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My teacher told yes that salt is paramagnetic and it is attracted by magnet slightly. But when I tried that at home it did not work.

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  • $\begingroup$ your teacher was wrong in this aspect $\endgroup$ – agha rehan abbas Jul 6 '14 at 18:25
  • $\begingroup$ could it be that the teacher said "some salts"? What is s/he a teacher of and at what level? $\endgroup$ – Silvio Levy Jul 27 '14 at 16:08
  • $\begingroup$ No actually he specified nacl, he is my chemistry teacher for jee coaching $\endgroup$ – Vidya Sagar V Jul 27 '14 at 16:20
  • $\begingroup$ Make its orbitals' diagram :) $\endgroup$ – ParaH2 Sep 17 '15 at 20:19
  • $\begingroup$ Nit picking here, but technically, you could generate a smaller magnetic moment due to nuclear magnetic moments, but I think it still is net zero in this example. $\endgroup$ – Zhe Jan 15 '19 at 23:34
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Table salt (NaCl) is diamagnetic. For paramagnetism, one needs free unpaired spins. The NaCl lattice is built up from closed shell cations and anion, there is no such unpaired spin pure NaCl.

Also: experimentally it is difficult to measure paramagnetism, and you generally cannot just try it with a magnet and observe it without some tricks.

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    $\begingroup$ What tricks are you referring to? $\endgroup$ – Dissenter Jul 6 '14 at 17:25
  • $\begingroup$ I would agree with it being difficult to "measure" paramagnetism in the sense of quantifying the strength of the resulting magnetization without specialized equipment (i.e., a magnetometer). However, I think it is possible to at least qualitatively observe and relatively rank the magnetization for many materials. Small samples of various paramagnetic salts, for example, will visibly respond to even small permanent magnets. On the other hand, diamagnetism, to my admittedly limited knowledge, is extremely difficult to observe without very powerful magnets. $\endgroup$ – Greg E. Jul 6 '14 at 19:12
  • $\begingroup$ In my experience, small samples of insulating materials (i.e. almost all the salts) can strongly influenced by electrostatic forces, wetting and other artifacts. Disclaimer: I never actually tried to do play with paramagnetic materials and small magnets, but worked with a classic Faraday balance for years on transition metal salts, and oh, boy.. Said balance used a huge, old fashioned electromagnet, and measurement at liq. N2 temperature, yet once could make gross mistakes even under those circumstances. $\endgroup$ – Greg Jul 7 '14 at 0:16
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NaCl is diamagentic. Every material is at least diamagnetic. The diamagnetic influences are very weak, so you can't see them.

Diamagentism: if you have a field H working on a materia, it will change the condition of the particles in the atom of the materia, to create a magentic moment, which is opposite to H. Sigma of all these induces fields is called B. If you have an inhomogene field you have to use a lot of work to create a higher field force of the diamagnet. Genereally the diamagnet want to have a low field force.

The quantumchemistry behind it is easy: the spin of electrons have a magentic moment and creates a field, but because of the Pauli Principle and the thermic move, you wouldn't see it on a macroscopi scale. First by a field outside you would see an induction of the magnetic dipole on an atomic basis.

So if your field is stron enough, you could actually see NaCL moving. But you need a very very very strong field to do it. Andre Geim for example used the diamagnetism of water to levitate frogs.

Paramagnetism: if you take have a magnetic field from outside, the microscopic magnetic moments will will all show in the same direction. But because of the thermic fluctuation the magnetic field would collaps as bald as the magnetic field collaps. The strongness of the field is calculated by:

$M= \chi H$,

where M is the magnetisation of the material, H is the magnetic field and $\chi$ is the suszeptibility.

Ferromagnetism: a material that will be attracted to a pol or is a magnetic itself. It's magnetic because of it's elementarmagnets inside it. The magnetic flux is: $\vec B = \mu \vec H = \mu_0 (\vec H + \chi \vec H) = \mu_0 (\vec H + \vec M)$.

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