
(a)
Interpretation: The electronic configuration for the given ions should be determined.
Concept Introduction:
Periodic Table: The available chemical elements are arranged considering their
In periodic table the horizontal rows are called periods and the vertical column are called group.
There are seven periods and 18 groups present in the table and some of those groups are given special name as follows,
Atomic Number: Atomic number of the element is equal to the number of protons present in the nucleus of the element which is denoted by symbol Z. The superscript presents on the left side of the
Cation: Removal of electron from the atom results to form positively charged ion called cation.
Anion: Addition of electron to atom results to form negatively charged ion called anion.
The net charge present in the element denotes the presence or absence of electrons in the element.
Electronic configuration: It is used to represent the distribution of electrons placed over orbitals that present in the atom.
The rules followed by the electrons are as follows,
The electrons gets distributed strictly following the order starting from lower energy orbital to higher energy orbital(Aufbau principle), pairing of electrons in subshell starts only when all the orbitals in the subshell are singly filled (Hund’s Rule) and finally, no two electrons that have same set of quantum numbers (Pauli’s Exclusion Principle).
(b)
Interpretation: The electronic configuration for the given ions should be determined.
Concept Introduction:
Periodic Table: The available chemical elements are arranged considering their atomic number, the electronic configuration and their properties. The elements placed on the left of the table are metals and non-metals are placed on right side of the table.
In periodic table the horizontal rows are called periods and the vertical column are called group.
There are seven periods and 18 groups present in the table and some of those groups are given special name as follows,
Atomic Number: Atomic number of the element is equal to the number of protons present in the nucleus of the element which is denoted by symbol Z. The superscript presents on the left side of the symbol of the element.
Cation: Removal of electron from the atom results to form positively charged ion called cation.
Anion: Addition of electron to atom results to form negatively charged ion called anion.
The net charge present in the element denotes the presence or absence of electrons in the element.
Electronic configuration: It is used to represent the distribution of electrons placed over orbitals that present in the atom.
The rules followed by the electrons are as follows,
The electrons gets distributed strictly following the order starting from lower energy orbital to higher energy orbital(Aufbau principle), pairing of electrons in subshell starts only when all the orbitals in the subshell are singly filled (Hund’s Rule) and finally, no two electrons that have same set of quantum numbers (Pauli’s Exclusion Principle).
(c)
Interpretation: The electronic configuration for the given ions should be determined.
Concept Introduction:
Periodic Table: The available chemical elements are arranged considering their atomic number, the electronic configuration and their properties. The elements placed on the left of the table are metals and non-metals are placed on right side of the table.
In periodic table the horizontal rows are called periods and the vertical column are called group.
There are seven periods and 18 groups present in the table and some of those groups are given special name as follows,
Atomic Number: Atomic number of the element is equal to the number of protons present in the nucleus of the element which is denoted by symbol Z. The superscript presents on the left side of the symbol of the element.
Cation: Removal of electron from the atom results to form positively charged ion called cation.
Anion: Addition of electron to atom results to form negatively charged ion called anion.
The net charge present in the element denotes the presence or absence of electrons in the element.
Electronic configuration: It is used to represent the distribution of electrons placed over orbitals that present in the atom.
The rules followed by the electrons are as follows,
The electrons gets distributed strictly following the order starting from lower energy orbital to higher energy orbital(Aufbau principle), pairing of electrons in subshell starts only when all the orbitals in the subshell are singly filled (Hund’s Rule) and finally, no two electrons that have same set of quantum numbers (Pauli’s Exclusion Principle).

Want to see the full answer?
Check out a sample textbook solution
Chapter 4 Solutions
Chemistry: Atoms First
- (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_forward8. (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_forward
- A 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_forward2. (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_forward
- From 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_forwardA solution contains 0.097 M Ce3+, 1.55x10-3 M Ce4+, 1.55x10-3 M Mn2+, 0.097 M MnO4-, and 1.00 M HClO4 (F= 9.649 x 104 C/mol). a) Write a balanced net reaction that can occur between species in this solution. b) Calculate deltaG0 and K for the reaction. c) Calculate E and deltaG for the conditions given. Ce4+ + e- = Ce3+ E0= 1.70 V MnO4- + 8H+ + 5e- = Mn2+ + 4H2O E0= 1.507 Varrow_forward
- 1. Provide a step-by-step mechanism for formation of ALL STEREOISOMERS in the following reaction. Na HCO3 (Sodium bicarbonate, baking soda) is not soluble in CH2Cl2. The powder is a weak base used to neutralize strong acid (pKa < 0) produced by the reaction. Redraw the product to show the configuration(s) that form at C-2 and C-4. Br2 OH CH2Cl2 Na* HCO3 Br HO OH + Na Br +arrow_forward2. Specify the solvent and reagent(s) required to carry out each of the following FGI. If two reagent sets must be used for the FGI, specify the solvent and reagent(s) for each reagent set. If a reaction cannot be carried out with reagents (sets) class, write NP (not possible) in the solvent box for reagent set #1. Use the letter abbreviation for each solvent; use a number abbreviation for reagent(s). Solvents: CH2Cl2 (A); H₂O (B); Reagents: HBr (1); R₂BH (6); H2SO4 (2); CH3OH (C); Br₂ (3); CH3CO₂H (D) NaHCO3 (4); Hg(OAc)2 (5); H₂O2/HO (7); NaBH4 (8) Reagent Set #1 Reagent Set #2 FGI + enant OH Solvent Reagent(s) Solvent Reagent(s)arrow_forwardGermanium (Ge) is a semiconductor with a bandgap of 2.2 eV. How could you dope Ge to make it a p-type semiconductor with a larger bandgap? Group of answer choices It is impossible to dope Ge and have this result in a larger bandgap. Dope the Ge with silicon (Si) Dope the Ge with gallium (Ga) Dope the Ge with phosphorus (P)arrow_forward
- Chemistry: The Molecular ScienceChemistryISBN:9781285199047Author:John W. Moore, Conrad L. StanitskiPublisher:Cengage LearningPhysical ChemistryChemistryISBN:9781133958437Author:Ball, David W. (david Warren), BAER, TomasPublisher:Wadsworth Cengage Learning,Chemistry & Chemical ReactivityChemistryISBN:9781337399074Author:John C. Kotz, Paul M. Treichel, John Townsend, David TreichelPublisher:Cengage Learning


