#Chem 161-2007 homework 11th week

#Chem 161-2007 homework 11th week - Chem 161-2007 11th week...

Info iconThis preview shows pages 1–3. Sign up to view the full content.

View Full Document Right Arrow Icon
Chem 161-2007 11 th Chapter 9 problems: 5,10,20,23,25,27,29,33,35,37,39,41,45,47,49,51,53,55,57,63,67,74,95,98,99,100 Self-Assessment Questions 5. Use Lewis symbols to represent formation of the ionic compound between (a) calcium and bromine atoms (b) barium and oxygen atoms (c) aluminum and sulfur atoms (a) . . . . . .-1 . .-1 :Br: :Br: :Br: :Br: . . . . . . . . Ca Ca 2+ . . . .-2 (b) :O: :O: . . . . Ba Ba 2+ (c) . . . . . . . .2- . .2- . .2- :S: :S: :S: :S: :S: :S: . . . . . . . . . . Al . . Al Al 3+ Al 3+ 10. Using only the periodic table (inside front cover), indicate which element in each set is more electronegative. (a) Br or F F (b) Br or Se Br (c) Cl or As Cl (d) N or H N 20. Given the bond energies, N-to-O bond in NO, 628 kJ/mol; H-H, 435 kJ/mol; N-H, 389 kJ/mol; O-H, 464 kJ/mol, calculate ∆H for the reaction 2 NO(g) + 5H 2 (g) → 2NH 3 (g) + 2H 2 O(g) Note: I do this problem in the same way that we have done ∆Hreaction = ΣH formationproducts - ΣH formationreactants but here we use “bond formation” rather than “formation”. ΔH reaction = ΣH bondformationproducts - ΣH bondformationreactants The enthalpy of bond formation is the reverse of enthalpy of bond energy. One is the energy to form a bond; the other is the energy to break a bond. 1
Background image of page 1

Info iconThis preview has intentionally blurred sections. Sign up to view the full version.

View Full DocumentRight Arrow Icon
Reactants Products ∆H bf ∆H bf 2 NO = 2 x -628 6 NH = 6 x -389 5 H 2 = 5 x -435 4 OH = 4 x -464 ((6 x -389) + (4 x -464)) – ((2 x -628) + (5 x -435)) = -759 kJ reaction = ΣH bondsbroken - ΣH bondsformed Reactants Products ∆H be ∆H be 2 NO = 2 x +628 6 NH = 6 x +389 5 H 2 = 5 x +435 4 OH = 4 x +464 ΔH reaction = ((2 x 628) + (5 x 435)) – ((6 x 389) + (4 x 464)) = -759 kJ Note: This answer is different than in the Solutions Manual. The Solutions Manual is using the formula backwards, and getting the reverse answer (+759 kJ). Ions and Ionic Bonding 23. Which of the following ions have noble-gas electron configurations? What are the electron configurations of the ions that do not? (a) Mo 6+ (b) Te 2- (c) Ti 4+ (d) Zn 2+ (e) La 3+ (f) Hg 2+ (a) 42 Mo 6+ = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 4 → 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 = electron configuration of Kr. (b) 52 Te 2- = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 4 → 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 = electron configuration of Xe. (c) 22 Ti 4+ = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 2 → 1s 2 2s 2 2p 6 3s 2 3p 6 = electron configuration of Ar. (d) 30 Zn 2+ = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 → 1s 2 2s 2 2p 6 3s 2 3p 6 3d 10 = not a noble-gas electron configuration. (e) 57 La 3+ = 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 6 4f 3 → 1s 2 2s 2 2p 6 3s 2 3p 6 4s 2 3d 10 4p 6 5s 2 4d 10 5p 3 4f 3 = not a noble-gas electron configuration. Note: The answer in the Solutions Manual is incorrect. To form a cation, first fill up all of the orbitals of the atom. Then remove the electrons from the valence shell, even
Background image of page 2
Image of page 3
This is the end of the preview. Sign up to access the rest of the document.

This homework help was uploaded on 04/12/2008 for the course CHEM 161 taught by Professor Vacillian during the Fall '08 term at Rutgers.

Page1 / 19

#Chem 161-2007 homework 11th week - Chem 161-2007 11th week...

This preview shows document pages 1 - 3. Sign up to view the full document.

View Full Document Right Arrow Icon
Ask a homework question - tutors are online