I notice that out of all of group 16 oxygen is the only element that can form a double bond. Why is that because shouldn't all of group 16 be able to form a double bond and be diatomic because they all have six valence electrons?
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Consider $\ce{CS2}$ – permeakra Dec 28 '15 at 06:08
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I removed your picture - it was terrible and misleading. Also tags were improper. – Mithoron Dec 28 '15 at 19:05
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@permeakra Why isn't it diatomic then? – Ruchir Baronia Jan 03 '16 at 02:58
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@Mithoron Thanks, but then why aren't the double bond molecules also diatomic – Ruchir Baronia Jan 03 '16 at 02:58
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It isn't proper place to describe bonds or multiatomic molecules, you should know basics when asking such question – Mithoron Jan 03 '16 at 13:58
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I notice that out of all of group 16 oxygen is the only element that can form a double bond.
This premise is not true. Sulfur in fact forms double bonds in many organic compounds. Selenium and tellurium both form double bonds with triphenylphosphine as triphenylphosphineselenide and triphenylphosphinetelluride.
The larger chalcogenides do have difficulty in forming double bonds in organic molecules due to lower bond strength, higher reactivity, preference of molecules for oxygen, and steric bulk. They can however indeed form double bonds; it's just not as common.
A.K.
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1I suppose in the gas phase they could be. In the solid phase though Sulfer is a 8 membered ring and selenium and tellurium are crystalline, but may for polymeric glasses. – A.K. Jan 03 '16 at 04:08
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Sulphur atoms are larger then oxygen atoms. This means that 3p orbitals in sulphur can't overlap efficiently or enough to create a pi bond (which is the "second" bond in the double bond) between two sulphur atoms. Oxygen, however, is small enough that its 2p orbitals can overlap to form a stable pi bond. The size of the atoms only increase down group 16, and so anything larger than oxygen will not form stable double bonds.
Georgeos Hardo
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When you say "second bond" what do you mean, because there is only one bond right? – Ruchir Baronia Dec 28 '15 at 06:09
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"Larger" alone does not quite answer the question. If S atoms were just like enlarged O atoms, then $\ce{S2}$ would be just like enlarged $\ce{O2}$ and would feel just as well. This is not the case, though. – Ivan Neretin Dec 28 '15 at 08:14
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@RuchirBaronia A double bond is made up of a sigma bond and a pi bond. The pi bond results from overlap of p orbitals. Because p orbital overlap would be poor in sulphur, the pi bond cannot form, and so the single sigma bond is all that remains. – Georgeos Hardo Dec 28 '15 at 15:36
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But each atom shares only one electron in a double bond, right? So you're saying that the size of the atom impacts whether or not that one atom will be shared? – Ruchir Baronia Dec 28 '15 at 17:24
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@RuchirBaronia if it is localised in a p orbital then yes it can. Overlapping p orbitals are what cause Pi bonds, – Georgeos Hardo Dec 28 '15 at 17:28
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@Ruchir Baronia a double bond is made up of two bonds, as Georgeos Hardo said. So, each atom shares two electrons – hence, a total of four electrons are shared in a double bond. The size of the bonding atom can inhibit the ability to form one of the two bonds making up a double bond, i.e. the π bond. Thus, they only form single bonds. – Corundum Dec 31 '15 at 08:59