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Sulphur is a diamagnetic element, meaning that it has no unpaired electrons. But its electronic configuration for valence shell is just like oxygen because they are in the same group i.e ns2, np4.

My question is that why is sulphur diamagnetic. If there are 4 electrons in p-orbital of its valence shell then shouldn't it have 2 unpaired electrons?

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You seem to be confusing the elements with the pure simple substances of the elements. In atomic terms, both the oxygen atom and the sulphur atom are paramagnetic, since they contain two unpaired electrons in the ground state. However, the substances oxygen ($\ce{O2}$) and sulphur (generally intended as $\ce{S8}$) are not the same as the atoms. In the simple picture of Lewis structures, both substances have all atoms forming two bonds, completing the octet rule and therefore without unpaired electrons. This would imply that both are diamagnetic. However, simple valence bond theory is inadequate to explain the bonding in oxygen. Simple molecular orbital theory suggests that there are in fact two unpaired electrons in the $\ce{O2}$ molecule. This gives rise to paramagnetism in the oxygen molecule, also.

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So, oxygen is paramagnetic weather it is in atomic state or molecular state. But in Sulphur is paramagnetic only in atomic state. Right? – Rafique Apr 21 '13 at 9:19
Correct. One must be take care to be clear when referencing the element itself or one of its allotropes. – Nicolau Saker Neto Apr 21 '13 at 14:21

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