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Experiment S4
Chemistry 114
Operation of the Magnet
The strength of the field between the poles of the electromagnet is determined by the current
that passes through the coils. An adjustable regulated power supply provides a highly stable
current; you must determine the relation between the current and the field strength by calibration
with a NiCl2 solution.
Sketched in Fig. 1 are the panels on the power supply:
Current Control
10-turn potentiometer
Regulation Meter
Power Switch
Before attempting to turn on the power supply, MAKE SURE THE VARIAC KNOB (large
knob) IS SET TO ZERO (fully counterclockwise). Always return the VARIAC KNOB to
ZERO before turning off the power supply.
With the Variac Knob at zero, turn the Power Switch to the ON position. (Both the regulation
meter and the Ammeter should read zero.)
The Current Control dial is a 10-turn potentiometer that reads in fractions of amperes. For
example, a reading of 250 (i.e., 2.5 turns) corresponds to a current slightly lower than 2.5 A. The
actual current value depends on the Variac knob position and is shown on the Ammeter. The
current control dial can be locked in position by moving the Locking Lever to the right.
After the current dial has been set, slowly increase the current in the coils by turning the
Variac knob clockwise. The ammeter will respond immediately, but at first the Regulation Meter
will not move. Continue to turn the current up until the needle of the Regulation Meter rises to the
part of the dial marked "Regulation." For the best control, the needle should be near the "Set"
mark on the dial. The magnetic field strength will now remain stable. The approximate value of
the field strength can be determined from the Field Strength vs. Current curve shown below in
Fig. 2. The precise value, however, is irrelevant for this experiment. The apparatus constant,
which is related to the magnetic field and the cross section of the Gouy tube, will be determined
by calibration of the instrument as described in GNS, pp.368-369.
The field strength may now be changed by setting the Current Control to a new value and
adjusting the Variac until the Regulation Meter is at the "Set" position. Do not turn down the
Revised Winter 2009
Experiment S4
page 2
Current Control without lowering the Variac setting. The Regulation Pointer should NEVER
be allowed to reach the Full-Scale position.
Field Strength vs. Current - 1" Gap
The magnet displays slight hysteresis and as a result the relation between the Current Control
setting and the magnetic field strength depends slightly on the path of magnetization. However,
the fields are highly reproducible if the same magnetization and demagnetization path is followed
each time. (For example, you might always start at zero field, increase the current in steps of 100
units and decrease it in steps of 200 units.) You can check the reproducibility by repeated
calibrations with NiCl2.
Do not operate the magnet at currents in excess of 7 A for long times. At high currents,
the temperature of the magnet rises sharply and may damage the coils. In addition, the high
temperature produces convection currents that will affect the weigh of your samples.
Experimental Procedure
Follow the procedure outlined in GNS, pp. 365-370. NiCl2 , KMnO4, MnSO4,
[Fe(H2O)6](NH4)2(SO4)2, K4Fe(CN)6 and K3Fe(CN)6 are available in the laboratory, some of
them as hydrated salts, keep this in mind when calculating concentrations. Do not use the Guoy
balance to prepare the solutions! Also, to avoid generating excess chemical waste, prepare only
10 mL solutions of each sample (except for NiCl2) using plastic scintillation vials (in S4 drawers)
and a graduated cylinder (The error in the volume measurement using a graduated cylinder is large
enough that using an analytical balance to weigh your samples is not necessary, the top loader,
with a +/- 5 mg uncertainty can be used without increasing the error of the concentration of your
solutions). The molar concentrations can be calculated after measuring the density of the different
solutions. For NiCl2 you should prepare 50 ml of solution in a volumetric flask. After you have
completed your experiment please clean up your work area, and be sure to properly dispose of
your solutions in the appropriate chemical waste container in the hood.
To calculate the solution density, add a precise volume of your solution to the Guoy tube (for
example 2.000ml). Be patient when you are weighing your sample in the Guoy tube, because the
balance readings will fluctuate until the tube is perfectly steady.
Do not remove the tube holder. Center the tube with respect to the magnet by adjusting its
vertical and horizontal position with the balance and tube holder respectively. Keep in mind that
calibration constants are specific for each Guoy tube.
Additional Theory (to derive Eq. (6) in GNS)
The interaction energy ε of a dipole µ with the applied magnetic field H is:
Experiment S4
page 3
ε = µ • µ0 H ,
where µ and H are vectors, and the dot signifies a scalar product. In the presence of a magnetic
field, all molecules have a diamagnetic contribution (α H) to the magnetic moment, where α is a
negative parameter characteristic of the molecule. Molecules that have at least one unpaired spin
have a paramagnetic dipolar contribution:
= 2 ms µe ,
µ(paramag. along H)
directed along the applied field H. (Here, ms is the component of spin along the applied field H;
ms = S,S-1,S-2,...,-S and µe is the Bohr magneton.) We now need the magnetic intensity I.
Since for a mole of substance IM/r is the sum of the dipole moments along the applied field H,
(I M/r) = No α µo H + S 2(µe ms) ,
where No is the Avogadro constant. In Eq. (C) and subsequently, all the vector quantities are
aligned along the magnetic field, so we drop the vector notation and use the scalar quantities I, H,
etc. The second term in this equation should be summed over all molecules but, since we might
naively expect that for every molecule with a given ms there will be one with -ms , we might
expect the sum to vanish. This is almost, but not quite, the case. From Eqs. (A) and (B), we see
that the energy for a molecule with ms > 0 is lower than that for a molecule with ms < 0. Since
systems with lower energy are more stable, there is a tendency for more molecules to have ms > 0
than ms < 0. The number Nms in each state with a given ms is described by the Boltzmann
Nms = (2S + 1) exp[-εms /kT] ,
where k is the Boltzmann constant, T is the absolute temperature, and εms is the interaction energy
corresponding to dipoles with a given ms. Thus, combining Eq.(A)-(D), the sum in Eq.(C) is:
Σ 2(µe ms)
µe No
2µemsµo H
= (2S + 1) Σ ms exp
In all cases of interest to us,
2µemsµo H
<< 1 ,
so that the exponential can be expanded in powers of [2µemsµoH/kT]. We retain the first two
terms of the expansion with the result that:
2Σ µems
Σ msµo H/kT + ...
(2S + 1) m
The sum is taken over all ms values; for S=1, ms=1,0,-1 , while for S=1/2, ms = 1/2 , - 1/2 .
(Explain) Consequently, the first sum in Eq.(G) vanishes and, if the second is correctly summed,
4µe2Noµo H
µems =
S(S + 1)
Check with S=1/2 and S=1 that the summation has been carried out correctly. If we now combine
Eqs. (H), (A), (C) and (D), we find that
Experiment S4
χM = Noµeα + 3kT
This is the equivalent to GNS's Eq. (6) (p.363)
page 4
NOTE: Much of this discussion would have been simpler if, in changing to SI units, GNS had
written B for µoH,
Additional Material for FIG.1 GNS (p.365)
Co(III) has six 3d electrons. If the energy correlating each of these electrons with the others
is neglected, the energy of each electron (called an orbital energy) can be determined. There are
five possible orbital energies for 3d electrons, and the Pauli Exclusion Principle states that no
more than 2 electrons can occupy an orbital, and double occupancy is permitted only if one
electron has ms = 1/2 and the other, ms = -1/2.
For a free Co3+ ion, all five orbitals have the same energy (see Fig.1-GNS). For an
octahedral arrangement of ligands, the energies are split; that is, three orbitals have lower energy
than those of Co3+ and two have higher energy. The "splitting" energy is labeled ∆. Thus, the
energy of the upper orbitals is +3∆/5, whereas that of the lower orbital is -2∆/5. If the energy of
the interacting spins could be neglected, all six electrons would reside in the three lower orbitals;
the energy of the system would be -12∆/5 and the net spin would be S=0.
Extra energy (P) is required to place two electrons within a single orbital. The total energy
of the six 3d electrons of Co(III) is equal to the sum of the six orbital energies plus the sum of the
energies required to form the number of electron pairs in the system. The most stable, or groundstate energy of the system is the lowest energy, and usually we study systems in the ground states.
If all six electrons are placed in the three lower orbitals, the energy of the system is (3P 12∆/5), and there are no unpaired spins (S=0). If four electrons are placed in the lower orbitals
and two in the upper ones, the lowest energy is (P - 2∆/5), which is the energy if two of the
electrons in the lower and the two in the upper orbitals are unpaired (S=2). If the splitting is large,
then the pairing energy in unimportant and the first (S=0) situation holds. If the splitting is weak,
the pairing is critical and the second (S=2) situation has a lower energy. (See Fig. 1, GNS). An
example of weak-field splitting is (Co(III)F6)3- , while Co(NH3)63+ is an example of strong-field
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