PROBLEMS SOLVED

1. Similar. Bismuth has a simple cubic unit cell. How many atoms of Bistmuth in each unit cell?

1 atom

FACE centered unit cell: 1. how many corner atoms> 2. what fraction of atom in each corner 3. how many face atoms> 4.what fraction of each atom or what volume is inside of each cube>

1. 8 2. 1/8 3.6 4. 1/2

12. [Ni(CN)4]2- is diamagnetic. Is the compound tetrahedral or square planar? Explain answer using diagram

1. diamagnetic- paired e 2. tetrahedral vs square planar diagrams 3. Ni2+ : [Ar]3d8 tetrahedral square planar -I---I- --I--- ---I-- ---------- EMPTY (dx2-y2) dxy dxz dyz ----I--I-- (dxy) there is NO Z ---I-I- ---I-I-- -I---I--- (dz2) dx2-y2 dz2 ---I-I-(dxz) ---II---(dyz) 4. Explain : square planner because diatomic means paired electrons. Since highest dx2-y2 empty then the others are paired. Cannot be tetrahedral because 2 e unpaired

6. Which octahedral complex below has its electronic transition at shortest wavelength ? a) [Ti(NH3)6]3+ b)[Ti(en)3]3+ c) [Ti(H2O6]3+ d)[TiCl6]3- e)[TiF6]3-

Cl-<F-<H20<NH3<en<NO2-<CN- Rule : 1. strength of legands rises to right 2.stronger ligand GREATER ENERGY gap 3.complex to right absorbs highest energy 4. left longest wavelength-right shortest wavelength answer: en

5. When the compound [Rh(Cl3)(NH3)4 is dissolved in water and treated with excess AgNO3(aq) one mole of AgCl(s) is formed for every mole of the complex what is the complex?

Rule:1. Write the initial complex given 2. look at the aquous soln added. AgNO3 in our case 3. So Ag+ is metal transition-no influence because excess 4. What

1. Iron is a body centerd cubic unit cell. How many atoms of iron in each unit cell?

body= 2 atoms of iron

What is the coordination nr for the following complexes [Co(NH3)4(H2O)2]3+ [Pb(EDTA)]2- [Zn(H2O)4]2+ [Ag(NH3)2]NO3

coordination nr depends on 1. #of ligands 2. if ligands bidentate: en, py-, gly 2 atoms; if hexadentate-6 atoms Answer a)6 b) EDTA=HEXADENTATE =6 atoms c) 4 d)2

1.hmw. you are given a small bar of an unknown metal x. you find the density of metal is 10.5 g/cm3. an edge is 409 pm. Assuming that the metal is face centered cubic, what is the metal most likely to be?

density=mass /volume volum is a^3= in cm ^3 409pm/10^12pm*10^2cm=4.09E-8 cmE3=6.84E-23 mass face centered= 7.18E-22 How do we find mass? [z times atomic weight/avogadro]/volume so not just mass for face is 4 4x/6.022E23 /volume =density so 4x/avogadr=volume times density 4x= 432.61 x=108.15 is silver

3. a) Determine the molar solubilityif PbI2(s) is in equilibrium with water. Ksp=9.8E-9. b) From part a it should be clear that pb2+ and i- aq will form and that solid precipitate PbI2 will form. What may be unexpected is waht happens when addition of iodide is added. One would expect the equilibrium to shift right and lead to a greater amount of precipitate formed. however the solution turns clear with no precipitate. The reaction is PbI2(s) +2I-aq--->[PbI4]2- . Determine the molar solubility of PbI2(s) if the iodide concentration is increased to 6.25 M. Kf[PbI4]2- =3.0E4 for the equation

determine molar solubility need ICE and Ktotal PbI2(s)---->Pb2+(aq) +2I- (aq) ksp=9.8E-9 Pb2+(aq)+ 2H20l---[PbO2]2- +2h2+ --------------------------------------------- PbI2( s)+2H20------- 2I- +complex +2h2+ I- - 0 0 0 C- - 2X X NOT INTEREST E- - k=(2x)2(x) k=4x^3 x=1.35E-3 b) we are given both equations use them both do ice table always x=.0115

1. Copper has a face-centered unit cell. how many atoms of Copper in each unit cell?

face=4 atoms

Rank these transition metal ions in order of decreasing number of unpaired electrons. Fe3+ Mn4+ Ni2+ V3+ Cu+

fe=ar 4s23d6, fe3+ =ar 3d5 mn-ar4s23d5- mn4+ arsd3 ni-ar 4s23d8 ni 2+ ar33d8 v-ar4s23d3- ar3+ ar3d2 cu is 29 so we use 4s1 ar4s13d10 so cu+ is ar3d10

ENANTIOMERS

if rotate you can get mirror image

Classify following as low spin or high spin [co(f6]3- high [mn(h20)6]2+ low [fe(cn)6]4- low [co(nh3)6]2+ high [fecn6]3- low

LEFT HIGH SPIN

KNOW THE 4 MAGNETISM

Paramagnetic-substance, element with one or more unpaired electrons( attracted to magnetic field) diamagnetic- all electrons paired, nonmagnetic ferromagnetism- unpaired e- influenced by neighbouring atoms or ions antiferromagnetism-unpaired e align so that their spins are oriented in the direction opposite the spin direction on neighboring atoms ferrimagnetism- both ferromagnetism and antiferromagnetism .unpaired e align so the spins in adjacent atoms or ions point in opposite directions , but the magnetic moments do not cancel each other. the properties of ferrimagnetic materials similar to properties of ferromagnetic materials.

[Pt(NH3)2(H2O)2]2+= diamminediaquaplatinum(II), [PtCl4]2−= tetrachloroplatinate(II), K2[PtCl4]= potassium tetrachloroplatinate(II), and [Pt(en)2]2+

RULE ENDS IN ATE IF is anion

What is the oxidation number of the central metal ion in each of the following complexes? [NiCl2Br2]2− [Cr(H2O)4(NH3)2]3+ Na[Ag(CN)2]

RULE: COUNT THE OUTSIDE OF COMPLEX ALSO. MAKE ALL EQUAL zero, unless given charge, if given element instead of charge count it as equal zero x-4=-2 so x=+2 of x+0+0=+3 x=+3 +1+x-2=0 then change signs x=+2-1=+1

From the following pairs of coordination complexes, identify the one that is a pair of coordination isomers (aka ionization isomers). Na2[NiBr2Cl2] and K2[NiBr2Cl2] [Ni(NH3)3Br]Cl and [Ni(NH3)3Cl]Br [Ni(NH3)3(H2O)]SO4 and [Ni(NH3)2(H2O)2]SO4

RULE: isomers when same # elements coordination isomers= [MX2Y] and [MXY]X switch out-in linkage isomers= *both structural isomers answer count elements left=right 3,2,2=no k smth else 1,3,1,1=1,3,1,1 yes

Consider the following vanadium species. Arrange them in order of strongest to weakest oxidizing agent. VO4 3- [V(H20)4O]2+ [V(H20)6]3+ [V(H20)6]2+

RULE: strongest oxidizing agent=highest oxidation charge answer: find oxidation # x-8=-3 x=+5 x-2=+2 x=+4 x=+3 x=+2 so strongest oxidizing will be a then b then c then d

Name the complex [Cr(CN)6]3− [Cu(NH3)2(H2O)4]2+ CoCl2(en)2 [Ni(H2O)3(CO)]SO4. K4[Pt(CO3)2F2]

RULES:\ 1. separate the compound into two sides: cation and anion. If after separation the coplex is the anion metal end in ate. 2. If only complex given and complex contains ONLY anion you metal ends in ate ex:[CoCl4]-cobaltate 3. If only complex and we have Zero charge ligand name NOT ATE ex[Co(NH3)5Cl]- cobalt not cobaltate 4.if no complex but only anion besides metal=end ATE

11. The absorbtion spectrum of the complex [Ti(NH3)6]3+ is shown to the right(Diagram 545nm wavelength) a)what color does the complex appear? why? b) determine the crystal field splitting energy, in (kj/mol) for this complex. c) if H2O were to replace NH3 as the ligand, would you expect the color of the complex to be more blue or more red. Explain your answer.

Rule a) spectrum colors circle learn 400-430 nm=violet 430-490nm=blue 490-560 nm=green ABSORBTION 560-580 nm=yellow COLORS 580-650 nm=orange 650-750nm=red Rule a) when one color absorbed the complementary color is reflected 3color bottom red,orange, yellow _________=______________ 3color top violet, blue, green so complementary will be violet-yellow, blue-orange,red-green. answer ) at 545 nm is between 490 and 560 nm so color absorbed is green, which means color reflected is red. B. Rule: given wavelength determine crystal field splitting energy 1. use formula: E= hc/wavelength 2. convert to kj and wavelength to meters from nm or pm 3. multiply *6.022*10^23 atoms/mol- to find kj/mol Answer: 1. E=[(6.626*10^-34)*(2.998*10^8)]/5.5*10^-9 m=3.61*10^-17 2. (3.61*10^-17)/10^3= 3.61*10^-20 3.(3.61*10^-20)*(6.022*10^23)= 21750 kj/mol c) Rule: left longer wavelength----right shorter wavelength red =shortest wavelength. 1. look at diagram what is the color absorbed 2. Look what wavelength is that 3. Look at the spectrum and see if longer wavelength for switch Cl-<F-<H2O<NH3<(en)<NO2-<CN- 4. Look color absorbed and complex compound 5. look complementary color of that -is the one we see. Answer: wavelength 560 color-green f0r NH3 H2O toward left which means long wavelength. If red longest wavelengtt(650-750nm) RED NO SHORT Wavele then reflected short wavelength . Blue is short wavelength so that's what we see. WE EXPECT COMPLEX COLOR BLUE.

Which of the following pairs of coordination complexes are linkage isomers? [Pt(Cl)2(SCN)4]4− and [Pt(Cl)2(NCS)4]4− [Pt(Cl)2(SCN)4]4− and [Pt(Cl)4(SCN)2]4− K4[Pt(Cl)2(SCN)4] and Na4[Pt(Cl)2(SCN)4]

Rule isomers=same nr of elements linkage isomers= [MONO] vs [MNO2] a

10 similar. Gallium crystallizes in a primitive cubic unit cell. The length of an edge of the cell of this cube is 362 pm. What is the radius of the Gallium atom?

Rule use the formula: density=mass/volume. 1. Rule: in a primitive cubic unit cell(simple)- 2 radii in one unit length. Answer: 362/2= 181 pm the radii

What is the charge on each of the following complexes: hexaaquachromium(II), [Cr(H2O)6]? tris(carbonato)ferrate(III), [Fe(CO3)3]? amminepentabromoplatinate(IV), [Pt(NH3)Br5]?

Rule. to see the charge of complex look at the roman charge of METAL ION a) chr +2, and water o so 2+ b)irom +3 and co3-6 so complex -3 c)plat 4 and nh30 and br-5 so complex -1

Arrange the metal oxides in order of increasing acidity ZnO, Cu2O, MnO, V2O3,Ni2O3

Rule1. More acidic when higher oxidation charge rule 2. if same oxidation charge. Acidity increases same as ionic radii increases except YOU DO FIRST OXIDATION\ 1. oxidation Cu2O, Zn and MnO, then V2O3 and Ni2O3 then ionic radii increases going up and right first mn then zn, first v then ni Cu2o, mno, zno,v203, ni2o3

9. Name the following compounds 1. [Fe(NH3)5Br]SO4 2. Cesium diamminetetracyanochromate(III) Li3[Fe(CN)6] hexaaquanickel(II) chloride

Rule: 1. Separate into cation vs anion 2. Ligand Named before the metal in [ ] 3. Ligands listed in order of alphabet prefixes not counted 4. Name anion ligand end in o if neutral ligands which have charge o have their own name. H2O- aqua NH3 ammine CO carbonyl en- ethylenediammine py- polydentate 5. ligands called di, tri,tetra, penta, hexa 6. etylenediamine, or polydentate- all use bis, tris, tetrakis, pentakis, hexakis 7. en and py in parenthesis 8. if end in cation- metal no change, if end in anion end in ate 9. oxidation nr always in roman letters. answer: pentaamminebromoferrate(III) chloride oxidation state: x-0-1-2=+3

13. What is the molar solubility of CdS(s) in 2.40 M NaCN(aq) if the complex ion [Cd(CN)4]2- forms? Ksp for CdS is 8.0*10^28 and Kf for Cd(CN)4 2- is 3.0*10^18 b) Will entropy increase or decrease when the complex [Cd(CN)4]2- forms from CdS(s)

Rule: VERY IMPORTANT BALANCE FOLLOW STEPS 1. Chose the solid one and dissociate it: 2. Chose the cation and add it to the aquous with given M --> into complex 3. balance because inside complex says diff amount M 4. add all together into the final equation 5. Kf*Ksp= k overall 6. find formula product /reactant^ moles CdS(s)-------------------------------->Cd+ CUT + S- Ksp Cd+(aq) Cut +4 NaCN (aq)-----> [Cd(CN)4]2- Kf *4CN ------------------------------------------------------------------- CdS(s) +4NaCN(aq)------>[Cd(CN)4]2- +S-(aq) K=ksp*kf= (3.0*10^18)(8.0*10^28)=2.4*10^-9 2.4*10^-9=prod/react= x*x/(2.40-x)^4 bcs greater ignore -x then x=square root of (2.4*10^-9)(2.4^4)=2.82*10^-4 b) Rule: entropy increases when # moles reactant<product Rule : also bidentate{en or py) twice more moles than monodentate Rule : solids or liquids count as moles ! answer: 5 moles reactant vs 2 moles product so entropy decreases :)

10. similar . The face centered gold crystal has an edge length of 407 pm. Based on the unit cell, calculate the density of the gold.

Rule: density =Mass/Volume Density =[(Z*atom weight)/avogadro's nr]/length edge^3(10^-m ) pm---m---cm---cm^3 mass: z- nr atoms: for face centered Four Faces-4 atoms Mass=(4*197g/mol)/6.022*10^23 atoms/mol=1.31*10^-21 g Volume: 407 pm=4.07*10^-8 cm volume=(edge length )^3=6.7*10^-23 cm^3 density=19.4g/cm^3

8. Which of the following compounds below are isomer of C3H6O

Rule: isomers-molecules with same composition but different arrangements of atoms. 1. Look at all compounds and chose one with same # of elements a) C3H6O b) C2H4O no c) C3H6O2 NO d) C3H8O no e) C3H5NH2O no bcs N

Arrange the metal oxides in order of decreasing acidity ZnO, Cu2O, MnO, V2O3,Ni2O3

Rule: small size and high charge =more acidic RULE 1. Determine oxidation states of all given elements Rule 2. Use periodic trends for ionic radii rule 3. put order of decreasing Answer: ZnO- (x-2=0) +2 Cu2O (2x-2=0) +1 MnO (x-2=0) +2 V2O3 (2x-6=0) +3 Ni2O3 (2x-6=0) +3 -Decreasing acidity means from more acidic to less acidic or from higher charge to lower charge. *THINK + acidic, -Basic ----> more + =more acidic Ni2O3 and V2O3, MnO and ZnO , then Cu2O -look at ionic radii. ionic radii decrease from right to left so Ni then V. Zn then Mn so Ni2O3, V2O3, ZnO, MnO, Cu2O

10. Vanadium cryystallizes in a body centered cubic structure and has an atomic radius of 131 pm. Determine the density (g/cm3) of vanadium, if the edge length of a bcc is 4r/sqrt3.

Rule: use formulas Densitiy =mass/volume RULE: unit cell density = [Z*(atom weight of element/avogadro constant)]/a^3 -Z- atoms per unit cell: simple cubic-1atom per unit cell body centered-2 atoms per unit c face-cubic--4 atoms (think Four=Face) -A avogadro's constant=6.022*10^23 a- side of a unit ell

What is the electron configuration of Titanium(II) Ti2+

[Ar]3d2

Predict based on electron configuration if elements are paramagnetic or dimagnetic? si al ar ca

rule write out the electron configuration. no matter if no d si 3s23p2 WHY SI IS PARAMAGNETIC al- 3s23p1 no ar- 3p6 yes ca-4s2 yes

1. CsCl crystallizes in a unit cell that contains the Cs+ ion at the center of a cube that has a Cl- at each corner. How many Cs+ and Cl- ions respectively, are in each unit cell? a)1and 8 b)2 and 1 c) 1 and 1 d)2 and 2 e) 2 and 4

rule: 3 types: 1. primitive cubic cell-corner -fraction 1/8 -atoms in corners - 1 atom per unit cell bcs 8 corners*1/8 2. body centered - fraction 1 -atoms in corner+one in center - 2 atoms per unit cell 3. face centered - fraction 1/2 - atoms in corners+on each face - 4 atoms per unit cell answer: CsCl is a compound. Cs+ crystalizes at center of a cube DOES NOT MEAN IT'S BODY CENTERED ! cl- at each corner means 1/8*8corners=1atom for simple cube. Look at the CHARGE NEGATIVE. so Cs is just ration of 1:1 1 mol of Cs+ =1 mol of Cl- So Cs depends on Cl- from the corners . i ANSWER IS C

6, similar. Which of the following complexes will absorb visible radiation of higher energy? a) [Ti(NH3)6]3+ b)[Ti(en)3]3+ c) [Ti(H2O6]3+ d)[TiCl6]3- e)[TiF6]3-

rule: absorb higher visible energy to the right answer: en

10 similar: Ni has a face centered cubic cell with a edge length of 352 pm. What is the density of this metal?

rule: density = mass/volume rule: density= [z-nr atoms* atomic weight/avogadro constant]/ edge length ^3 mass: face centered=4 atoms 4*58.69 g/6.022*10^23=3.9*10^-22 edge=352 pm=3.52*10^-8cm=4.36*10^-23 cm^3 density=8.94g/cm^3

HOW many outershell d e- are in each of the transition metals? mn(4) co(3) fe(2) ni(ii)

rule: outer shell is subtract NO ADD. EX: Mn=[Ar] 4s23d5. Mn4+ =[Ar]3d3 NOT [Ar]3d7 3 6 6 8

7. How many atoms are in the simple hexagonal structure shown? a)2 b) 3 c)6 d)7 e)14

rule: the hexagonal closest packed has a coordination nr 12 and contains 6 atoms per unit cell. the face centered cubic has coordination nr of 12 and contains 4 atoms per unit cell. the body centered cubic has a coordination nr of 8 and contains 2 atoms per unit cell. b why?

DIASTOMERS

rule> Can you rotate the pair in such a way to get a mirror image? if not you have a diastomer

12. similar. Nickel(II) complexes in which the metal coordination nr is 4 can have either square-planar or tetrahedral geometry. [NiCl4]2- is paramagnetic, and [Ni(CN)4]2- is diamagnetic. One of these complexes is square planar, and the other is tetrahedral. Use the relevant crystal-field splitting diagrams in the text to determine which complex has which geometry.

tetrahedral square planar -------- (dx2-y2) EMPTY --II- -I--- -I--- --II-----dxy --II-- ---II--- ----II---dz2 -----II---dzy -----II--dxz- Nicl4- paramagnetic=unpaired- 2 unpaired is in tetrahedral nicn4-diamagnetic=paired- paired e is the square planar Ni(II)---Ni 2+: [Ar]3d8

1o. similar. Radius of a single atom of a generic element X is 139 picometers and a crystal of X has a unit cell that is body centered cubic. Find the volume.

volume =a^3 for face centered PM= 10^-12 METERS how many radii in body centered? NO NEED TO KNOW use formula: 4r/srt3 is equal to edge length. then ^3 =3.31*10^-29