Chapter 22 - Metallurgy a. More active = more it wants to...

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I. Metallurgy a. More active = more it wants to be in a compound and harder to get into elemental form b. Most use electrolysis c. Less active metals chemical reduction i. Need effective cheap reducing agent ii. Graphite (C) iii. Fossil fuel iv. SnO 2 + C yields Sn + CO 2 v. PbO 2 + C yields Pb + CO 1. H = very unfavorable = very endothermic at room temp 2. Go to high temp because S is favorable 3. Find critical temperature vi. Often start with sulfide then convert to oxide 1. Roasting = convert sulfide into oxide 2. 2PbS + 3O 2 yields 2PbO + 2SO 2 II. Reactions of Metals a. Single displacement b. Table 4.6 = activity series (to displace/produce H 2 ) i. Most active (I and II metals) react w/ water ii. Least Active (Cu, Ag, Hg, Au) don’t react with water iii. Intermediate react with acid c. Combination (metal and nonmetal) i. 2Al + 3Cl 2 yields 2AlCl 3 ii. Oxidation state change III. Oxidation states of Metals a. S block: group number b. P block: towards bottom of G III-V i. G# - 2 = key ii. Sn: [Kr]5s 2 4d 10 5p 2 iii. Sn 2+ : [Kr]5s 2 4d 10 iv. Sn 4+ : [Kr]4d 10 v. Because remove higher energy electrons vi. Remove p electrons before s electrons vii.Pb = +2 only c. D block: +2 very important i. Fe: [Ar]4s 2 3d 6 ii. Fe 2+ : [Ar]3d 6 iii. So is 4s higher in energy than 3d? because of the 4? Then why do we write 3d after 4s? then doesn’t it have higher energy? If it’s filled after? I guess it all depends on so location and energy level are not proportional (wrong word but you know what I’m trying to say) iv. Very high oxid n states 1. Cr +6 2. Mn +7
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3. May be tempted to want to come down in oxidation state = REDUCTION 4. Good oxidizing agents IV.Acidic/Basic Properties of Metals a. Acid i. NaCl vs. AlCl 3 1. NaCl = dissolves a. Blue = neutral 2. AlCl 3 = SMOKY! Very violent reaction a. Yellow (bromothymol) = acidic
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Chapter 22 - Metallurgy a. More active = more it wants to...

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