Chemistry 201: General Chemistry II - Lecture

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1 Chemistry 201: General Chemistry II - Lecture Dr. Namphol Sinkaset Chapter 23 Study Guide Concepts 1. In the transition metals, the ns orbital fills before the (n-1)d orbitals. However, the ns orbital empties before the (n-1)d orbitals. 2. Atomic size of the transition metals generally decreases across the period as expected (due to increasing nuclear charge). Also as expected, atoms of the 2nd transition row are larger than atoms of the 1st transition row (due to opening of a new shell). However, atoms in the 3rd transition row are the same size as those in the 2nd transition row due to the lanthanide contraction. 3. As expected, the first ionization energy of the transition metals increases across a period. 3rd row transition metals have higher first ionization energies because they are the same size as the 2nd row transition metals, but have higher nuclear charge. 4. In general, the electronegativity of transition metals increases across a period. However, electronegativity increases between the 1st and 2nd period transition metals (opposite of what s seen in main group elements). Note that Au has an exceptionally high electronegativity. 5. The transition metals have a wide range of possible oxidation states. 6. A complex ion has a central metal ion bound to ligands. 7. A complex ion and its counterions form a coordination compound. 8. The primary valence of a coordination compound refers to the oxidation state of the central metal. 9. The secondary valence of a coordination compound refers to the number of ligands attached to the cental metal. The secondary valence is also known as the coordination number. 10. The bond between the central metal and a ligand is called a coordinate covalent bond. 11. Ligands can be monodentate, bidentate, or polydentate. These terms refer to the number of lone pairs on the ligand that can coordinate to a metal. 12. Coordination compounds have a complex set of naming rules. (1) Like ionic compounds, the cation is named first followed by the anion. (2) Ligands are named in alphabetical order followed by the metal. Neutral ligands keep their name, but there 1

2 are some special names like aqua for the H 2 O ligand. -ide ligands become -o. -ate ligands become -ato. -ite ligands become -ito. (3) The number of ligands is denoted with a Greek prefix. However, if a ligand has a Greek prefix in its name, use bis-, tris-, or tetrakis- to indicate number. (4) If the metal is in the cation, the metal is named with its elemental name. If the metal is in the anion, the -ate suffix is added to the root of the metal s name. (5) The oxidation state of the metal is indicated with a Roman numeral in parentheses after the metal name. 13. Because of the many bonding sites on the central metal, there is a lot of variation in structures of coordination compounds. 14. Isomers are compounds that have the same formula, but different structures. 15. Coordination isomers occur when a coordinated ligand exchanges places with an uncoordinated counterion. 16. Some monodentate ligands have more than one possible coordination site. Linkage isomers occur when a ligand coordinates one way in one isomer and a different way in another isomer. 17. Geometric isomers occur when ligands bond to different coordination sites, leading to distinct compounds. In square planar MA 2 B 2 and octahedral MA 4 B 2 complexes, there are cis-trans isomers. In octahedral MA 3 B 3 complexes, there are fac-mer isomers. 18. Optical isomers are nonsuperimposable mirror images of one another. Your right hand and left hand are nonsuperimposable mirror images. Molecules or ions that have this quality are called chiral, and the isomers are called enantiomers. 19. Geometries of coordination compounds can be described with valence bond theory. However, valence bond theory cannot be used to describe color and magnetic properties of coordination compounds. 20. Crystal field theory focuses on what happens when e s on the ligands approach the central metal. As the ligands approach, they repel e s in the unhybridized orbitals of the central metal. This repulsion destabilizes the metal s unhybridized d orbitals. 21. Orbitals in direct line with the incoming ligands are destabilized most. 22. In an octahedral complex, the d orbitals are split into two sets. The d x 2 y 2 and d z 2 orbitals form one set at a higher energy level than the d xy, d yz, and d xz orbitals. The energy difference between the two sets is called the crystal field splitting energy (CFSE), represented by the variable. 23. Different ligands result in greater energy difference between the two sets of d orbitals. If the energy difference is large, it s a strong-field complex. If the energy difference is small, it s a weak-field complex. 24. The spectrochemical series enables one to predict whether a metal complex will be strong-field or weak-field. In order of decreasing ability to force strong-field complexes: CN > NO 2 > en > NH 3 > H 2 O > OH > F > Cl > Br > I. 2

3 25. Note that high-charge metal cations promote strong-field complexes. 26. Colors of complexes arise from electrons transitioning between the two different sets of d orbitals. 27. Absorbance spectra can be used to calculate the splitting between d orbitals in a metal complex. 28. Recal that Hund s rule states that electrons will maximize spin before pairing in the same orbital because occupying the same orbital costs energy. If the CFSE is small enough, the energy cost of pairing is higher than the energy cost of moving to the slightly higher energy d orbitals. 29. Low-spin complexes have CFSE s that are larger than the electron pairing energy. High-spin complexes have CFSE s that are smaller than the electron pairing energy. 30. In a tetrahedral complex, the d orbitals are split into two sets as well. The d xy, d yz, and d xz orbitals form one set at a higher energy than the d x 2 y 2 and d z 2 orbitals. 31. Almost all tetrahedral complexes are high-spin (weak-field) due to less ligand-metal interactions. 32. Square planar complexes have a complicated splitting pattern. The d x 2 y2 orbital lies at the highest energy, followed by the d xy orbital, the d z 2 orbital, and the d xz and d yz orbitals which lie at lowest energy. 33. Normally, square planar complexes are low-spin. 34. Metal complexes are found in many important biomolecules. Equations 1. = E photon = hν = hc λ Representative Problems (Equation to calculate crystal field splitting energy) Give the name or formula of the following complexes. 1. [CoI(NH 3 ) 5 ]Cl 2 2. Na[Rh(EDTA)] 3. [RuCl 5 (H 2 O)] 2 4. (NH 4 ) 2 [CuBr 4 ] 5. ammineaquadicarbonyldicyanoiron(iii) 6. tetraamminedichloroplatinum(iv) tetrachloroplatinate(ii) 7. cobalt(ii) hexanitrocobaltate(iii) 8. sodium tetrathiocyanatoosmiate(iii) 9. potassium ethylenediaminetetraacetatoferrate(ii) 1. [CoI(NH 3 ) 5 ]Cl 2 3

4 In naming metal complexes, we always need to identify the cation and the anion. The cation is listed first, and if it s in square brackets, then some sort of complex forms that cation. In this case, the Co occurs in the cation. We name the ligands first in alphabetical order then state the name of the metal, followed by its oxidation state. Here, we see that the anion is 2 Cl which must carry a 2 charge; the cation as a whole must carry a + 2 charge. The cobalt must then be in the + 3 state since the one I has a negative charge and NH 3 is neutral. The proper name for the complex is: pentaammineiodocobalt(iii) chloride. 2. Na[Rh(EDTA)] Here, sodium is the cation and the metal, rhodium, occurs in the anion. The cation carries a + 1 charge, so the anion must carry a 1 charge. Fully deprotonated EDTA carries a 4 charge which means Rh is in the + 3 oxidation state. The name of the complex is: sodium ethylenediaminetetraacetatorhodate(iii). Note that we must use the -ate ending on the metal because it occurs in the anion. 3. [RuCl 5 (H 2 O)] 2 In this complex, there is no counterion. All we have is an anion. The overall charge on the anion is 2. Water is neutral and the 5 Cl contribute a total charge of 5. Therefore, Ru must be + 3. It s an anion, so the metal must take the -ate ending. The complex is named: aquapentachlororuthenate(iii). 4. (NH 4 ) 2 [CuBr 4 ] The ammonium serves as the cation in this salt, and taken twice, it has a charge of + 2. The anion must have a total charge of 2. 4 Br have a total 4 charge, so the copper must be + 2. Notice the special -ate name for copper in the correct name for the complex: ammonium tetrabromocuprate(ii). 5. ammineaquadicarbonyldicyanoiron(iii) We know this must be a cation because the metal does not have an -ate ending. A cation must have an overall positive charge, so we calculate that first. Ammine, aqua, and carbonyl ligands carry 0 charge, so they don t count. We only have a dicyano, 2 CN, which carries 2 charge. Note that CN can coordinate through the C (cyano) or through the N (isocyano). Since cyano is being used, we have to write the ligand as CN, not NC. When the 2 is added to the + 3 of the iron, we get an overall charge of + 1. The formula is: [Fe(H 2 O)(NH 3 )(CO) 2 (CN) 2 ] +. Note that the order of the ligands does not matter. 6. tetraamminedichloroplatinum(iv) tetrachloroplatinate(ii) 4

5 We know we have a cation and an anion because we have two words which make up the name. Of course, the first is the cation and the second is the anion. When forming a salt (containing both cation and anion), we must balance the charge when putting them together. To do that, we need to know the charge on each. We start with the cation. The 4 ammines (NH 3 ) have 0 charge, but the 2 chloros (Cl ) contribute a total of 2. When added to the + 4 of the platinum, we get a charge on the cation of + 2. We form the cation as: [Pt(NH 3 ) 4 Cl 2 ] 2+. Now we form the anion. Here, we have 4 chloros which contribute a total of 4. When added to the + 2 of the platinum, we get an overall charge of 2. The anion is then: [PtCl 4 ] 2. Since the cation and anion have equal but opposite charges, we can combine them directly to get: [Pt(NH 3 ) 4 Cl 2 ][PtCl 4 ]. 7. cobalt(ii) hexanitrocobaltate(iii) Again, two words, so we have both a cation and anion. The cation is simply: Co 2+. In the anion, we have 6 nitro ligands (NO 2 ), which contribute a 6 charge. When combined with the + 3 of the cobalt, we get an overall anionic charge of 3. The anion is then: [Co(NO 2 ) 6 ] 3. To combine the cation and anion correctly to form a neutral salt, we need 3 of the former and 2 of the latter. The correct formula is then: Co 3 [Co(NO 2 ) 6 ] sodium tetrathiocyanatoosmiate(iii) We have a cation and an anion. The cation is just: Na +. We focus on the anion. We have four SCN ligands which yield a charge of 4. The osmium has a + 3 charge, so the overall charge on the anion is 1. Therefore, we just need one sodium cation to give: Na[Os(NCS) 4 ]. Note that SCN is a monodentate ligand that can coordinate through the S or the N. The thiocyanato ligand coordinates through the N and is thus written as NCS in the formula. The isothiocyanato ligand coordinates through the S and would be written as SCN in the formula. 9. potassium ethylenediaminetetraacetatoferrate(ii) Two words means a cation and an anion. The cation is just: K +. In the anion, we have EDTA which contributes a 4 charge. The iron has a + 2 charge, so the overall charge on the anion is 2. The formula is: K 2 [Fe(EDTA)]. 10. Predict the magnetic properties and colors of [Co(NH 3 ) 6 ] 3+ and [CoF 6 ] 3 using Crystal Field Theory. We begin by finding the number of d electrons we re working with. Co is a d 9, and Co 3+ (the oxidation state in both these complexes) is therefore a d 6. Let s start with [Co(NH 3 ) 6 ] 3+. We consult the spectrochemical series to determine whether this complex has a strong or weak field. NH 3 occurs towards the left, so we determine that it will favor a strong field (low-spin) complex. The d orbitals split in an octahedral field into a lower set (d xy, d yz, and d xz ) and an upper set (d x 2 y 2 and d z2). Since this is a strong-field (low-spin) complex, all six 5

6 electrons will occupy the lower set of orbitals as shown in the diagram below. Since all the spins are paired, the complex is diamagnetic. If a complex is low spin, that means is large. This means that high energy light will be necessary to promote electrons between the two levels. Higher energy light is in the blue region, which means the color of the complex will be in the red region. For [CoF 6 ] 3, we have a weak-field (high-spin) complex because F is more towards the right of the spectrochemical series. The d orbitals are still split into a lower and an upper set as in the previous complex, but the separation is much smaller. Thus, the electrons will maximize spin as shown in the diagram below. Since we have unpaired spins, this complex is paramagnetic. Since there s a smaller, low energy light is needed for transitions between the two levels. Lower energy light is in the red region, which means the color of the complex will be in the blue region. z 2 x 2 -y 2 z 2 x 2 -y 2 xy yz xz xy yz xz [Co(NH 3 ) 6 ] 3+ [CoF 6 ] 3- diamagnetic (all spins paired) paramagnetic (unpaired spins) large, so color in red region small, so color in blue region 6

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