1. Use the expression for the Coulomb potential energy to calculate the energy for formation of 1 mole of sodium chloride ion-pairs, that is, the energy
... [Show More] change for the following reaction:
Na+(g) + Cl(g) Na+Cl(g) Use r12 = 283 pm.
Ans: 491 kJmol1
2. If 491 kJmol1 is released in the reaction Na+(g) + Cl(g) Na+Cl(g), what is the energy change for the reaction Na(g) + Cl(g) Na+Cl(g)? (Hint: See the discussion in the text and apply Hess's Law.)
Ans: 346 kJmol1
3. If 346 kJmol1 is released in the reaction Na(g) + Cl(g) Na+Cl(g), is the energy change for the reaction Na+Cl(g) NaCl(s) endothermic or exothermic?
Ans: Exothermic
4. The Madelung constant is different for all crystals. True or false? Ans: True
5. Use the expression for the Coulomb potential energy to calculate the energy for formation of 1 mole of rubidium chloride ion-pairs, that is, the energy change for the following reaction:
Rb+(g) + Cl(g) Rb+Cl(g) Use r12 = 330 pm.
Ans: 421 kJmol1
6. Which of the following has the lowest lattice energy?
A) KCl B) LiCl C) KBr D) NaCl E) KI Ans: E
7. Which of the following has the highest lattice energy?
A) NaCl B) KI C) MgO D) BaO E) CaO Ans: C
8. Which of the following has the highest melting point?
A) KF B) KI C) RbF D) KBr E) KCl Ans: A
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9. Metals rarely lose electrons in chemical reactions because
A) their electron affinities are too high. D) their size is too small.
B) their ionic radii become too small. E) their ionization energies are too high.
C) their ionization energies are too small.
Ans: E
10. An element, E, has the electronic configuration [Ne] 3s23p1. Write the formula of its compound with sulfate.
Ans: E2(SO4)3
11. Predict the electronic configuration in the oxide ion in CaO.
A) [He]2s22p6 or [Ne] D) [Ne]3s13p3
B) [He]2s22p5 E) [Ne]3s23p3
C) [He]2s22p63s2
Ans: A
12. Write the formula of magnesium phosphide. Ans. Mg3P2
13. Which of the following metal ions has the ground-state electron configuration [Ar]3d6?
A) Ni3+ B) Fe2+ C) Mn2+ D) Cu+ E) Ca2+ Ans: B
14. For the ground-state ion Pb2+, what type of orbital do the electrons with highest energy reside in?
A) 6p B) 5p C) 4f D) 6s E) 5d Ans: D
15. For the ground-state ion Sn4+, what type of orbital do the electrons with highest energy reside in?
A) 4p B) 5p C) 4f D) 4d E) 5s Ans: D
16. For the ground-state ion Bi3+, what type of orbital do the electrons with highest energy reside in?
A) 5d B) 6s C) 4f D) 5p E) 6p Ans: B
17. For the ground-state ion I, what type of orbital do the electrons with highest energy reside in?
A) 4d B) 6s C) 5p D) 5d E) 5s Ans: C
18. Because of the octet rule, the gaseous O2 ion is stable. True or false? Ans: False
19. All the following elements exist as diatomic gases at room temperature and atmospheric pressure except
A) H. B) Ar. C) N. D) Cl. E) O. Ans: B
20. How many lone pairs of electrons are found in the Lewis structure of the interhalogen compound ICl3?
A) 10 B) 4 C) 8 D) 6 E) 7
Ans: A
21. How many lone pairs of electrons are found in the Lewis structure of urea, (NH2)2CO? A) 2 B) 3 C) 6 D) 4 E) 8
Ans: D
22. How many lone pairs of electrons are found in the Lewis structure of hydrazine, H2NNH2?
A) 8 B) 4 C) 1 D) 0 E) 2
Ans: E
23. Draw the Lewis structure of xenon difluoride and give the number of lone pairs electrons around the central atom.
Ans: Three
24. Draw the Lewis structure of the formate ion and indicate whether resonance forms are possible.
Ans: Two resonance forms are possible.
25. Draw the “best” Lewis structures of hydrogen azide, HN1N2N3, and the azide ion,
N N N . The subscripts are used for identification. For each, match the following bond lengths to the correct N–N bond. The bond lengths can be used more than once.
N–N bond Bond length, pm hydrogen azide N1–N2 113
N2–N3 116
azide ion N1–N2 124
N2–N3
Ans: hydrogen azide: N1–N2, 124 pm; N2–N3, 113 pm; azide ion: N1–N2, 116 pm; N2– N3, 116 pm
26. Which of the following do not have resonance structures?
A) CH3CONH B) CH2COCH3 C) H2CO D) All have resonance structures. Ans: C
27. For dinitrogen monoxide, the arrangement of the atoms is N-N-O. In the Lewis structure with a double bond between NN and NO, the formal charges on N, N, and O, respectively, are
A) 0, 1, +1. B) 1, +1, 0. C) 0, +1, 1. D) 0, 0, 0. E) 2, +1, +1.
Ans: B
28. For dinitrogen monoxide, the arrangement of the atoms is N-N-O. In the Lewis structure with a single bond between NN and a triple bond between NO, the formal charges on N, N, and O, respectively, are
A) 1, +1, 0. B) 0, 0, 0. C) 0, +1, 1. D) 0, 1, +1. E) 2, +1, +1.
Ans: E
29. In the “best” Lewis structure of XeO4, there are two double bonds and the formal charge on Xe is zero. True or false?
Ans: False
30. Write three Lewis structures for the cyanate ion, NCO, where the arrangement of atoms is N-C-O. In the most plausible structure,
A) there is a triple bond between N and C.
B) there are two double bonds.
C) there is a triple bond between C and O.
D) the formal charge on O is +1.
E) the formal charge on N is 1. Ans: A
31. Predict the N-O bond lengths in NO2, given the N-O and N=O bond lengths of 140 and 120 pm, respectively.
Ans: Both ~ 130 pm
32. Why are the N-O bond lengths in NO3 the same? Ans: The explanation is resonance.
33. Which of the following species are radicals?
A) CO2 B) HNO3 C) NO2 D) NO3 E) HNO3 Ans: C only.
34. Which of the following species are radicals? [Show Less]