Chem 27 Chapter 14 & 19
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Substances having only paired electrons are called: A. Paramagnetic B. Diamagnetic C. Ferromagnetic D. None of the above
B. Diamagnetic Explanation: Substances with only paired electrons are called diamagnetic.
Knewton (Lewis Acids and Bases) Which of the following substances is the strongest Lewis base? A. HF B. F- C. NH4+ D. Ag+
B. F- Explanation: The fluoride ion has a formal negative charge, which makes it possible for it to donate an electron pair, as opposed to the other species which are neutral or positively charged.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) The compound [Fe(H2O)6]SO2 has a pale green color. The compound K4[Fe(CN)6] has a bright yellow color. Which of the following statements are NOT true? A. The compound K4[Fe(CN)6] absorbs higher energy light. B. H2O produces a stronger ligand field than CN−. C. The compound K4[Fe(CN)6] is likely a low-spin complex because it contains strong-field ligands. D. The compound [Fe(H2O)6]SO2 is likely a high-spin complex because it contains weak-field ligands.
B. H2O produces a stronger ligand field than CN−. Explanation: The complex K4[Fe(CN)6] absorbs higher energy light than the complex [Fe(H2O)6]SO2, so CN− is a stronger ligand than H2O. The complex [Fe(H2O)6]SO2 is a high-spin complex.
Knewton (Coordination Chemistry of Transition Metals) What is the geometry of the complex [Ag(NH3)2]Cl? A. octahedral B. linear C. square planar D. trigonal planar
B. linear Explanation: The complex [Ag(NH3)2]Cl is a 2-coordinate structure. 2-coordinate structures form linear complexes.
Knewton (Hybridization) Which molecule has a central atom that is sp2 hybridized? A. NH3 B. PCl3 C. BF3 D. CH4
C. BF3 Explanation: The boron atom has three electron domains, so it will be sp2 hybridized.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) The complex [Cu(NH3)4]2+ absorbs orange light. Therefore, the complex will likely appear to be what color? A. Red B. Yellow C. Blue D. Orange
C. Blue Explanation: A complex is generally the complement of the color it absorbs. A complex that absorbs orange light will most likely appear blue.
Knewton (Hybridization) Which of the following exhibits sp3 hybridization? A. BH3 B. CH2O C. CCl4 D. all of the above
C. CCl4 Explanation: CCl4 exhibits sp3 hybridization because the carbon atom has 4 electron domains.
Knewton (Hybrid Atomic Orbitals) What is the hybridization of the xenon atom in XeF6? A. sp3 B. sp3d C. sp3d2 D. sp3d3
D. sp3d3 Explanation: With six atoms and one lone pair, the seven total electron domains make it sp3d3 hybridized.
Quiz 14 (reading) The second and third bond between multiply bonded atoms is formed between. A. unhybridized s orbitals B. sp hybrids C. orbital theories cannot account for multiple bonding D. unhybridized p orbitals
D. unhybridized p orbitals
Knewton (Molecular Orbital Theory: Energy Diagrams) Which of the following describes a covalent bond as the overlap of half-filled atomic orbitals that yields a pair of electrons shared between two bonded atoms? A. atomic orbital theory B. covalent bond theory C. electron pair theory D. valence bond theory
D. valence bond theory Explanation: The overlap of half filled atomic orbitals in the same region of space allows for the attraction of positively charged nuclei and negatively charged electrons to form a strong electron bonded pair, resulting in a covalent bond. This concept is described by valence bond theory.
Knewton (Molecular Orbit Theory Bond Order) What is the highest occupied molecular orbital for O2? A. σ2p bonding B. σ∗2p antibonding C. π2p bonding D. π∗2p antibonding
D. π∗2p antibonding Explanation: The π∗ orbitals will each have one electron in them.
Knewton (Coordination Chemistry of Transition Metals) In square planar complexes, ligand pairs form what bond angles? A. 120∘ B. 90∘ C. 180∘ D. 109.5∘
B. 90∘ C. 180∘ Explanation: In square planar complexes, each ligand has two other ligands at 90∘ angles (called the cis positions) and one additional ligand at an 180∘ angle, in the trans position.
Knewton (Molecular Orbit Theory Bond Order) What is the maximum bond order that is possible for a bond between elements in the second row of the periodic table? A. 1 B. 2 C. 3 D. 4
C. 3 Explanation: No element in the second row can form more than a triple bond.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Sapphire is aluminum oxide. Aluminum oxide crystallizes with aluminum ions in two-thirds of the octahedral holes in a closest-packed array of oxide ions. What is the formula of aluminum oxide? A. AlO B. Al3O2 C. Al2O3 D. Al2NO3
C. Al2O3 Explanation: Because there is one octahedral hole per anion (oxide ion) and only two-thirds of these holes are occupied, the ratio of aluminum to oxygen must be 2/3:1, which would give Al23O. The simplest whole number ratio is 2:3, so the formula is Al2O3. We also know that the ions in this crystal must be present in a ratio that will make the solid electrically neutral (its net charge must be zero). To cancel out the charge on two Al3+ cations, we need three O2− anions. This, too, gives us the formula Al2O3.
Knewton (Molecular Orbit Theory Bond Order) What will be the bond order for C2?
2 Explanation: There will be a double bond in a molecule of C2.
Which bond is strongest in a carbon-carbon triple bond? A. the sigma bond B. the first pi bond C. the second pi bond D. these are all the same strength
A. the sigma bond Explanation: The sigma bond involves direct orbital overlap, so that will be the strongest bond.
Knewton (Coordination Chemistry of Transition Metals) What is the coordination number of the compound [Ru(NH3)2(H2O)ClBr]Br?
5 Explanation: The coordination number is determined by counting the number of ligands inside the coordination sphere, or inside the brackets of the molecular formula. Inside the brackets there are two NH3 ligands, one H2O ligand, one Cl ligand, and one Br ligand, making five ligands. Each ligand is monodentate, so the coordination number is 5. The counter ion outside the brackets is Br.
Knewton (Molecular Orbital Theory: Energy Diagrams) How many filled molecular orbitals are present in the molecular orbital diagram for H2? A. 1 B. 2 C. 4 D. 8
A. 1 Explanation: Molecular hydrogen consists of two hydrogen atoms, each containing 1 electron in a 1s atomic orbital. Therefore, the molecular orbital diagram for H2 will have 2 electrons, both of which are placed in the lowest energy sigma bonding molecular orbital.
Knewton (Lewis Acids and Bases) A. 2Cl− in BeCl2+2Cl−→BeCl2−4 B. BF3 in F−+BF3→BF−4 C. BF−4 in F−+BF3→BF−4 D. CaCO3 in CaO+CO2→CaCO3
A. 2Cl− in BeCl2+2Cl−→BeCl2−4 Explanation: In a Lewis acid-base reaction, the Lewis base is the species that donates a pair of electrons. The chloride ions are the electrons that are donated.
Quiz 14 (lecture) Use MO theory to compute the bond order of N2. In the subtractions below, consider only pairs in bonding/antibonding MOs constructed from valence atomic orbitals. A. 4-1=3 B. 4-3=1 C. 5-0=5 D. 5-2=3
A. 4-1=3
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) What is a high-spin complex? A. A high-spin complex is a coordination compound with weak-field ligands that doesn't follow the Aufbau principle. B. A high-spin complex is a coordination compound with strong-field ligands that doesn't follow the Aufbau principle. C. A high-spin complex is a coordination compound with weak-field ligands that follows the Aufbau principle. D. A high-spin complex is a coordination compound with strong-field ligands that follows the Aufbau principle.
A. A high-spin complex is a coordination compound with weak-field ligands that doesn't follow the Aufbau principle. Explanation: A high-spin complex is a complex with weak-field ligands that does not follow the Aufbau principle.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) A tetrahedral hole is described by which of the following? A. A tetrahedral hole is a space formed by four atoms or ions in a crystal. B. A tetrahedral hole is the simplest repeating unit of a face-centered cubic crystal. C. A tetrahedral hole is an open space in a crystal at the center of six particles located at the corners of a tetrahedron. D. none of the above
A. A tetrahedral hole is a space formed by four atoms or ions in a crystal. Explanation: A "hole" is a space between atoms within a crystal. Therefore, a tetrahedral hole is defined as a space formed by four atoms or ions in a crystal. For example, a cation in a tetrahedral hole will be adjacent to four anions. One anion will be in the plane above and three anions in the plane below the cation.
Quiz 14 (lecture) Match each diatomic molecule/ion to one that it is isoelectronic with. A. CN- B. C2 C. NeO (in outer space) D. NO
A. CN-: N2 B. C2: BN C. NeO (in outer space): F2 D. NO: CF
Knewton (Molecular Orbit Theory Bond Order) A switch in orbital ordering occurs because of A. s-p mixing B. bond order C. degenerate atomic orbitals D. decreasing orbital energies
A. s-p mixing Explanation: s-p mixing causes a switch in orbital ordering. It does not create new orbitals; it merely influences the energies of the existing molecular orbitals.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Which statements are true about constructive interference? A. Constructive interference is a process by which waves combine to yield an increase in amplitude B. Constructive interference results when waves are diffracted at Bragg angle C. Destructive interference results in an increase in the amplitude of both interacting waves D. The Bragg equation relies on the perfectly destructive interference of two waves
A. Constructive interference is a process by which waves combine to yield an increase in amplitude B. Constructive interference results when waves are diffracted at Bragg angle Explanation: When scattered waves traveling in the same direction encounter each other, they undergo interference, a process by which the waves combine to yield either an increase (constructive interference) or a decrease (destructive interference) in amplitude (intensity) depending upon the extent to which the combining waves' maxima are separated. When the waves are diffracted at the Bragg angle, constructive interference results. When diffraction occurs at a different angle that does not satisfy the Bragg condition, destructive interference results.
Knewton (Coordination Chemistry of Transition Metals) Which atom/group of atoms is NOT in the coordination sphere of K[Co(NH3)2Cl2Br2]? A. K B. NH3 C. Br D. Cl
A. K Explanation: In the compound K[Co(NH3)2Cl2Br2], potassium (K) is not in the coordination sphere. It is a counter ion that balances the charge of the coordination sphere.
Knewton (Hybrid Atomic Orbitals) Which of the following molecules has a nitrogen atom with an sp2 hybridization? A. NO3- B. HCN C. N2 D. NH3
A. NO3- Explanation: The NO3− molecule contains a nitrogen with 3 electron regions so its hybridization is sp2.
Knewton (Molecular Orbital Theory: Energy Diagrams) Which of the following molecules will contain a node, or nodal plane, along the internuclear axis? A. O2 B. H2 C. N2 D. HCl E. Br2
A. O2 C. N2 Explanation: In a π bond, the regions of orbital overlap lie on opposite sides of the internuclear axis. Along the axis itself, there is a node: a plane in which there is no probability of finding an electron. Of the molecules listed above, only O2 and N2 will contain π bonds, and so only O2 and N2 will contain a node along the internuclear axis.
Quiz 14 (lecture) Which aspects of covalent bonding are clarified by working through MO theory for diatomics? A. That multiple bonding involves π orbitals (e− not directly between nuclei) B. Why C2 is reactive instead of extremely stable. C. What it means for electrons to be "shared" D. Why O2 is paramagnetic, even though it has an even number of electrons E. How to predict the geometries of general polyatomic molecules F. How "sharing" electrons results in a bonding curve (bond energy) G. Where the numbers in the expression "4 × 2(pair) = 8(octet)" physically come from.
A. That multiple bonding involves π orbitals (e− not directly between nuclei) B. Why C2 is reactive instead of extremely stable. C. What it means for electrons to be "shared" D. Why O2 is paramagnetic, even though it has an even number of electrons F. How "sharing" electrons results in a bonding curve (bond energy) G. Where the numbers in the expression "4 × 2(pair) = 8(octet)" physically come from.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) What is a true statement about a coordination complex where the crystal field splitting is less than the electron-pairing energy? A. The complex is a high-spin complex. B. The complex is a low-spin complex. C. The complex has strong-field ligands. D. The complex will follow the Aufbau principle when filling electrons.
A. The complex is a high-spin complex. Explanation: When the crystal field splitting is less than the electron-pairing energy, the electrons occupy the higher-energy orbitals before pairing up in lower-energy degenerate orbitals.
Knewton (Hybrid Atomic Orbitals) Hybridization predicts geometry best for: A. small atoms B. large atoms C. electronegative atoms D. gaseous atoms
A. small atoms Explanation: Hybridization begins to break down in its ability to predict geometry as atoms get larger.
Quiz 14 (reading) Match each hybridization scheme to the relevant electronic geometry (all numerals should be superscripted, but Canvas doesn't support it for matching questions). A. sp B. sp2 C. sp3 D. dsp3 E. d2sp3
A. sp: linear B. sp2: trigonal planar C. sp3: tetrahedral D. dsp3: trigonal bipyramidal E. d2sp3: octahedral
Knewton (Coordination Chemistry of Transition Metals) What is the coordination number of a coordination compound? A. the number of donor atoms that form coordinate covalent bonds B. the number of ligands bonded to the central metal or ion C. the charge of the coordination sphere D. the oxidation state of the central metal ion or atom
A. the number of donor atoms that form coordinate covalent bonds Explanation: The coordination number is the number of coordinate covalent bonds in a coordination compound. This is equal to the number of donor atoms.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) The complex [Co(H2O)6]Cl3 has weak-field ligands. What will occur if the ligands are replaced with stronger CN− ligands? A. The complex will go from being paramagnetic to diamagnetic. B. The complex will absorb longer wavelengths of light. C. The color will change, going from yellow in the weak-field complex to blue in the strong-field complex. D. The complex with strong-field ligands will absorb more light than the complex with weak-field ligands.
A. The complex will go from being paramagnetic to diamagnetic. Explanation: The complex [Co(H2O)6]Cl3 has six weak-field ligands. The complex has an octahedral geometry and is a high-spin complex. The charge on the coordination sphere must be 3+ in order to balance the charge of the three chloride counter ions. The H2O ligands are uncharged, and so cobalt must be in a +3 oxidation state. Cobalt is a member of group 9, so cobalt(III) will have 6 d electrons. The high-spin complex will have 4 unpaired electrons, so it will be paramagnetic. If the ligands are replaced with strong-field ligands, then the complex will be in a low-spin state, and the 6 electrons will pair in the t2g orbitals, making the complex diamagnetic. The complex with weak-field ligands will absorb lower energy light, red and yellow, and transmit higher energy light, causing it to appear blue or green. If the ligands are replaced with strong-field ligands, the complex will absorb higher energy blue and green light and transmit higher energy light, making them appear red or yellow.
Knewton (Molecular Orbital Theory: Energy Diagrams) Which of the following is included in molecular orbital diagrams? A. Valence electrons B. Core electrons C. Excited electrons D. All of the above
A. Valence electrons Explanation: Molecular orbital diagrams include only valence electrons, core electrons can be omitted.
Knewton (Coordination Chemistry of Transition Metals) What is a ligand? A. an atom, molecule, or ion that can form a covalent bond to a metal atom by donating electrons B. an atom, molecule, or ion that can form a covalent bond to a metal atom by accepting electrons C. an atom, molecule, or ion that can form an ionic a bond to a metal atom through strong electrostatic interactions D. an atom, molecule, or ion that can form a covalent bond to a metal atom by sharing electrons
A. an atom, molecule, or ion that can form a covalent bond to a metal atom by donating electrons Explanation: A ligand is an atom, molecule, or ion that can form a covalent bond to a metal atom by donating electron density. Ligands are generally considered Lewis bases.
Knewton (Molecular Orbit Theory Bond Order) Which orbitals result in bond stabilization when they are occupied? A. bonding orbitals B. antibonding orbitals C. sigma orbitals D. pi orbitals
A. bonding orbitals Explanation: A bonding orbital is a molecular orbital located between two nuclei; electrons in a bonding orbital stabilize a molecule. An antibonding orbital is a molecular orbital located outside of the region between two nuclei; electrons in an antibonding orbital destabilize the molecule.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Diffraction occurs when electromagnetic waves: A. change direction due to a physical barrier B. change wavelength due to a physical barrier C. change direction due to an electromagnetic field D. change frequency due to an electromagnetic field
A. change direction due to a physical barrier Explanation: Diffraction is defined as the direction change of electromagnetic waves occurring when the wave encounters a particular physical barrier.
Knewton (Molecular Orbital Theory: Energy Diagrams) According to molecular orbital theory, the regions of the wave function with the highest probability of finding electrons are areas with _______. A. constructive interference B. destructive interference C. nodes D. ψ=0
A. constructive interference Explanation: The wave function ψ is used in molecular orbital theory to describe the behavior of electrons in a molecule. We use LCAO to combine the wave functions of atomic orbitals to generate molecular orbitals. Combination of in-phase waves produces constructive interference, generating a wave with a greater amplitude and a higher probability of electron density. Combination of out-of-phase waves produces destructive interference, and describes areas with a low probability of electron density.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) In an octahedral complex, what symbol is used to refer to the set of d orbitals with increased potential energy? A. eg orbitals B. t2g orbitals C. e orbitals D. t2 orbitals
A. eg orbitals Explanation: In an octahedral field, the dz2 and dx2−y2 orbitals have lobes pointing directly on the axes and at the ligands; they are therefore raised in energy and are referred to as eg orbitals. t2g refers to the lower-energy dxy, dxz, and dyz orbitals in an octahedral field.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Transition metal complexes with weak-field ligands are generally what colors? A. green B. yellow C. violet D. red E. blue F. orange
A. green C. violet E. blue Explanation: A complex with weak-field ligands will have a small crystal field splitting. Therefore, they will generally absorb the lower energy red, orange, or yellow light and appear to be green, blue, or violet.
Quiz 14 (lecture) Which of the following are characteristics of bonding orbitals? A. has a longer approximate/effective wavelength than its parent atomic orbitals B. lower energy than its parent atomic orbitals C. decreases electron density between nuclei D. has a new node between nuclei E. increased electron density between the nuclei F. retains nodes of its parent atomic orbitals G. higher energy than its parent atomic orbitals
A. has a longer approximate/effective wavelength than its parent atomic orbitals B. lower energy than its parent atomic orbitals E. increased electron density between the nuclei F. retains nodes of its parent atomic orbitals
Knewton (Coordination Chemistry of Transition Metals) What is the name of the following coordination compound: [Co(NH3)5Br]Br2? A. pentaamminebromocobalt(III) bromide B. pentaamminebromocobalt(II) bromide C. pentaammoniumbromocobalt(II) bromide D. bromine pentaamminebromocobaltate(III)
A. pentaamminebromocobalt(III) bromide Explanation: The complex ion is the cation and the bromine outside of the brackets is the anion. The two ligands are NH3, named ammine, and Br, named bromo. There are five amine ligands, so ammine gets the prefix penta-. The central metal is cobalt. The two bromine anions have a charge of 2−, the NH3 is neutral, and the single Br ligand has a charge of 1−, so the cobalt must have an oxidation state of 3+ to make the compound neutral. The ligands are written in alphabetical order, ignoring the prefixes. The name of the compound is pentaamminebromocobalt(III) bromide.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) When x-rays of wavelength λ are scattered by atoms in adjacent crystal planes separated by a distance d, what will occur if the difference between the distances traveled by the x-rays is an integer factor of the wavelength? A. the x-rays will undergo constructive interference. B. The x-rays will undergo destructive interference. C. The x-rays will not undergo interference. D. The x-rays will combine resulting in an increased amplitude.
A. the x-rays will undergo constructive interference. D. The x-rays will combine resulting in an increased amplitude. Explanation: When X-rays of a certain wavelength λ are scattered by atoms in adjacent crystal planes separated by a distance d, they may undergo constructive interference when the difference between the distances traveled by the two waves prior to their combination is an integer factor of the wavelength. This condition is satisfied when the angle of the diffracted beam, θ, is related to the wavelength and to the interatomic distance by the equation: nλ=2dsinθ
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) When x-rays of a certain wavelength λ are scattered by atoms in adjacent crystal planes separated by a distance, d, they may undergo constructive interference. This occurs when the difference between the distances traveled by the two waves prior to their combination is an integer factor, n, of the wavelength. An integer factor is any whole number greater than zero, 1, 2, 3, etc. The conditions that support constructive interference are satisfied when the angle of the diffracted beam, θ, is related to the wavelength and to the interatomic distance by the Bragg equation: nλ=2dsinθ The Bragg equation includes: (select all that apply) A. wavelength of X-ray B. distance between two waves C. angle of diffracted beam D. integer factor
A. wavelength of X-ray B. distance between two waves C. angle of diffracted beam D. integer factor The Bragg equation is nλ=2dsinθ, which includes all of the elements above.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Hexaaquacopper(II), [Cu(H2O)6]2+ absorbs light from 560 to 580nm. This substance is most likely what color? A. Red B. Blue C. Green D. Yellow
B. Blue Explanation: The color of a compound is generally the complement of the color it absorbs. Light from 560 to 580nm is yellow light. Thus, the compound [Cu(H2O)6]2+ is blue, as blue is the complement of yellow. The color of a compound is generally the complement of the color it absorbs. Light from 560 to 580nm is yellow light, which is not the complement of green.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Why do compounds containing copper(I) appear colorless, while compounds containing copper(II) almost always appear to be colored? A. Complexes of copper(I) have very large crystal field splittings and absorb light with higher energy than visible light. Thus, complexes of copper(I) appear to be colorless. Complexes of copper(II) have smaller crystal field splittings that cause them to absorb visible light. B. Complexes of copper(I) have completely filled d orbitals, and require high energy photons to excite electrons. Complexes of copper(II) has a vacancy in the d orbitals, which allows for d-d transitions that occur when the complex absorbs visible light. C. Complexes of copper(I) have a crystal field splitting energy that is smaller than the energy of visible light, thus these complexes appear to be colorless. Complexes of copper(II) have a larger crystal field splitting energy that causes them to absorb visible light. D. Complexes of copper(I) have all paired electrons, so electrons cannot be excited to higher energy orbitals. The unpaired electrons in complexes of copper(II) allow for electronic transition that result in color.
B. Complexes of copper(I) have completely filled d orbitals, and require high energy photons to excite electrons. Complexes of copper(II) has a vacancy in the d orbitals, which allows for d-d transitions that occur when the complex absorbs visible light. Explanation: Copper(I) has 10 d electrons, so all the d orbitals are fully occupied. Absorption of light will promote electrons to orbitals of much higher energy (4p orbitals). Thus, complexes of copper(I) absorb in the UV-region and do not generally form colored complexes. Copper(II) complexes have a vacancy in the d orbitals, allowing electrons to be promoted between the split d orbitals, so complexes of copper(II) absorb in the visible region of the spectra, forming colored complexes. The number of unpaired electrons will determine the magnetic properties of a complex but is unrelated to the perceived color.
Knewton (Hybridization) Which of the following statements about hybridization are true? A. Hybrid orbitals exist in isolated atoms B. Hybrid orbitals with the same atom have the same energy and shape C. Hybrid orbitals are described mathematically as linear combination of atomic orbitals D. An atom can have both hybridized and unhybridized orbitals at the same time
B. Hybrid orbitals with the same atom have the same energy and shape C. Hybrid orbitals are described mathematically as linear combination of atomic orbitals D. An atom can have both hybridized and unhybridized orbitals at the same time Explanation: Hybridization is a model used to describe the observed bonding in molecules. These hybrid orbitals are described mathematically by a process called LCAO, or the linear combination of atomic orbitals. The hybrid orbitals are equal in size, shape, and energy. sp and sp2 hybridization mixes 1 or 2 p orbitals with the s orbitals, while the remaining p orbital(s) do not form hybrid orbitals.
Knewton (Molecular Orbit Theory Bond Order) Which molecule will have the same bond order as H2? A. He2 B. Li2 C. C2 D. O2
B. Li2 Explanation: Both H2 and Li2 will have a bond order of 1, corresponding to a single covalent bond.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) A. Strong-field ligands generally form high-spin complexes. B. Strong-field ligands generally form coordination compounds with a large Δoct. C. Strong-field ligands generally form coordination compounds with a small Δoct. D. Coordination compounds with strong-field ligands do not obey the Aufbau principle when filling orbitals with electrons.
B. Strong-field ligands generally form coordination compounds with a large Δoct. Explanation: Strong-field ligands generally form low-spin complexes and have a large value of Δoct.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Transition metal complexes with strong-field ligands are generally what colors? A. Green B. Yellow C. Violet D. Red E. Red F. Blue G. Orange
B. Yellow D. Red G. Orange Explanation: A complex with strong-field ligands will have a large crystal field splitting. Therefore, they will generally absorb the higher energy violet, blue, or green light and appear to be yellow, orange, or red.
Knewton (Coordination Chemistry of Transition Metals) Which of the following compounds will show geometric isomerism? A. [Fe(CO)5NH3]SO4 B. [Fe(CO)4(NH3)2]SO4 C. [Fe(CO)(NH3)5]SO4 D. [Fe(NH3)6]SO4
B. [Fe(CO)4(NH3)2]SO4 Explanation: The compound [Fe(CO)4(NH3)2]SO4 has two possible geometric isomers, one with two adjacent NH3 ligands (cis) and one with the NH3 ligands directly across from one another (trans).
Knewton (Coordination Chemistry of Transition Metals) Square planar complexes will likely have what d configuration? A. d0 B. d8 C. d10 D. d5
B. d8 Explanation: The central metal generally has a d8 electron configuration in square planar complexes.
Knewton (Molecular Orbit Theory Bond Order) The energies of the molecular orbitals for homonuclear diatomic molecules from the second period will decrease as: A. effective nuclear charge decreases B. effective nuclear charge increases C. bond order decreases D. bond order increases
B. effective nuclear charge increases Explanation: As effective nuclear charge increases, the orbitals are pulled closer to the nucleus and are thus lower in energy.
Knewton (Coordination Chemistry of Transition Metals) Complete the following statement. In the formula for a complex ion, the symbol indicating the central metal ion comes __________, and in the name for a complex ion, the name indicating the central metal ion comes __________. A. first; first B. first; last C. last; first D. last; last
B. first; last Explanation: In the formula for a complex ion, the symbol indicating the central metal ion comes first, and in the name for a complex ion, the name indicating the central metal ion comes last.For example, in the ion [Co(NH3)6]3+, the central metal is cobalt. It comes first in the formula. In the name of the ion, which is hexaamminecobalt(III), it comes last, after the ligands.
Knewton (Molecular Orbital Theory: Energy Diagrams) A pi bond involves: A. direct orbital overlap B. lateral orbital overlap C. a node D. none of the above
B. lateral orbital overlap C. a node Explanation: Pi bonds involve the lateral overlap of p orbitals, in which the electron density is on either side of the internuclear axis. Along the axis, there is a node in which there is no probability of finding an electron.
Knewton (Hybridization) In order for a pi bond to form, there must be: A. direct overlap of p orbitals B. lateral overlap of p orbitals C. direct overlap of hybrid orbitals D. lateral overlap of hybrid orbitals
B. lateral overlap of p orbitals Explanation: Any direct overlap is a sigma bond, and pi bonds are specifically the lateral overlap of unhybridized p orbitals.
Knewton (Hybridization) In order for a pi bond to form, there must be: A. overlap of s orbitals B. overlap of unhybridized orbitals C. overlap of hybrid orbitals D. hybrid orbitals with different energies
B. overlap of unhybridized orbitals Explanation: Pi bonds are created by the lateral overlap of unhybridized orbitals that contain at least one node along the internuclear plane, while sigma bonds occur through overlap of hybridized orbitals. Since s orbitals do not contain nodes, they cannot be involved in making pi bonds.
Knewton (Molecular Orbital Theory: Energy Diagrams) A pi bond involves: A. a direct orbital overlap with the electron density concentrated along the internuclear axis B. side-by-side, or lateral orbital overlap with the electron density concentrated on opposite sides of the internuclear axis C. side-by-side, or lateral orbital overlap with the electron density concentrated along the internuclear axis D. none of the above
B. side-by-side, or lateral orbital overlap with the electron density concentrated on opposite sides of the internuclear axis Explanation: π bonds involve side-by-side or lateral orbital overlap, in which the electron density is on either side of the internuclear axis. Along the axis itself, there is a node: a plane in which there is no probability of finding an electron.
Knewton (Hybrid Atomic Orbitals) What is the hybridization of C in CN−? A. sp3 B. sp C. sp3d D. sp3d2
B. sp Explanation: C in CN− has an sp hybridization because of its 2 regions of electron density (1 bonding, 1 non bonding).
Knewton (Coordination Chemistry of Transition Metals) What geometry is expected for a 4-coordinate complex that contains a central metal with a d0 or d10 configuration? A. octahedral B. tetrahedral C. square planar D. trigonal bipyramidal
B. tetrahedral Explanation: Generally, 4-coordinate complexes with metal atoms that have d0 or d10 configurations form tetrahedral geometry.
Knewton (Molecular Orbital Theory: Energy Diagrams) Which of the following determines the strength of a covalent bond? A. the charges of the ions involved B. the amount of overlap for the orbitals involved C. the mass of the atoms involved D. all of the above
B. the amount of overlap for the orbitals involved Explanation: When overlap occurs between two atomic orbitals, the positively charged nuclei and the negatively charged electrons for the atoms involved create a covalent bond. The more overlap that occurs between two orbitals, the stronger the attraction between the positively charged nuclei and the negatively charged electrons. Therefore, the more overlap that occurs, the stronger the covalent bond.
Knewton (Coordination Chemistry of Transition Metals) What generally acts as a Lewis acid in a coordination compound? A. the ligands B. the central metal ion C. the counter ion D. coordinatioon compounds do not involve Lewis acid base base reaction
B. the central metal ion Explanation: The central metal ion or atom generally acts as a Lewis acid and accepts an electron pair from a donor atom within a ligand.
Knewton (Molecular Orbit Theory Bond Order) Bond order allows us to identify: A. the mass of the electrons based on bond strength B. the net combination of electrons for the bond strength of a molecule C. both A and B D. neither A or B
B. the net combination of electrons for the bond strength of a molecule Explanation: Bond order allows us to identify the net combination of electrons in a bond between two atoms. We can determine the bond order of a molecule by subtracting the antibonding (destablizing electrons) from the bonding (stabilizing electrons) and dividing this number by two. The larger the bond order, the stronger the bond between the two given atoms. The bond order is the number of bonding pairs of electrons between two atoms, when using Lewis structures.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Which of the following is an experimental technique for determining distances between atoms in a crystal? A. deposition B. x-ray crstallography C. titration D. capillary action
B. x-ray crstallography Explanation: X-ray crystallography is an experimental technique for determining distances between atoms in a crystal.
Quiz 14 (reading) What kind of diatomic molecular orbital is doubly degenerate (comes as a degenerate pair of orbitals) and has a node between the nuclei? A. π B. π* C. σ* D. σ
B. π*
Knewton (Molecular Orbital Theory: Energy Diagrams) How does molecular orbital theory differ from valence bond theory? A. Only valence bond theory can be used to describe the structure of the molecule B. Molecular orbital theory uses hybridized orbitals, while valence bond theory uses Lewis structures C. Molecular orbital theory considered electrons to be distributed over the entire molecule, while valence bond theory considers electrons to be localized to a bond D. They are essentially interchangeable
C. Molecular orbital theory considered electrons to be distributed over the entire molecule, while valence bond theory considers electrons to be localized to a bond Explanation: In molecular orbital theory, electrons are found in molecular orbitals that are distributed over the molecule according to the wave function ψ, and form bonding and antibonding orbitals. Valence bond theory localizes electrons to hybridized atomic orbitals around each atom, and uses the orbitals' overlap to describe intramolecular bonds. Both theories can be used to describe a molecule's structure.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) What is true of a compound that is weakly attracted to an external magnetic field? A. The compound will have weak-field ligands. B. The compound is diamagnetic. C. The compound contains at least one unpaired electron. D. The compound will likely have strong field ligands.
C. The compound contains at least one unpaired electron. Explanation: A compound that is weakly attracted to an external magnetic field will have at least one unpaired electron. The ligand field strength is generally unrelated to the magnetic properties of a complex.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) I−<Br−<Cl−<F−<H2O<C2O2−4<NH3<en<NO−2<CN −−→−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−−− a few ligands of the spectrochemical series in order of increasing field strength of the ligand Consider the complex[Fe(CN)6]4−. Based on the spectrochemical series, which of the following is true? A. [Fe(CN)6]4− is a high-spin complex because CN− is a weak-field ligand. B. [Fe(CN)6]4− is a high-spin complex because CN− is a strong-field ligand. C. [Fe(CN)6]4− is a low-spin complex because CN− is a strong-field ligand. D. [Fe(CN)6]4− is a low-spin complex because CN− is a weak-field ligand.
C. [Fe(CN)6]4− is a low-spin complex because CN− is a strong-field ligand. Explanation: CN− is a strong-field ligand, so [Fe(CN)6]4−, which has an octahedral geometry, likely has a large crystal field splitting energy. Thus, the electrons will pair up in the lower-energy t2g orbitals before occupying the higher-energy eg orbitals, making [Fe(CN)6]4− a low-spin complex.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) The primary data an x-ray diffractometer collects are the: A. wavelength of incident x-ray B. percentage of waves that are deflected C. angles at which rays are diffracted D. the number of collisions a wave experiences in traversing a crystal
C. angles at which rays are diffracted Explanation: Diffractometers measure diffraction angles.
Knewton (Molecular Orbital Theory: Energy Diagrams) Which bond is the shortest? A. carbon-oxygen single bond B. carbon-oxygen double bond C. carbon-oxygen triple bond D. these are all the same length
C. carbon-oxygen triple bond Explanation: With each additional bond, the bond length contracts, so the triple bond will be the shortest.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Diffraction will occur when the physical barrier is: A. dramatically larger than the wavelength of incident light B. dramatically smaller than the wavelength of incident light C. comparable in size to the wavelength of incident light D. depends on other factors
C. comparable in size to the wavelength of incident light Explanation: Diffraction occurs when the physical barrier encountered has dimensions comparable to those of the incoming wavelength. In this case, the dimensions would be the distance between neighboring atoms in crystals.
Knewton (Lewis Acids and Bases) Under the Lewis model, a base is a(n): A. proton donor B. proton acceptor C. electron pair donor D. electron pair acceptor
C. electron pair donor Explanation: Under the Lewis model of acids and bases, a base is an electron pair donor.
Knewton (Hybridization) The orbitals within a set of hybrid orbitals will have: A. equivalent shape but different energy B. equivalent energy but differing shape C. equivalent shape and energy D. depends on the type of hybrid orbitals
C. equivalent shape and energy Explanation: All orbitals within a set of hybrid orbitals will have identical shape and energy.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) During diffraction, constructive interference can happen at: A. any angle B. any integer angle C. only angles that satisfy the Bragg equation D. constructive interference
C. only angles that satisfy the Bragg equation Explanation: Only certain angles, which are dictated by the Bragg equation, will result in constructive interference.
Knewton (Coordination Chemistry of Transition Metals) What is the name of the following complex ion: [Co(SO4)(H2O)5]+? A. pentaaquasulfatocobaltate(III) ion B. cobalt(III) pentaaquasulfate ion C. pentaaquasulfatocobalt(III) ion D. pentaaquacobalt(II) sulfate ion
C. pentaaquasulfatocobalt(III) ion Explanation: The complex ion is a cation, with two ligands. They are H2O, named aqua, and SO4, named sulfato. In the name, the aqua will come before the sulfato because the ligands are named in alphabetical order. There are five aqua ligands, so aqua gets the prefix penta-, and there is one sulfato ligand, so sulfato will have no prefix. The central metal is cobalt. Since the charge on the ion is 1+, the aqua ligands are neutral, and the sulfato ligand has a charge of 2− (the charge on a sulfate anion), the oxidation number on the cobalt must be +3, and it will have a roman numeral (III) next to the metal name. Since the complex ion is a cation, there will be no suffix -ate added to the metal name. Remember that the metal is placed at the end of the name of the complex ion. So the name is the pentaaquasulfatocobalt(III)ion.
Knewton (Hybridization) What is the hybridization of the oxygen atom in a water molecule? A. sp B. sp2 C. sp3 D. sp3d
C. sp3 Explanation: Oxygen will be sp3 hybridized in a water molecule, because it is surrounded by four electron domains, and will thus need to hybridize all four of the orbitals in its valence shell.
Knewton (Coordination Chemistry of Transition Metals) What geometry is expected for a 4-coordinate complex that contains a central metal with a d8 configuration? A. octahedral B. tetrahedral C. square planar D. trigonal bipyramidal
C. square planar Explanation: Generally, 4-coordinate complexes with metal atoms that have d8 configurations form square planar geometry. Tetrahedral geometry is also a 4-coordinate structure, but it is not usually formed in metals with a d8 configuration. Trigonal bipyramidal geometry is a 5-coordinate structure, and octahedral geometry is a 6-coordinate structure.
Knewton (Hybridization) Hybridization was developed to explain observations about: A. the energy contained in certain covalent bonds B. the order in which orbitals are filled C. the bond angles in certain molecules D. none of the above
C. the bond angles in certain molecules Explanation: From valence bond theory, bond angles in a molecule like water would be expected to be 90∘ rather than the observed 104.5∘, so hybridization was developed to understand these observations.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) When a beam of x-rays of a particular wavelength strikes a crystal, the rays within the beam: A. will continue in a straight line B. will redirect at a particular angle from the incident beam C. will be scattered in all directions D. depends on the wavelength
C. will be scattered in all directions Explanation: Diffraction is occurring, so the x-rays are being scattered in all directions.
Knewton (Coordination Chemistry of Transition Metals) In the complex [Ni(NH3)2(en)]2+, the en refers to a bidentate ligand ethylenediamine. Write the formula for this chemical.
C2H8N2 Explanation: Bidentate ligands are those in which two atoms in the ligand coordinate to the metal center. The abbreviation en refers to ethylenediamine, which is a common bidentate ligand. The structure of ethylenediame is H2NCH2CH2NH2. Each of the nitrogen atoms has a lone pair of electrons and can coordinate individually to the nickel ion in the center of the complex.
Knewton (Coordination Chemistry of Transition Metals) Write the chemical formula for the ligand in the coordination compound tetracarbonylplatinum(IV) chloride. *Do not include the phase abbreviation.
CO Explanation: In the coordination compound tetracarbonylplatinum(iv) chloride, the chloride is the noncomplex anion and the tetracarbonylplatinum(iv) is the complex cation. The name platinum(iv) refers to the metal ion in the complex cation, and carbonyl refers to the ligand. (There is only one type of ligand in this compound.) The ligand named carbonyl is carbon monoxide (CO). The complete formula for the compound is [Pt(CO)4]Cl4.
Knewton (Molecular Orbital Theory: Energy Diagrams) How many molecular orbitals will have an n value of 2? A. 2 B. 4 C. 6 D. 8
D. 8 Explanation: The atomic orbitals with an n value of 2 are the 2s orbital and the three 2p orbitals, so when four orbitals from one atom interact with four orbitals from another atom, we must generate 8 molecular orbitals, half of which will be bonding and the other half of which will be antibonding.
Knewton (Hybridization) Which of the following exhibits sp2 hybridization? A. CCl4 B. BeCl2 C. CO2 D. BH3
D. BH3 Explanation: BH3 exhibits sp2 hybridization because the boron atom has 3 electron domains.
Knewton (Lewis Acids and Bases) Consider the reactions below. Which of the following correctly identifies the coordinate complex? A. SO3 in O2−+SO3→SO2−4 B. O2− in O2−+SO3→SO2−4 C. BH3 in (CH3)2S+BH3→H3BS(CH3)2 D. BeCl2−4 in BeCl2+2Cl−→BeCl2−4
D. BeCl2−4 in BeCl2+2Cl−→BeCl2−4 Explanation: In a Lewis acid-base reaction, the coordinate complex is the compound that is generated by the formation of coordinate covalent bond(s) between the Lewis acid and the Lewis base.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Which of the following is the redirection of electromagnetic radiation that occurs when it encounters a physical barrier of appropriate dimensions? A. Capillary action B. Titration C. Deposition D. Diffraction
D. Diffraction Explanation: Diffraction is defined as the redirection of electromagnetic radiation that occurs when it encounters a physical barrier of appropriate dimensions. It is used in X-ray crystallography to determine the distances between the atoms in a crystal.
Knewton (Molecular Orbit Theory Bond Order) Why doesn't an oxygen molecule exhibit significant s−p mixing? A. It has a large effective nuclear charge B. It has too many valence electrons C. It has paired electrons D. It has paired p electrons
D. It has paired p electrons Explanation: For oxygen through neon, there will be pairs of p electrons. Because the electrons are paired, this raises the energy of the p orbitals away from the energy of the s orbital, making s−p mixing negligible.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Which of the following statements is true? A. Diamagnetic compounds have one or more unpaired electron. B. Transition metal complexes rarely absorb light in the visible region of the electromagnetic spectrum. C. Transition metal complex are almost always paramagnetic. D. The color of a transition metal complex is generally the complement of the color of light absorbed.
D. The color of a transition metal complex is generally the complement of the color of light absorbed. Explanation: The color of a transition metal complex generally is the complement of the color of light absorbed.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Determine whether the high-spin complex [Mn(H2O)6]Cl2 is paramagnetic or diamagnetic by determining the number of unpaired electrons. A. The complex is diamagnetic; it has no unpaired electrons. B. The complex is paramagnetic; it has 1 unpaired electron. C. The complex is paramagnetic; it has 3 unpaired electrons. D. The complex is paramagnetic; it has 5 unpaired electrons.
D. The complex is paramagnetic; it has 5 unpaired electrons. Explanation: The complex [Mn(H2O)6]Cl2 has six H2O ligands and exists in an octahedral geometry. For octahedral complexes, the d orbitals split into two higher energy eg orbitals and three lower energy t2g orbitals. The coordination sphere must have a charge of 2+ to balance the two chloride counter ions. The H2O ligands are uncharged, so manganese must be in a +2 oxidation state. Manganese is a member of group 7, so Mn2+ will have 5 d electrons. It is given that the complex is in a high-spin state, so there will be 5 unpaired electrons, one in each of the d orbitals.
Knewton (Molecular Orbital Theory: Energy Diagrams) Which kind of orbitals can s orbitals overlap with? A. s B. p C. d D. all of the above
D. all of the above Explanation: Any orbital type can overlap with any other orbital type to generate covalent bonds
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) The Bragg equation is as follows: nλ=2dsinθ It contains a term that corresponds to: A. the angle of the diffracted beam B. the wavelength of the ray C. the distance between two crystal planes D. all of the above
D. all of the above Explanation: These are all terms that are contained in the Bragg equation.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) Which ligand is a strong field ligand? A. bromide B. fluoride C. water D. cyanide
D. cyanide Explanation: Cyanide is a strong field ligand, the other options are weak field ligands.
Knewton (Coordination Chemistry of Transition Metals) Which type of isomerism describes the transfer of an anionic ligand from the coordination sphere and the counter ion? A. geometric isomerism B. optical isomerism C. linkage isomerism D. ionization isomerism
D. ionization isomerism Explanation: Ionization isomers (or coordination isomers) occur when one anionic ligand in the inner coordination sphere is replaced with the counter ion from the outer coordination sphere. A simple example of two ionization isomers are [CoCl6]Br and [CoCl5Br]Cl.
Knewton (Hybridization) The energy of an sp orbital will be: A. less than that of an s or p orbital B. greater than that of an s or p orbital C. less than that of an s orbital but greater than that of a p orbital D. less than that of a p orbital but greater than that of an s orbital
D. less than that of a p orbital but greater than that of an s orbital Explanation: An sp orbital will have an energy in between those of the unhybridized s and p orbitals.
Knewton (Coordination Compounds: Properties and Structures Using Spectroscopy) In a tetrahedral complex, what symbol is used to refer to the d orbitals with increased potential energy? A. eg orbitals B. t2g orbitals C. e orbitals D. t2 orbitals
D. t2 orbitals Explanation: In a tetrahedral field, the off-axis dxy, dxz, and dyz orbitals are raised in energy and are referred to as t2 orbitals.
Knewton (Coordination Chemistry of Transition Metals) Write the formula for the ionization isomer of [Co(NH3)4Cl2]Br. Remember that the inner coordination sphere is written inside brackets and the counter ion is outside the brackets.
[Co(NH3)4ClBr]Cl Explanation: In an ionization isomer, an anionic ligand is exchanged for the counter ion. In the formula [Co(NH3)4Cl2]Br, Cl atoms are ligands and Br is the counter ion. In this ionization isomer, Br becomes a member of the inner coordination sphere and one Cl ligand becomes the counter ion. The formula is [Co(NH3)4BrCl]Cl.
Knewton (Coordination Chemistry of Transition Metals) Write the chemical formula for pentaamminechlorocobalt(III) chloride. *Do not include the phase abbreviation.
[Co(NH3)5Cl]Cl2 Explanation: To write the formula for a coordination compound, first identify the ions from the name. The second part of the name, chloride, indicates that Cl− will be the anion and it is not a part of the coordination sphere. The pentaamminechlorocobalt(III) will make up the complex ion, which in this case will be a cation because it is the first part of name. The central metal ion is cobalt(III), indicated by cobalt(III). The ligands precede it in the name of the ion, indicated by ammine for NH3 and chloro for Cl−. The penta- indicates that there will be five NH3 ligands, and there will be one Cl− ligand because there is no prefix on the chloro. Since cobalt is in a 3+ oxidation state, the NH3 ligands have no charge. The Cl− ligand has a charge of 1−, so the charge on the complex anion will be 2+. Remember that brackets are used to identify the coordination sphere. The formula for the complex ion is [Co(NH3)5Cl]2+. Since the anion is Cl−, there must be two chloride ions for each complex ion. The chloride symbol will be written last in the formula, just like it would be in an ionic compound. It is written outside of the brackets because it is not part of the coordination sphere. The complete chemical formula is [Co(NH3)5Cl]Cl2.
Knewton (Coordination Chemistry of Transition Metals) Write the chemical formula for the complex ion hexacyanoferrate(III). *Do not include the phase abbreviation. *Include the charge on the complex ion.
[Fe(CN)₆]³⁻ Explanation: To write the formula for the complex ion in a coordination compound, first identify the ions from the name. The first part of the name, potassium, indicates that K+ will be the cation, acting as a counter ion and not as a ligand. Hexacyanoferrate(III) is the name of the complex ion, which in this case will be an anion. It follows the cation in the name and includes the suffix -ate at the end of the name. The only ligand is the cyanide group, indicated by cyano, which includes six cyanides, indicated by hexa-. The central metal ion is iron(III), indicated by ferrate(III). The formula for a cyanide ion is CN−. Since the iron is in an oxidation state of 3+ and the six cyanide ions will have a charge adding up to 6−, the charge on the complex anion will be 3−. Remember that brackets are used to identify the coordination sphere. The complete formula for the complex ion is [Fe(CN)6]3−.
