General Chemistry: Atoms First
General Chemistry: Atoms First
2nd Edition
ISBN: 9780321809261
Author: John E. McMurry, Robert C. Fay
Publisher: Prentice Hall
Question
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Chapter 21, Problem 21.64SP
Interpretation Introduction

Interpretation:

The increase and decrease in melting point of second series of transition metals has to be explained using band energy.

Concept introduction:

Band theory:

Molecular orbital theory deals molecules with few atoms whereas cubic metallic crystal contains more atoms.  When atomic orbital combines to form molecular orbital, the number of molecular orbital is same as number of atomic orbital.  Therefore, number of molecular orbital of Na3 is three, number of molecular orbital of Na4 is four and so on.

In a cubic crystal of metal larger number of metal atoms forms molecular orbital in which electrons are delocalized throughout the crystal.  The energy difference between molecular orbitals in metal crystal decreases as metal atom increases which leads to merging of molecular orbital to form continuous energy band.  Thus, molecular orbital theory for metals are known as band theory.

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Chapter 21 Solutions

General Chemistry: Atoms First

Ch. 21.8 - Prob. 21.11PCh. 21.9 - Prob. 21.12PCh. 21.9 - Prob. 21.13PCh. 21.9 - Prob. 21.14PCh. 21 - Prob. 21.15CPCh. 21 - Prob. 21.16CPCh. 21 - Prob. 21.17CPCh. 21 - Prob. 21.18CPCh. 21 - Prob. 21.19CPCh. 21 - Prob. 21.20CPCh. 21 - Prob. 21.21CPCh. 21 - Prob. 21.22SPCh. 21 - Prob. 21.23SPCh. 21 - Prob. 21.24SPCh. 21 - Prob. 21.25SPCh. 21 - Prob. 21.26SPCh. 21 - Prob. 21.27SPCh. 21 - Prob. 21.28SPCh. 21 - Prob. 21.29SPCh. 21 - Prob. 21.30SPCh. 21 - Prob. 21.31SPCh. 21 - Prob. 21.32SPCh. 21 - Prob. 21.33SPCh. 21 - Prob. 21.34SPCh. 21 - Prob. 21.35SPCh. 21 - Prob. 21.36SPCh. 21 - Prob. 21.37SPCh. 21 - Prob. 21.38SPCh. 21 - Prob. 21.39SPCh. 21 - Prob. 21.40SPCh. 21 - Prob. 21.41SPCh. 21 - Prob. 21.42SPCh. 21 - Prob. 21.43SPCh. 21 - Prob. 21.44SPCh. 21 - Prob. 21.45SPCh. 21 - Prob. 21.46SPCh. 21 - Prob. 21.47SPCh. 21 - Prob. 21.48SPCh. 21 - Prob. 21.49SPCh. 21 - Prob. 21.50SPCh. 21 - Prob. 21.51SPCh. 21 - Prob. 21.52SPCh. 21 - Prob. 21.53SPCh. 21 - Prob. 21.54SPCh. 21 - Prob. 21.55SPCh. 21 - Prob. 21.56SPCh. 21 - Prob. 21.57SPCh. 21 - Prob. 21.58SPCh. 21 - Prob. 21.59SPCh. 21 - Prob. 21.60SPCh. 21 - Prob. 21.61SPCh. 21 - Prob. 21.62SPCh. 21 - Prob. 21.63SPCh. 21 - Prob. 21.64SPCh. 21 - Prob. 21.65SPCh. 21 - Prob. 21.66SPCh. 21 - Prob. 21.67SPCh. 21 - Prob. 21.68SPCh. 21 - Prob. 21.69SPCh. 21 - Prob. 21.70SPCh. 21 - Prob. 21.71SPCh. 21 - Prob. 21.72SPCh. 21 - Prob. 21.73SPCh. 21 - Prob. 21.74SPCh. 21 - Prob. 21.75SPCh. 21 - Prob. 21.76SPCh. 21 - Prob. 21.77SPCh. 21 - Prob. 21.78SPCh. 21 - Prob. 21.79SPCh. 21 - Prob. 21.80SPCh. 21 - Prob. 21.81SPCh. 21 - Prob. 21.82SPCh. 21 - Prob. 21.83SPCh. 21 - Prob. 21.84SPCh. 21 - Prob. 21.85SPCh. 21 - Prob. 21.86SPCh. 21 - Prob. 21.87SPCh. 21 - Prob. 21.88SPCh. 21 - Prob. 21.89SPCh. 21 - Prob. 21.90SPCh. 21 - Prob. 21.92SPCh. 21 - Prob. 21.93SPCh. 21 - Prob. 21.94SPCh. 21 - Prob. 21.95SPCh. 21 - Prob. 21.96SPCh. 21 - Prob. 21.97SPCh. 21 - Prob. 21.98SPCh. 21 - Prob. 21.99SPCh. 21 - Prob. 21.100SPCh. 21 - Prob. 21.101SPCh. 21 - Prob. 21.102SPCh. 21 - Prob. 21.103SPCh. 21 - Prob. 21.104SPCh. 21 - Prob. 21.105SPCh. 21 - Prob. 21.106SPCh. 21 - Prob. 21.107SPCh. 21 - Prob. 21.108SPCh. 21 - Prob. 21.109SPCh. 21 - Prob. 21.110CHPCh. 21 - Prob. 21.111CHPCh. 21 - Prob. 21.112CHPCh. 21 - Prob. 21.113CHPCh. 21 - Prob. 21.114CHPCh. 21 - Prob. 21.115CHPCh. 21 - Prob. 21.116CHPCh. 21 - Prob. 21.117CHPCh. 21 - Prob. 21.118CHPCh. 21 - Prob. 21.119CHPCh. 21 - Prob. 21.120CHPCh. 21 - Prob. 21.121CHPCh. 21 - Prob. 21.122CHPCh. 21 - Prob. 21.123CHPCh. 21 - Prob. 21.124CHPCh. 21 - Prob. 21.125CHPCh. 21 - Prob. 21.127CHPCh. 21 - Prob. 21.128CHPCh. 21 - Prob. 21.129CHPCh. 21 - Prob. 21.130MPCh. 21 - Prob. 21.131MPCh. 21 - Prob. 21.132MPCh. 21 - Prob. 21.133MPCh. 21 - Prob. 21.134MPCh. 21 - At high temperatures, coke reduces silica...Ch. 21 - Prob. 21.136MPCh. 21 - Zinc chromite (ZnCr2O4), once used to make...Ch. 21 - Prob. 21.140MP
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