SampleEx4S - Sample final exam answers 1 For both these...

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Sample final exam answers 1. For both these compounds, the formal oxidation state is Mn(II), which is d 5 . Since NCS is a weak-field, high-spin ligand (from the spectrochemical series), the electron configuration for the octahedral complex will be (t 2g ) 3 (e g ) 2 , and it will have 5 unpaired electrons. The tetrahedral complex, which is also high-spin, will have electron configuration (e) 2 (t 2 ) 3 and it will also have 5 unpaired electrons. Since the magnetic susceptibility is a function of the number of unpaired electrons, you can't tell these compounds apart in this way . 2. (a) The Cu(I) complex is d 10 and colorless; the Cu(II) complex is d 9 and blue. The colors are a result of d-d electronic transitions; none of these are possible in the d 10 case, since there are no empty orbitals into which an electron can go. However, for the d 9 case, such an excitation is possible. (b) For 1 FeBr 6 3– , this is Fe(III), d 5 . Since Br is a weak-field ligand, this is a high-spin complex with electron configuration (t 2g ) 3 (e g ) 2 and 5 unpaired electrons . It is paramagnetic and has CFSE = 3(–2/5 o ) + 2(3/5 o ) = 0. For 2 [FeCl 4 2– ], this is Fe(II), d 6 . Since the complex is tetrahedral, it is high-spin, with electron configuration (e) 3 (t 2 ) 3 and 4 unpaired electrons. It is paramagnetic and has CFSE = 3(–3/5 t ) + 3(2/5 t ) = –3/5 t . 3. The required Born-Haber cycle is sketched below.
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From this cycle, H f o (MgO) = H sublimation (Mg) + 1/2D(O-O) + IE 1 (Mg) + IE 2 (Mg) +EA(O) +EA(O ) + U lattice (MgO)
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SampleEx4S - Sample final exam answers 1 For both these...

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