Inorganic Exam 1

The steps to the Born-Haber cycle for RbCl are outlined. Identify each step as endothermic or exothermic.

1. sublimation of Rb - endothermic 2. dissociation of chlorine - endothermic 3. ionization of rubidium - endothermic 4. electron affinity of chlorine - exothermic 5. formation of RbCl - exothermic 6. overall - exothermic

H2SO4

12 atoms surround S

Determine the total number of valence electrons in bromine pentafloride, BrF5.

42 valence electrons square pyramidal 90 polar

Identify the coordination number for an atom in a simple (primitive) cubic crystal lattice.

6

The density of solid FeFe is 7.87 g/cm3.. How many atoms are present per cubic centimeter (cm3) of Fe?

7.87g/cm^3(1mol/55.85gFe)(6.02x10^23) = 8.5x10^22 atoms As a solid, FeFe adopts a body‑centered cubic unit cell. How many unit cells are present per cubic centimeter (cm3) of Fe? in a bcc, there are 2 atoms 8.5x10^22 atoms (1 unit cell/2 Fe atoms) = 4.25 x10^22 unit cells/cm^3 What is the volume of a unit cell of this metal? V = 1cm^3/4.25 x10^22 unit cells = 2.4x10^-23 What is the edge length of a unit cell of Fe? cube root of the volume = 2.89x10^-8 cm

Consider the body‑centered cubic unit cell shown in this image.

8 corner atoms 1 body atom 1/8 2 total

Consider the simple cubic (primitive cubic) unit cell shown in this image.

8 corner atoms 1/8 1 total atom

Consider the face‑centered cubic unit cell shown in this image.

8 corner atoms 1/8 6 face atoms 1/2 4 total atoms

Identify the formula for each precipitate that forms.

A + C = Ag3AsO4 A + D = AlAsO4 B + C = Ag2C2O4 B + D = Al2(C2O4)3 C + D = AgBr

Use the concept of hard and soft acids and bases to predict the order of solubility in water of the salts of fluoride. Arrange the compounds according to their relative solubilities.

A compound consisting of a hard acid and a hard base will have a high formation constant and be less apt to dissolve in water. Soft acids and soft bases will form similarly strong bonds and will be relatively insoluble. Most > least soluble: cadmium(II) fluoride iron (II) fluoride magnesium fluoride

Arrange the metal ions by their acidity in aqueous solution.

A weaker O−H bond leads to a greater acidity. To weaken the O−H bond, the metal ion must attract the electron density from the oxygen atom. Small, highly charged ions attrct electrons most strongly from the O−H bond in the hydrated complex, which weakens the bond and makes it easier to release H+. Ga3+ and In3+ have the highest charges of the given ions, but Ga3+ is smaller, so it is likely to be more acidic. Cd2+ is approximately the same size as In3+ but Cd2+ has a lower charge, making it less acidic. order from most to least: Ga3+, In3+, Cd2+

Study the chemical equations that show how each substance behaves when dissolved in water. Then classify each substance as a Brønsted-Lowry acid, Brønsted-Lowry base, or as neither an acid nor a base.

Acids : HClO2, HF, NH4+ Bases: NH3, CO32- neither: CaCl2

Select the molecules that are polar.

AsF3, IOF5, Ch2Cl2

For each molecule, specify the polarity of the bonds and the overall polarity of the molecule.

BeCl2 : polar bonds, non polar molecule H2O : polar bonds, polar molecule

Lewis Structure and Bond Order of NO3-, NO2-,NO2+

Bond order = #of bonds/#of outer atoms NO3- : 1.33 NO2- : 1.5 NO2+ : 2

Where, approximately, is the negative pole on each of these molecules?

COF2 : between the F atoms COFH : between the O and F atoms Which molecule should have the higher dipole moment, and why? COFH because the polar bonds in COF2 nearly cancel each other out.

Which dopant added to Si will create a p‑type semiconductor?

Ga

Complete these Brønsted-Lowry reactions.

HCO3- + H+ = H2CO3 HCO3- + OH = H2O + CO32-

Identify the reactions that will occur.

Hard acids tend to pair with hard bases. Soft acids tend to pair with soft bases. For a reaction to occur spontaneously, the pairings in the products must be more favorable than the pairings in the reactants. of the choices: CdF2+Ca(CN)2⟶Cd(CN)2+CaF2

The steps of the Born-Haber cycle are shown for the ionic compound MX, made up of the generic metal M and the generic, gaseous halogen X.

Identify the overall reaction: M(s)+1/2X2(g)⟶MX(s) Step 1 : enthalpy of sublimation (enthalpy of atomization) Step 2 : 1/2 bond energy X-X Step 3: ionization energy Step 4: electron affinity Step 5: lattice energy enthalpy of formation of MX

The steps of the Born-Haber cycle are shown for the ionic compound MX2, made up of generic metal M and generic, gaseous halogen X. Identify the overall reaction and identify the name of the enthalpy change for each reaction.

Identify the overall reaction: M(s)+X2(g)⟶MX2(s) Step 1: enthalpy of sublimation (enthalpy of atomization) Step 2: bond energy of X-X Step 3: 1st + 2nd ionization energies Step 4: 2 x electron affinity Step 5: lattice energy enthalpy of formation of MX2

Classify these factors by how they affect the acidity of a metal cation.

Increases acidity : increase in charge, increased electronegativity Decreases acidity : increase in ion size

Classify each of these solids as ionic, molecular, or metallic.

Ionic : KCl Molecular: CCl4 Metallic: Na

Predict the order of solubility in water of the salts of Pb(II).Pb(II). Arrange the compounds according to their relative solubilities.

Ionic compounds whose constituent ions have widely different radii are more soluble in water than ionic compounds with similar-sized cations and anions. Polyatomic ions are larger than monoatomic ions, so you can conclude that the sulfide anion is closest in radius to Pb(II).Pb(II). Thus, the sulfide salt has the lowest solubility. Similarly, anions that contain larger atoms have larger radii. Therefore, you can conclude that lead(II) selenate has a larger anion-cation size ratio than lead(II) sulfate, and that lead(II) selenate is most soluble. Most > least soluble lead (II) selenate lead (II) sulfate lead (II) sulfide

Classify each of these solids as ionic, molecular, metallic, or covalent (also known as covalent‑network solids or macromolecular solids).

Ionic: Na2S Molecular: FCl Metallic: Ca Covalent: diamond

How does the lattice energy for an ionic solid relate to the size of the ions in the ionic solid?

Lattice energy becomes less exothermic (less negative) with increasing ionic radius. How does the lattice energy for an ionic solid relate to the magnitude of the charge of the ions in the ionic solid? Lattice energy becomes more exothermic (more negative) with increasing magnitude of ionic charge.

Calculate the enthalpies of formation, Δ𝐻∘f,ΔHf∘, of group 1 chloride compounds from their elements by using the Born-Haber cycle.

LiCl: -405 RbCl: -433

Determine the empirical formula of the compound with a crystal structure where magnesium ions occupy all of the octahedral holes in a cubic close‑packed array of sulfur anions.

MgS cubic close pack = fcc the number of sulfur anions will be equal to the number of octahedral holes; if all of the holes are occupied by Mg, #S=#Mg = 1:1 ratio

Calculate the enthalpies of formation, Δ𝐻f∘,ΔHf∘, of the group 1 fluoride compounds from their elements using the Born-Haber cycle.

NaF = -565 RbF = -543

Which dopant added to Si will create a n‑type semiconductor?

Sb

ClBr3

T shaped 90, 180 polar

Unknown element XX forms a binary compound with fluorine, FF . The compound melts at 900 °C and conducts electricity in the liquid phase.

The EN value for FF is 4.194.19 , according to Figure 4.8a of your textbook. The minimum average EN for a FF compound is about 2.422.42 , and the maximum average EN for a FF compound is about 2.902.90 . Thus, minimum =2(2.42)−4.19=0.65 maximum: 2(2.90)−4.19=1.61

Select the correct statement about polar bonds and dipole moments.

The dipole moment for a polyatomic molecule equals the vector sum of the individual bond moments. Where, approximately, is the negative pole on COF2? between the F atoms Where, approximately, is the negative pole on CHOF? between the O and F atoms Which molecule should have the higher dipole moment, and why? CHOF because the bond moments in COF2 nearly cancel each other out.

Cesium bromide adopts the CsCl(BCC) structure type. Estimate the lattice energy, ΔU, for CsBrCsBr using the Born-Landé equation. The bond length for Cs−BrCs−Br is 3.70×102 pm and the Madelung constant is 1.7627. The Born exponents of the cation and anion are 10 and 10, respectively.

U = -596

The radius of a single atom of a generic element X is 133 picometers (pm)133 picometers (pm) and a crystal of X has a unit cell that is body-centered cubic. Calculate the volume of the unit cell.

V = (cell edge)^3 body centered cubic: cell edge = 4r/sqrt3 = 4(133)/sqrt3 = 307.15 pm (1x10^-12 m) = 3.07x10^-10 V = (3.07 x 10^10)^3 = 2.89x10^-29 m^3

The radius of a single atom of a generic element X is 171 and a crystal of X has a unit cell that is face‑centered cubic. Calculate the volume of the unit cell.

V = (cell edge)^3 fcc: edge length = 2sqrt2(r)=2sqrt2(171) = 483.66 pm = 4.836 x 10^-10 m (4.836 x 10^-10 m)^3 = V V = 1.131 x 10^-28 m^3

The relative ionic radii of four ions, A, B, X, and Y, are shown. The ions shown in teal carry positive charges: a +2 charge for A and a +1 charge for B. The ions shown in blue carry negative charges: a −1 charge for X and a −2 charge for Y.

Which combinations of these ions will produce ionic compounds where there is a 1:1 ratio of cations and anions? A and Y. B and X Of the combinations that produce an ionic compound with a 1:1 ratio of cations and anions, which combination results in the ionic compound with the largest lattice energy? A and Y Of the combinations that produce an ionic compound with a 1:1 ratio of cations and anions, which combination results in the ionic compound with the smallest lattice energy? B and X

Identify the products formed in this Brønsted-Lowry reaction. HSO4- + NO2-↽−−⇀acid+base

acid: HNO2 base: SO42-

In an ionic compound, the size of the ions affects the internuclear distance (the distance between the centers of adjacent ions), which affects lattice energy (a measure of the attractive force holding those ions together). Based on ion sizes, arrange these compounds by their expected lattice energy.

big charge, small radius - radius increases as you move down the periodic chart Greatest > least LiF NaF KF RbF CsF

Given that the density of TlCl(s) is 7.00 g/cm3 and that the length of an edge of a unit cell is 385 pm, determine how many formula units of TlCl there are in a unit cell.

edge length: 385 pm = 3.85 x 10^-8 cm V = (3.85 x 10^-8)^3 = 5.707 x10^-23 mass in unit cell = d(V) = 7.00 (5.707 x10^-23) = 3.995 x 10^-22 g mol TlCl = mass in unit cell/molar mass = 3.995 x 10^-22 g/239.84 = 1.666 x 10^-24 mol # of units = 1.666 x 10^-24 mol (6.02x10^23) = 1 formula unit same as CsCl lattice

What is the major impurity in silicon used to make semiconductors?

elements with one more or one fewer valence electron than silicon

Determine whether the equilibrium for the reaction lies to the left or to the right.

for Br3In−As(CH3)3+H3B−NH3 ↽−−⇀ Br3In−NH3+H3B−As(CH3)3 the reaction goes to the left. for F3B−SbH3+I3Ga−N(CH3)3 ↽−−⇀ F3B−N(CH3)3+I3Ga−SbH3 it goes to the right Hard acids tend to pair with hard bases. Soft acids tend to pair with soft bases. In the first reaction, the hard-hard and soft-soft combinations are in the reactants, so the equilibrium lies to the left. In the second reaction, the hard-hard and soft-soft combinations are in the products, so the equilibrium lies to the right.

Determine the empirical formula of the compound with a crystal structure where copper ions occupy one‑half of the tetrahedral holes in a cubic close‑packed (ccp) array of chlorine anions.

in cubic closed packed, the number of atoms forming lattice = 4 = 4 Cl- tetrahedral holes = 2(#atoms forming lattice) = 2(4)=8 half of the 8 holes are occupied by Cl- so 4Cl-, 4Cu+ = CuCl (answer)

The atomic radius of metal X is 1.30×102 picometers1.30×102 picometers (pm) and a crystal of metal X has a unit cell that is face-centered cubic. Calculate the density of metal X (atomic weight = 42.3 g/mol).

m = 4(42.3)/6.023x10^23 = 2.81x10^-22 V=(cell edge)^3 edge = 2sqrt2(r)=2sqrt2(1.30x10^2) = 367.69 pm = 3.67x10^-8 cm (3.67x10^-8)^3 = 4.97 x 10^-23 d = 2.81x10^-22/4.97x10^-23 = 5.65

A semiconductor doped with donor impurities is a:

n type A semiconductor doped with acceptor impurities is a: p type

What type of semiconductor would be formed if SiSi were doped with Al?

p type

For each description of the band structure of a solid material, select the best category for that material.

partially filled conduction band: conductor energy gap of about 1 eV between the valence and conduction bands: semiconductors energy gap of about 5 eV or more between the valence and conduction bands: insulators

Calculate the radius ratio for CuO if the ionic radii of Cu2+and O2− are 73 pm and 140 pm , respectively.

radius ratio: cation/anion = .5214 Based on the radius ratio, what is the expected coordination number of CuO? The range of radius ratios for each coordination number is given in parentheses. 6 (.414 - .732)

XeCl4

square planar, 90 and 180, non polar

Determine the empirical formula of the compound with a crystal structure where lithium ions occupy all of the tetrahedral holes in a cubic close‑packed (ccp) array of selenium anions.

the selenium forms the lattice.. tetrahedral holes = 2(#of Se atoms) Tetrahedral holes are all occupied by Li; # of Li = # tetrahedral holes = 2 (Se) : 2:1 ratio Li2Se