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Introduction To Chemistry I

by: Austen Pollich

Introduction To Chemistry I CHM 12500

Marketplace > Purdue University > Chemistry > CHM 12500 > Introduction To Chemistry I
Austen Pollich
GPA 3.75

John Nash

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John Nash
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This 2 page Class Notes was uploaded by Austen Pollich on Saturday September 19, 2015. The Class Notes belongs to CHM 12500 at Purdue University taught by John Nash in Fall. Since its upload, it has received 98 views. For similar materials see /class/207968/chm-12500-purdue-university in Chemistry at Purdue University.


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Date Created: 09/19/15
Electron Configurations for Neutral Transition Metal Atoms Ar 1s2 2s2 2p6 3s2 3p6 neutral atom electron SC 4s Ti 4s V 4s Cr 4s Mn 45 Fe 4s C0 4s 3d Ni 4s Cu 4s Z11 4s aThe 4s and 3d orbitals are very nearly degenerate electronelectron repulsion is minimized by placing one electron in each orbital bThe energies of the 3d orbitals are slightly less than the energy of the 4s orbital lling the 3d subshell provides a more energetically favorable con guration Electron Configurations for Positive Transition Metal Ions 0 the energies of the 3d orbitals are signi cantly lower than the energy of the 4s orbital 0 electron con gurations usually written by lling in the electrons from the bottom ie ls2 2s2 2p6 etc electron con gurations for positive transition metal ions can be derived by removing electrons from the top of the neutral transition metal atom Le remove as many electrons as needed starting with the 4s electrons Examples OLE CL Co Ar 4s2 3d7 Co Ar 4s2 3d7 002 Ar 3d7 003 Ar 3d6 Rules for Balancing REDOX Equations HalfReaction Method Acidic or Neutral Solution 1 Divide the skeleton equation into two separate reduction and oxidation halfreactions For each halfreaction Balance all elements except hydrogen and oxygen Balance oxygen by adding water HzOl Balance hydrogen by adding Haq ion Balance charge by adding electrons 909quot If required multiply each balanced halfreaction by the smallest whole number necessary to equalize the number of electrons in the two half reactions Sum the two halfreactions to obtain the overall equation If possible simplify the overall equation by canceling like species Check that the elements and charges balance on both sides of the equation Basic Solution N 4 Complete steps 15 described above for an acidic solution Add one OH39aq ion to both sides of the overall equation for every Haq ion present The Haq ions on one side combine with the added OH39aq ions to form HzOl and OH39aq ions appear on the other side of the equation Haq OH39aq a H20l If possible simplify the resulting overall equation by canceling like species Check that the elements and charges balance on both sides of the equation


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