Consider the following table of ionization energies in kJ/mole. 2nd 3rd 4th Element Na 1 st 496 5th 13353 6th 4562 6912 9544 16610 Mg 738 1451 7733 10540 13630 17995 18378 Al 578 1817 2745 11577 14831 Si 786 1577 3232 4356 16091 19785 1012 1903 2912 4957 6274 21269 a. Explain why the first ionization energy generally increases as one goes down the group of elements listed above. b. Explain why for a particular element, the second ionization energy is greater than the first ionizatior energy, the third is greater than the second etc. c. Explain why Mg has a higher first ionization energy than does Na, but a lower second ionization energy than does Na. d. Explain why for aluminum there is a large increase in ionization from 3rd to 4th

Consider the following table of ionization energies in kJ/mole. 2nd 3rd 4th Element Na 1 st 496 5th 13353 6th 4562 6912 9544 16610 Mg 738 1451 7733 10540 13630 17995 18378 Al 578 1817 2745 11577 14831 Si 786 1577 3232 4356 16091 19785 1012 1903 2912 4957 6274 21269 a. Explain why the first ionization energy generally increases as one goes down the group of elements listed above. b. Explain why for a particular element, the second ionization energy is greater than the first ionizatior energy, the third is greater than the second etc. c. Explain why Mg has a higher first ionization energy than does Na, but a lower second ionization energy than does Na. d. Explain why for aluminum there is a large increase in ionization from 3rd to 4th

Chemistry

9th Edition

ISBN:9781133611097

Author:Steven S. Zumdahl

Publisher:Steven S. Zumdahl

Chapter8: Bonding: General Concepts

Section: Chapter Questions

Problem 152CP: Think of forming an ionic compound as three steps (this is a simplification, as with all models):...

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Consider an ionic compound, MX2MX2, composed of generic metal MM and generic, gaseous halogen XX. The enthalpy of formation of MX2MX2 is Δ?∘f=−923ΔHf∘=−923 kJ/mol. The enthalpy of sublimation of MM is Δ?sub=131ΔHsub=131 kJ/mol. The first and second ionization energies of MM are IE1=755IE1=755 kJ/mol and IE2=1364IE2=1364 kJ/mol. The electron affinity of XX is Δ?EA=−329ΔHEA=−329 kJ/mol. (Refer to the hint). The bond energy of X2X2 is BE=153BE=153 kJ/mol. Determine the lattice energy of MX2MX2. Δ?lattice=ΔHlattice= kJ/mol
Why does Mg have a relatively low first and second ionization energies? A) Because the first and second ionization energies remove the valence electrons to form the stable noble gas configuration B) Because Mg is a nonmetal, and nonmetals are electronegative C) Because the first and second ionization energies remove the electrons that are the closest to the nucleus D) Because Mg does not like to lose electrons E) Because the first and second ionization energies remove the core electrons that are stable
Suppose a chemist discovers a new metallic element and names it "Xtrinsium" (Xt). Xt exhibits chemical behaviour similar to an alkaline earth. Xt(s) + F2(g) → XtF2(s) Lattice energy for XtF2 -2360. kJ/mol First Ionization energy of Xt 520. kJ/mol Second Ionization energy of Xt 936 kJ/mol Electron affinity of F -327.8 kJ/mol Bond energy of F2 154 kJ/mol Enthalpy of sublimation (atomization) of Xt 180. kJ/mol Use the above data to calculate ΔH°f for Xtrinsium fluoride.
Consider an ionic compound, MX3, composed of generic metal M and generic gaseous halogen X. The enthalpy of formation of MX3 is Δ?∘f=−925 kJ/mol. The enthalpy of sublimation of M is Δ?sub=175 kJ/mol. The first, second, and third ionization energies of M are IE1=579 kJ/mol, IE2=1677 kJ/mol, and IE3=2479 kJ/mol. The electron affinity of X is Δ?EA=−369 kJ/mol. (Refer to the hint). The bond energy of X2 is BE=179 kJ/mol. Determine the lattice energy of MX3.
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Consider an ionic compound, MX3, composed of generic metal M and generic gaseous halogen X. The enthalpy of formation of MX3 is ΔHf∘=−965 kJ/mol. The enthalpy of sublimation of M is ΔHsub=123 kJ/mol. The first, second, and third ionization energies of M are IE1=557 kJ?mol, IE2=1751 kJ/mol, and IE3=2731 kJ/mol. The electron affinity of X is ΔHEA=−339 kJ/mol The bond energy of X2 is BE=235 kJ/mol. Determine the lattice energy of MX3.
Suppose a chemist discovers a new metallic element and names it "Xhaustium" (Xh).Xh exhibits chemical behaviour similar to an alkaline earth. Xh(s) + F2(g) → XhF2(s) Lattice energy for XhF2 -2140. kJ/mol First Ionization energy of Xh 310. kJ/mol Second Ionization energy of Xh 589 kJ/mol Electron affinity of F -327.8 kJ/mol Bond energy of F2 154 kJ/mol Enthalpy of sublimation (atomization) of Xh 150. kJ/mol Use the above data to calculate ΔH°f for Xhaustium fluoride. Your answer must be accurate and precise to the nearest 1 kJ/mol, as are the given parameters.
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