1. LORA Applied chemistry CHEM44415

2. * To explain how the mass susceptibility can be calculated.? * To explain how the molar susceptibility can be calculated from the mass susceptibility.? * To explain how the diamagnetic correction is carried out on x M to calculate x para..? * To predict the geometries and the spin state of the Co(II) complexes using magnetic moments.? * To explain how the magnetic moment can be calculated.?

4. 3.9506 g Weight of empty tube, field off 3.9493 g Weight of empty tube, field on 4.4319 g Weight of tube filled to line with water, field off 1 2 3

5. Weight of tube filled to line with solid, field off Weight of tube filled to line with solid, field on Standard 4.6186 4.6704 Sample 1 4.2371 4.2751 Sample 2 4.2507 4.2899 4 5 4.2439 4.2825 6 7

6. Where m = tube filled with solid – empty tube  = empty tube field on – empty tube field off  = tube filled with solid, field on – tube filled with solid, field off V = (tube filled with water – empty tube) / density of water  g of standard = 16.44E-6 cgs  (m)- (0.029  10 -6 )V =  (  -  )

8. we calculate  (calibration constant) by using m,  and X g (mass susceptiblity )of the standard : X(g)= 16.44x10 -6 cm 3 /g(emu/g)

9. By using the value of  we calculate X (g) of the sample :  (m)- (0.029  10 -6 )V =  (  -  )

10.  para =  M -  dia

11. atom  A (  10 -6 ) cm 3 /mol atom  A (  10 -6 ) cm 3 /mol H-2.93N-5.57 C( aliphatic) -6.00N( aromatic) -4.61 C( aromatic) -6.24Co 2+ -12.00 S-15.0

12. Type of atom# of atoms xgTotal XDIA Co 2+ 1-12.8 x 10 -6 H14 -2.93  10 -6 -4.102x10-5 C(aromatic)10 -6.24  10 -6 -6.24x10-5 C(aliphatic)4 -6.00  10 -6 -2.40x10-5 N(aromatic)2 -4.61  10 -6 -9.22x10-6 N (aliphatic)2 -5.57  10 -6 -1.114x10-5 S2 -15.00  10 -6 -3.00x10-5 Total= -1.9058x10 -4 cm 3 /mol. AND Remember that this unit = emu/mol

13. From this value. How can you determine the geometry and the spin state of the complex..?

14. Co 2+ (d 7 ) High Spin Co 2+ (d 7 ) Low Spin egeg t 2g egeg

15. Tetrahedral Co 2+ (d7) only High Spin t2t2 e

16. Tetrahedral geometry μ eff (B.M) Color (solid) Complexes 4.42Blue Co(py) 2 Cl 2 4.47Blue Co(py) 2 I 2 4.50Blue Co(Py) 2 Br 2 4.47Blue Co(2-Me-py) 2 (Cl) 2 4.30Blue Co(2-Me-py) 2 (NCS) 2 4.49Blue Co(3-Me-py) 2 Cl 2 4.48Blue Co(3-Me-py) 2 Br 2 4.30Blue Co(3-Me-py) 2 (NCS) 2

17. μ eff (B.M) Color (solid) Complexes 4.41Blue Co(4-Me-py) 2 Br 2 4.30Blue Co(4-Me-py) 2 (NCS) 2 4.70Blue Co(3-Et-py) 2 Br 2 4.46Blue Co(3-Et-py) 2 (NCS) 2 4.45blue Co(4-Et-py) 2 Br 2

18. The magnetic moment of tetrahedral geometry is in range(4.30 – 4.74 B.M.) and absorb light strongly at range (580-780) nm.

19. Octahedral geometry μ eff (B.M) Color (solid) Complexes 5.15Pink Co(py) 4 Cl 2 5.10Purple-red Co(Py) 4 (NCS) 2 4.94Pink Co(3-Me-py) 4 Cl 2 5.07Pink Co(3-Me-py) 4 Br 2 5.03Pink Co(3-Me-py) 4 (NCS) 2 4.92Lilac Co(4-Me-py) 4 (NCS) 2 5.02Pink Co(3-Et-py) 4 Br 2 5.15lilac Co(3-Et-py) 4 (NCS) 2

20. The magnetic moment of octahedral geometry is in range(4.90 – 5.40 B.M.) and absorb light weakly in the range (640-600 nm)

21. High λ Low ∆ 0 HS. u eff = 4.95 BM Tetrahedral (4.30-4.74 ) BM

22. 1.Magnetic moments are used to determine the spin state (high spin or low spin). 2.Octahdral complexes can be either high spin or low spin. 3.Tetrahedral complexes can only be high spin. 4.Experimental magnetic moments for Co(II) Complexes are always higher than the spin-only magnetic moments because of the significant Orbital contribution.

23. 5.The magnitude of the orbital contribution differ for tetrahedral and octahedral,it is greater for octahedral than for tetrahedral. Therefore,we can distinguish between tetrahedral and octahedral. Tetrahedral; 4.30 – 4.72 B.M. Octahedral; 4.90 – 5.40 B.M.