
Concept explainers
(a)
Interpretation: The reason behind equal potential energies for all four
Concept introduction: Two or more of orbitals undergo redistributions of their different energies so as to form mathematically averaged orbitals in terms of energy although they may differ in shape and orientation. This phenomenon is referred as hybridization.
(b)
Interpretation: The reason behind energy level line for
Concept introduction: Two or more of orbitals undergo redistributions of their different energies so as to form mathematically averaged orbitals in terms of energy although they may differ in shape and orientation. This phenomenon is referred as hybridization. The new hybrid orbitals are always equal in number to number of atomic orbitals that combine.
(c)
Interpretation: The diagram given below for
Concept introduction: Two or more of orbitals undergo redistributions of their different energies so as to form mathematically averaged orbitals in terms of energy although they may differ in shape and orientation.
The energy and orientation of the new hybrid orbital depends upon by the kind and number of orbitals used in the hybridization. For instance, when one
(d)
Interpretation: The second-row element shown in left diagram of below image and hybridization state in right diagram of below image should be identified.
Concept introduction: Hund’s rule of maximum multiplicity states that electrons are not allowed to be paired up until each degenerate set of orbital has got at least one electron.
(d)
Interpretation: Whether the unpaired electrons in below figure are in accordance with Hund’s rule or not should be identified.
Concept introduction: Hund’s rule of maximum multiplicity states that electrons are not allowed to be paired up until each degenerate set of orbital has got at least one electron.

Want to see the full answer?
Check out a sample textbook solution
Chapter 3 Solutions
Organic Chemistry: A Guided Inquiry
- (3 pts) Use the Kapustinskii equation to calculate the lattice enthalpy for MgBr2 anddiscuss any differences between this result and that from #4.arrow_forward(3 pts) Silver metal adopts a fcc unit cell structure and has an atomic radius of 144 pm. Fromthis information, calculate the density of silver. Show all work.arrow_forward4. (3 pts) From the information below, determine the lattice enthalpy for MgBr2. Show all work. AH/(kJ mol-¹) Sublimation of Mg(s) +148 lonization of Mg(g) +2187 to Mg2+(g) Vaporization of Br₂(1) +31 Dissociation of Br,(g) +193 Electron gain by Br(g) -331 Formation of MgBr₂(s) -524arrow_forward
- 1. (4 pts-2 pts each part) Consider the crystal structures of NaCl, ZnS, and CsCl (not necessarily shown in this order). a. For one of the three compounds, justify that the unit cell is consistent with stoichiometry of the compound. b. In each of the crystal structures, the cations reside in certain holes in the anions' packing structures. For each compound, what type of holes are occupied by the cations and explain why those particular types of holes are preferred.arrow_forward(2 pts) What do you expect to happen in a Na2O crystal if a Cl− ion replaces one of the O2−ions in the lattice?arrow_forward(2 pts) WSe2 is an ionic compound semiconductor that can be made to be p-type or n-type.What must happen to the chemical composition for it to be p-type? What must happen tothe chemical composition for it to be n-type?arrow_forward
- 8. (2 pts) Silicon semiconductors have a bandgap of 1.11 eV. What is the longest photon wavelength that can promote an electron from the valence band to the conduction band in a silicon-based photovoltaic solar cell? Show all work. E = hv = hc/λ h = 6.626 x 10-34 Js c = 3.00 x 108 m/s 1 eV 1.602 x 10-19 Jarrow_forwardA solution containing 100.0 mL of 0.155 M EDTA buffered to pH 10.00 was titrated with 100.0 mL of 0.0152 M Hg(ClO4)2 in a cell: calomel electrode (saturated)//titration solution/Hg(l) Given the formation constant of Hg(EDTA)2-, logKf= 21.5, and alphaY4-=0.30, find out the cell voltage E. Hg2+(aq) + 2e- = Hg(l) E0= 0.852 V E' (calomel electrode, saturated KCl) = 0.241 Varrow_forwardFrom the following reduction potentials I2 (s) + 2e- = 2I- (aq) E0= 0.535 V I2 (aq) + 2e- = 2I- (aq) E0= 0.620 V I3- (aq) + 2e- = 3I- (aq) E0= 0.535 V a) Calculate the equilibrium constant for I2 (aq) + I- (aq) = I3- (aq). b) Calculate the equilibrium constant for I2 (s) + I- (aq) = I3- (aq). c) Calculate the solubility of I2 (s) in water.arrow_forward
- 2. (3 pts) Consider the unit cell for the spinel compound, CrFe204. How many total particles are in the unit cell? Also, show how the number of particles and their positions are consistent with the CrFe204 stoichiometry - this may or may not be reflected by the particle colors in the diagram. (HINT: In the diagram, the blue particle is in an interior position while each red particle is either in a corner or face position.)arrow_forwardFrom the following potentials, calculate the activity of Cl- in saturated KCl. E0 (calomel electrode)= 0.268 V E (calomel electrode, saturated KCl)= 0.241 Varrow_forwardCalculate the voltage of each of the following cells. a) Fe(s)/Fe2+ (1.55 x 10-2 M)//Cu2+ (6.55 x 10-3 M)/Cu(s) b) Pt, H2 (0.255 bar)/HCl (4.55 x 10-4 M), AgCl (sat'd)/Ag Fe2+ +2e- = Fe E0= -0.44 V Cu2+ + 2e- = Cu E0= 0.337 V Ag+ + e- = Ag E0= 0.799 V AgCl(s) + e- = Ag(s) + Cl- E0= 0.222 V 2H+ + 2e- = H2 E0= 0.000 Varrow_forward
- Organic Chemistry: A Guided InquiryChemistryISBN:9780618974122Author:Andrei StraumanisPublisher:Cengage LearningChemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningChemistry: Principles and PracticeChemistryISBN:9780534420123Author:Daniel L. Reger, Scott R. Goode, David W. Ball, Edward MercerPublisher:Cengage Learning


