(a)
Interpretation:
The ground-state electronic configuration of isoelectronic
Concept Introduction:
The fundamental principles that are followed to write an electronic configuration include three rules as follows:
Electron in a
Hund’s rule suggests electrons are not allowed to be paired up until each degenerate set of orbital has got at least one electron.
Pauli Exclusion Principle states two electrons within the same orbital cannot possess same set for four possible quantum numbers. Quantum-mechanical equivalent for this state means that two electrons are not allowed to have the same space simultaneously. Since one electron may certain discrete same set of n, l and m values, however, the fourth spin quantum number should be different. This provides a unique address to every electron. Since the possible magnitudes of spin can be either +1/2 or -1/2 thus at maximum two electrons can occupy any given orbital.
(b)
Interpretation:
Number of unpaired electrons in
Concept Introduction:
Refer to part (a).
(c)
Interpretation:
Neutral atom that has the same electron configuration as
Concept Introduction:
Refer to part (a).
Want to see the full answer?
Check out a sample textbook solutionChapter 2 Solutions
Chemical Principles: The Quest for Insight
- (a) Describe the molecule xenon trioxide, XeO3, using four possible Lewis structures, one each with zero, one, two, or three Xe—O double bonds. (b) Do any of these resonance structures satisfy the octet rule for every atom in the molecule? (c) Do any of the four Lewis structures have multiple resonance structures? If so, how many resonance structures do you find? (d) Which of the Lewis structures in (a) yields the most favorable formal charges for the molecule?arrow_forwardAn important starting material for the manufacture ofpolyphosphazenes is the cyclic molecule (NPCl₂)₃. The mol-ecule has a symmetrical six-membered ring of alternating N and P atoms, with the Cl atoms bonded to the P atoms. The nitrogen-phosphorus bond length is significantly less than that expectedfor an N−P single bond.(a) Draw a likely Lewis structure for the molecule.(b) How many lone pairs of electrons do the ring atoms have?(c) What is the order of the nitrogen-phosphorus bond?arrow_forwardWhich of the following bonds are polar: (a) P—O; (b) S—F; (c) Br—Br; (d) O—Cl? Which is the more electronegative atom in each polar bond?arrow_forward
- Write the ground-state electron configuration for each atomand ion pair-Zr, Zr2+, Co, Co2+, Tc, Tc3+, Os, Os4+?arrow_forwardIf an element is bonded to 4 other atoms and has a formal charge of +1, what group must the element be in? I know that group 3A atoms are elctron deficient, and that period 3 elements and below, except for group 3A elements like Aluminum, can expand their octet because of their available d-orbital, which may not be relevant to this problem. I don't understand this question, or why the answer would be 5A. Is it because 5A have odd valence electrons, and can form free radicals, like NO?arrow_forwardThere are two main types of covalent bond breakage. In ho-molytic breakage , each atom in the bond gets one of the shared electrons. In some cases, the electronega-tivity of adjacent atoms affects the bond energy. In heterolytic breakage, one atom gets both electrons and the other gets none;thus, a cation and an anion form. (a) Why is the C−C bond in H₃C−CF₃(423 kJ/mol) strongerthan that in H₃C−CH₃(376 kJ/mol)?(b) Use bond energy and any other data to calculate the heat ofreaction for the heterolytic cleavage of O₂.arrow_forward
- Common exceptions to the octet rule are compounds and polyatomic ions with central atoms having more than 8 electrons around them. Phosphorus pentafluoride, PF5; sulfur tetrafluoride, SF4; xenon tetrafluoride, XeF4; and tri-iodide ion, I3, are all examples of exceptions to the octet rule. (a) Draw the Lewis structures of these substances.(b) For which elements in these substances can theatoms have more than 8 electrons around them?(c) How can the atoms of the elements youidentified in Part (b) be surrounded by morethan 8 electrons?arrow_forward(c) Draw the orbital diagrams and Lewis symbols to depict the formation of Na* and CI ions from the atoms. Give the formula of the compound formed. (d) The predicted bond length for HF is 109 pm (the sum of the covalent radii of H, 37 pm and F. 72 pm), however the actual bond length for HF is shorter (92 pm). It was observed that the difference between predicted and actual bond lengths becomes smalleor going down the halogen group from HF to HI Describe these observationsarrow_forwardWhich of these statements about resonance is true?(a) When you draw resonance structures, it is permissibleto alter the way atoms are connected.(b) The nitrate ion has one long N¬O bond and two shortN¬O bonds.(c) “Resonance” refers to the idea that molecules areresonating rapidly between different bonding patterns.(d) The cyanide ion has only one dominant resonancestructure.(e) All of the above are true.arrow_forward
- Three main group elements, X, Y and Z, from three different groups inthe Periodic Table, each form a neutral trifluoride, i.e. XF3, YF3 andZF3. (a) To which three groups of the Periodic Table could the elementsX, Y and Z belong? Are any of these atoms X, Y or Zhypervalent in the trifluorides formed? (b) Use VSEPR theory to deduce the shapes of the molecules XF3,YF3 and ZF3, showing full working, and sketch these shapes (c) Only two of the three trifluorides have a non-zero dipole moment.For these molecules to which groups of the Periodic Table do thecentral atoms belong?arrow_forwardAmmonia reacts with boron trifluoride to form a stablecompound, as we saw in Section 8.7. (a) Draw the Lewisstructure of the ammonia–boron trifluoride reaction product.(b) The B—N bond is obviously more polar than theC—C bond. Draw the charge distribution you expect on theB—N bond within the molecule (using the delta plus anddelta minus symbols mentioned in Section 8.4). (c) Borontrichloride also reacts with ammonia in a similar way tothe trifluoride. Predict whether the B—N bond in the trichloridereaction product would be more or less polar thanthe B—N bond in the trifluoride product, and justify yourreasoningarrow_forwardTwo substances with empirical formula HNO are hyponi-trous acid ( μ=62.04 g /mol) and nitroxyl (μ=31.02 g/mol).(a) What is the molecular formula of each species?(b) For each species, draw the Lewis structure having the lowestformal charges. (Hint:Hyponitrous acid has an N=N bond.)(c) Predict the shape around the N atoms of each species.(d) When hyponitrous acid loses two protons, it forms the hy-ponitrite ion. Draw cisand transforms of this ion.arrow_forward
- ChemistryChemistryISBN:9781305957404Author:Steven S. Zumdahl, Susan A. Zumdahl, Donald J. DeCostePublisher:Cengage LearningChemistryChemistryISBN:9781259911156Author:Raymond Chang Dr., Jason Overby ProfessorPublisher:McGraw-Hill EducationPrinciples of Instrumental AnalysisChemistryISBN:9781305577213Author:Douglas A. Skoog, F. James Holler, Stanley R. CrouchPublisher:Cengage Learning
- Organic ChemistryChemistryISBN:9780078021558Author:Janice Gorzynski Smith Dr.Publisher:McGraw-Hill EducationChemistry: Principles and ReactionsChemistryISBN:9781305079373Author:William L. Masterton, Cecile N. HurleyPublisher:Cengage LearningElementary Principles of Chemical Processes, Bind...ChemistryISBN:9781118431221Author:Richard M. Felder, Ronald W. Rousseau, Lisa G. BullardPublisher:WILEY