chemistry assignments, Some normally nonmagnetic substances are attracted by a magnetic field and studies of these "paramagnetic" substances give information about the number of unpaired electrons in the atoms, molecules, or ions of the substance.
The paramagnetic effect can be introduced problem of the magnetic behavior of an electron revolving about a nucleus. A classical treatment is easily made and yields a result which can then be converted to the correct quantum mechanical result. The motion of an electron in an orbit corresponds, in this connection, to the passage of a current through a coil of wire. A current in a coil of wire of ordinary dimensions produces a coil of wire. A current in a coil of wire of ordinary dimensions produces a magnetic field perpendicular to the coil. The magnetic field so produced is equal, according to Ampere's law, to that of a magnet with magnetic moment given by the product of the current and the cross section area fo the loop of wire.
μM = iA
The current corresponding to an electron in orbit is obtained by multiplying the number of times the electrons passes any point on the orbit by its electronic charge. Thus, with this classical picture of an electron in an atomic orbit, I = [v/ (2∏r)]e, where the electron velocity is v and the orbit has the radius r. the cross section area is A = ∏r2. The magnetic moment μM is expressed as
μM = vre/2 or μm/mvr = e/2m
This final form expresses the result of this classical derivation that can be carried over into quantum mechanical systems, namely, that the ratio of the magnetic moment to the orbital angular moment due to the orbital angular moment is equal to e/(2m).
The orbital angular momentum of an electron of an atom depends, as shown on the quantum number l and is given by the expression √l (l + l) h. we can express the magnetic moment due to the orbital motion of the electrons as:
μM = h√l (l + l) (e/2m) = eh/2m √l (l + l)
The constant factor in this equation provides a convenient unit in which to express the magnetic moment of atoms and molecules, and one therefore introduces the symbol μB = eh/2m
With this unit, the orbital magnetic moment of an electron of an atom is given by:
μM = μB √ l (l + l)
When a similar approach is extended to molecules and ions, rather than free atoms, it would seem reasonable to except the orbital motions of the electrons to contribute a magnetic moment of the order of an electronic Bohr magneton. This expectation is not generally borne out, and it appears that the orbitals motions of the electrons in a polyatomic system are tried into the nuclear configuration of the magnetic field and are therefore ineffective. Even for single atom ion is solution, the interaction of the orbitals of the ion with the solvating molecules is apparently sufficient to prevent orbitals from being oriented so that their magnetic moment contributes in the directions of the field. Thus the orbital magnetic moment contribution to the magnetic susceptibility is generally quite small.
We must look to electron spin to explain the larger part of the magnetic moment of those molecules and ions which have magnetic moments. The association of a spin angular momentum of √S (S+ 1)h, where s has the value of ½, leads, according to the equation above to the expectation of a spin magnetic moment expected on the basis of the ratio of the magnetic moment to the angular momentum implied by the equation. Therefore, for the spin magnetic moment due to the electron spin, expressed in terms of the spin quantum number S of an atom or molecule, we have:
μM = 2μB √S (S + 1)
For one, two, three..... unpaired electrons, the spin, angular momentum quantum number S is ½, 2/2, 3/2... with the above equation and the assumption that the magnetic polarizability μM contribution has been taken care of and that the orbital contribution to x is negligible, the magnetic susceptibility of the eq, is related to the total electron spin by the relation:
X = (4N µ0 µ2B/3kT) S (S + 1) = (6.29 × 10-6/T) S (S +1)
At 25° C this expression gives:
X = (2.11 × 10-8) S (S + 1) 25° C