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# Magnetic Dipole Moment

Magnetic dipole moment of the nucleus arises due to the motion of charged particles. Orbitals and spin angular momentum of protons produce magnetic field within the nucleus. This field can be described in terms of resultant magnetic dipole moment located at the center of the nucleus.

In a covalent bond present between two different atoms (hetero atomic molecule) the bond pair of electrons are attracted towards atom with higher electronegativity.

Hence it acquires partial negative charge and the other atom acquires partial positive charge. Such a molecule with two different poles (one with partial positive charge and the other with partial negative charge) is called polar molecule. Such bonds are called polar bonds and behaves as a dipole with positive and negative charges separated by a distance (bond length).

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## Origin of Magnetic Dipole Moment

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Basically the magnetic materials are classified on the basis of presence of magnetic dipole moments in the materials. A charged particle with angular momentum always contributes to the permanent magnetic dipole moments. In general there are three important contributions to the angular moment of an atom.
1. Orbital magnetic dipole moment
2. Electron spin magnetic moment and
3. Nuclear spin magnetic moment

The angular momentum of an electron is called spin of the electrons. As electron is a charged particle, the spin of the electron produces magnetic dipole moment. In an atom with completely filled orbitals the contribution in spin magnetic moment is zero. In other words the spins of the electrons in in completely filled shells contribute more in the resultant spin magnetic moment.

The total magnetic dipole moment of an atom can be calculated by summing up all the above mentioned magnetic dipole moments in appropriate manner.

## Dipole Moment

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Dipole moment is a vector quantity and is represented by an arrow pointing towards more electronegative atom. In hetero diatomic molecules the dipole moment is never equal to zero. Greater the electronegative difference between the bonded atoms greater will be the dipole moment.

 H - F H - Cl H - Br HI 1.78D 1.07D 0.79D 0.38

## Shapes of the Molecules

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### 1. Triatomic molecules (AB2)

In triatomic molecules if $\mu$= 0, it will possess linear structure and $\mu$ not equal to zero, then it will have V shape.

Example: Molecules like CO2, BeF2 have zero dipole moment hence they have linear shape. In molecules like H2O, H2S the
$\mu$ $\neq$ 0 hence they have V shape. ### 2. Tetra atomic molecules (AB3)

Tetra atomic molecules with planar shape have $\mu$ = 0 and those with pyramidal shape do not have zero dipole moment ii.e. molecules with a lone pair of electrons. Example NH3, NF3.The dipole moment in ammonia is 1.47D while NF3 has 0.24D.This is because in NH3 the bond moments are in same direction as that of lone pair,in NF3 the bond moments are in different direction with that of lone pair of electron. ### 3. Penta atomic molecules (AB4)

Penta atomic molecules with tetrahedral shape ,with similar atoms bonded to central atom have zero dipole moment.

Example: CH4, CCl4 etc.If one of the atom is replaced by a different atom the dipole moment is not zero.

Example:CHCl3 ## Importance of Magnetic Dipole Moment

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1. Polar and non-polar molecules can be distinguished .Non polar molecules have zero dipole moment.
2. Shapes of molecules can be predicted
3. Ionic character in the molecule can be known

## Magnetic Dipole Moment Problems

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Below you could see problems

### Solved Examples

Question 1: A paramagnetic substance is composed of atoms with a magnetic dipole moment of 3.3$\mu$B. It is placed in a magnetic field of strength 5.2T. To what temperature must the substance be cooled so that the magnetic energy of each atom would be as large as the mean translational kinetic energy per atom?
Solution:

The magnetic energy of a dipole in an external field is U = -$\mu$.B, and the mean translational kinetic energy per atom is (3/2)kT.

These are equal in magnitude when the temperature is

T = $\frac{\mu B}{3/2\ k}$

$\frac{3.3\ X\ 9.27\ X\ 10^{-24}J/T\ X\ 5.2T}{1.5\ X\ 1.38\ X\ 10^{-23}j/k}$

= 7.7K

Question 2: Calculate M for a material in which
1. $\mu$R = 1.4 and H = 350ax A/M
2. $\mu$R = 6 and there are 2.5 x 1029 atoms/m3, each having a magnetic dipole moment of 2.8 x 10-30 ax A.m2

Solution:

1. Given $\mu$R = 1.4 and H = 350ax A/m
The magnetic susceptibility,
xm = $\mu$R = 1.4-1 = 0.4
The magnetization vector for the material,
M = xm H = 0.4 x 350ax = 140A/m

2. Here $\mu$R = 6, n = 2.5 x 1029 atoms/m3 magnetic dipole moment is m= 2.8 x 10-30 ax Am2
The magnetization, M = n x m = 2.5 x 1029 x 2.8 x 10-30ax = 0.7ax A/m

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