. They are = as energetic as the AOs that create them. For example, adding 4 electrons to the mixed p-orbitals will create full nonbonding/bonding orbs. The antibonding orbs are left empty. (In this example, the molecule is σ bonded and the remaining 4 e⁻ are being placed in the unhybridized p-orbitals left over) |
| Nonbonding Electron Location (9.6) | The pair of nonbonding electrons is delocalized. It is factored in to the bonding order of the entire molecule and divided by the number of atoms. This provides an explanation for resonance structures with fractional bonds. |
| Characteristics of Metallic Bonds (9.6) | Cu atoms in solid copper are surrounded by a total of 12 other Cu atoms and each is bonded to all 12 neighbors. This means each Cu atom must share its 4s electron with many atoms. Inevitably, the dispersion of bonding electrons weakens the Cu―Cu bond between each pair of Cu atoms. Metallic elements have low ionization energies, which reflects the diffuse nature of this bonding. Many metallic atoms emit photoelectrons when illuminated by photons of ultraviolet radiation or even lower-energy visible light. |
| Band Theory (9.6) | When the 4s atomic orbitals on two Cu atoms overlap to form Cu2, the atomic orbitals combine to form two molecular orbitals with different energies, equally spaced above and below the initial energy value. (Almost like the formation of lower energy bonding orbitals and higher energy antibonding orbitals). |
| Valance Band Explanation of Conductivity (9.6) | No gap exists between the energy of the occupied lower portion of the valence band and the empty upper portion. Electrons can be promoted to the upper level of the band, where they are free to migrate through delocalized empty orbitals throughout the solid. |
| Semiconductors (9.6) | Have a limited ability to conduct electricity (ex. silicon). |
| Doping/Dopant (9.6) | To enhance the conductivity of any elemental metalloid by replacing some of the atoms with another species. The dopant (replacement atoms) must have a similar atomic radius and often the presence of extra electrons populates the band gap preventing efficient conductivity (ex. Nitrogen-Doped Mesoporous Graphitic Carbon Spheres. Their synthesis was the subject of a paper by Dr. Dassanayake, assisted by Shen group). |
| n-Type Semiconductor (9.6) | The dopant contributes extra negative charges (electrons) to the structure of the host element. |
| p-Type Semiconductor (9.6) | A reduction in the number of negatively charged electrons in the valence band is equivalent to the presence of positively charged "holes" in the material. |
Created by:
bluonk
Popular Chemistry sets