LORA
Applied chemistry
CHEM44415
* 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 xM to calculate xpara..?
* To explain how the magnetic moment can
be calculated.?
* To predict the geometries and the spin state of
the Co(II) complexes using magnetic
moments.?
Weight of empty
tube, field off
• 3.9506 g
1
2
Weight of empty
tube, field on
Weight of tube
filled to line with
water, field off
• 3.9493 g
3
• 4.4319 g
Standard
Sample 1
Sample 2
4.6186
4.2371
4.2507
Weight of tube filled to
line with solid, field off
4.2439
4
6
Weight of tube filled to
line with solid, field on
4.6704
4.2899
4.2751
7
5
4.2825
 (m)- (0.029  10-6 )V =  (- )
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
V= (3-1) / d (water)
V= 0.4813 cm3(mL)
= (2-1)= -0.0013 g
Standard
Sample
 = (5-4)= 0.0518 g
 = (7-6)= 0.0386 g
m = (4-1)= 0.6680 g
m = (6-1)= 0.2933 g
we calculate  (calibration constant) by using m,
 and Xg (mass susceptiblity )of the standard :
 (m)- (0.029  10-6 )V =  (- )
X(g)= 16.44x10-6
cm3/g(emu/g)
 (m)  (0.029  10 ) V

 
4
  2.0655 10
6
 (m)- (0.029  10-6 )V =  (- )
By using the value of  we calculate X(g) of
the sample :
 (   )  (0.029  10 6 ) V
g 
m
5
 g  2.819 10 emu / g
( Molar suscebilit y ) :
 M   g  molar mass
 M  (2.819 10 5 ) emu / g  361.362 g / mol
 0.01018 emu / mol
 para =  M -  dia
atom
 A(10-6)
atom
cm3/mol
 A(10-6)
cm3/mol
H
-2.93
N
-5.57
C(
aliphatic)
C(
aromatic)
S
-6.00
N(
aromatic)
Co2+
-4.61
-6.24
-15.0
-12.00
Type of atom # of
atoms
xg
Total XDIA
Co2+
-12.8 x 10-6
-12.8 x 10-6
1
H
14
-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
S
2
-15.00 10-6
-3.00x10-5
Total=
-1.9058x10-4
cm3/mol.
AND
Remember
that this unit
= emu/mol
x para  xM  xDIA



x para  0.01018   1.9058  10  4 emu / mol
 0.01037 emu / mol ( BM 2 / K
 eff  2.828 T X para
 eff  2.828 295 K  0.01037 BM 2 / K
 eff  4.95 BM
From this value.
How can you
determine the
geometry and the
spin state of the
complex..?
Co2+ (d7)
Co2+ (d7)
High Spin
Low Spin
eg
eg
t2g
t2g
n  1, S  1 / 2, L  5
n  3, S  3 / 2 , L  3
us  4S S  1
u s   4S S  1
1/ 2
1/ 2
 1.73BM
 3.87 BM
1


u S  L  2 S S  1  LL  1
4


 5.19
1/ 2
1


uS  L  2 S S  1  LL  1
4


 6.16 BM
1/ 2
Tetrahedral
Co2+ (d7) only
High Spin
t2
e
n  3, S  3 / 2 , L  3
u s   4S S  1
1/ 2
 3.87 BM
1


u S  L  2 S S  1  LL  1
4


 5.19
1/ 2
Tetrahedral geometry
Complexes
Co(py)2Cl2
Co(py)2I2
Color (solid)
Blue
Blue
μeff (B.M)
4.42
4.47
Co(Py)2Br2
Co(2-Me-py)2 (Cl)2
Blue
Blue
4.50
4.47
Co(2-Me-py)2 (NCS)2
Blue
4.30
Co(3-Me-py)2Cl2
Blue
4.49
Co(3-Me-py)2Br2
Blue
4.48
Co(3-Me-py)2(NCS)2
Blue
4.30
Complexes
Color (solid)
μeff (B.M)
Co(4-Me-py)2Br2
Blue
4.41
Co(4-Me-py)2(NCS)2
Blue
4.30
Co(3-Et-py)2Br2
Blue
4.70
Co(3-Et-py)2(NCS)2
Blue
4.46
Co(4-Et-py)2Br2
blue
4.45
The magnetic moment of
tetrahedral geometry is
in range(4.30 – 4.74 B.M.) and
absorb light strongly
at range (580-780) nm.
Octahedral geometry
Complexes
Co(py)4Cl2
Co(Py)4(NCS)2
Color (solid)
Pink
Purple-red
μeff (B.M)
5.15
5.10
Co(3-Me-py)4Cl2
Co(3-Me-py)4Br2
Pink
Pink
4.94
5.07
Co(3-Me-py)4(NCS)2
Pink
5.03
Co(4-Me-py)4(NCS)2
Lilac
4.92
Co(3-Et-py)4Br2
Pink
5.02
Co(3-Et-py)4(NCS)2
lilac
5.15
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)
High λ
Low ∆0
HS.
ueff= 4.95 BM
Tetrahedral
(4.30-4.74 ) BM
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 .
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.
2-
.