Two lone 3d electrons pair up which frees one 3d orbital for hybridization As a

Two lone 3d electrons pair up which frees one 3d

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However, it’s a little more complicated than that. Two lone 3d electrons pair up, which frees one 3d orbital for hybridization. As a result, four dsp 2 orbitals are used to accommodate the four CN - ligands, leaving one empty 4p orbital unused. This arrangement is consistent with the fact that the complex is diamagnetic (no unpaired electrons). - [Zn(OH) 4 ] 2- forms a tetrahedral complex. Zn 2+ has the Aufbau structure of [ 18 Ar]4s 2 3d 10 → [ 18 Ar]3d 10 . This can be drawn as an empty 4s orbital, five completely filled 3d orbitals, and three empty 4 p orbitals. Hence, there are four available empty orbitals for hybridization (sp 3 ) to accomodate four OH - ligands in [Zn(OH) 4 ] 2- tetrahedral structure. Based on the inability to match up the coordination numbers of complex ions in B&O Table 17.5, with the number of available orbitals identified by the above valence bond theory, one cannot well-predict the complex ion coordination number. But at least this theory describes the home of the bonding ligands. Chem 162-2015 Chapter 18 Acid-base & solubility equilibria lecture notes 63
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Z&Z 15-99 (modified with fictitious data) ET: Complex ion formation is sequential reactions with ligands on primarily transition metal ions, giving large K’s. Complex ion: Species where a transition metal ion is bonded to ligands. This is a Lewis acid – Lewis base reaction When aqueous KI is added gradually to a mercury(II) nitrate solution, an orange precipitate forms. Continued addition of KI causes the precipitate to dissolve. Write balanced equations to explain these observations. (Hint: Hg 2+ reacts with I - to form HgI 4 2- .) What is the overall formation constant for HgI 4 2- ? Use K f1 = 0.2 x 10 4 , K f2 = 6.5 x 10 10 , K f3 = 7.6 x 10 7 and K f4 = 0.1 x 10 9 . Hg(NO 3 ) 2 → Hg 2+ + 2NO 3 - = a solution Hg 2+ + I - [HgI + ](aq), a soluble complex ion K f1 = 0.2 x 10 4 [HgI + ] + I - HgI 2 (s)*, an orange precipitate K f2 = 6.5 x 10 10 *Note no charge in HgI 2 . HgI 2 + I - [HgI 3 - ](aq), a soluble complex ion K f3 = 7.6 x 10 7 [HgI 3 - ] + I - [HgI 4 2- ](aq), a soluble complex ion K f4 = 0.1 x 10 9 Hg 2+ (aq) + 4I - (aq) [HgI 4 2- ](aq) K f = 9.9 x 10 29 Chem 162-2015 Chapter 18 Acid-base & solubility equilibria lecture notes 64
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CHEM 162-2010 Final exam Chapter 22 - Transition Metals and Coordination Chemistry Complex Ion Equilibria Given K f and initial concentrations, find equilibrium concentration. 28. 0.40 moles of AgNO 3 and 2.5 moles of Na 2 S 2 O 3 (sodium thiosulfate) are dissolved in 1.00L of aqueous solution. A coordination complex forms. Ag + (aq) + 2S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq) K f = 1.7 x 10 13 Calculate [Ag + ] in this solution. A . 8.1 x 10 -15 M B. 5.3 x 10 -15 M C. 1.7 x 10 -13 M D. 2.0 x 10 -12 M E. 1.7 x 10 -13 M Ag + (aq) + 2S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq) K f = 1.7 x 10 13 Ag + (aq) + 2S 2 O 3 2- (aq) [Ag(S 2 O 3 ) 2 ] 3- (aq) Initial 0.40 2.5 0 Change -X -2X +X Equilibrium 0.40-X 2.5-2X +X [Ag(S 2 O 3 ) 2 ] 3- /([Ag + ][S 2 O 3 2- ] 2 )= K f = 1.7 x 10 13 ([X])/([0.40-X] x ([2.5-2X] 2 )) = K f = 1.7 x 10 13 Quadratic equation. Use the large K rule. Bring the reaction to completion. Ag + is the limiting reactant.
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