which element exhibits the highest oxidtionstate?
491963
it's either copper or iron.  i don't remem ber exactly which one.
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Hydrogen\nOxygen\nCarbon\nNitrogen\nPhosphorus\nSulphur
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Osmium is the densest element. 22.6 g/ml\nhttp://www.science.uwaterloo.ca/~cchieh/cact/c120/qzelemnt.html
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Carbon (C), Hydrogen (H), and Oxygen (O)
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It´s 10. Neon. \nIt has two energy levels, filled with 2 and 8 electrons respectively.
68791
Reactivity in goup 1 increses down the group from Lithium to Sodium to Potassium. Therefore, Sodium would the closest. They are all group 1 metals, just in different periods. All have one vacant electron which makes them very reactive.
149295
carbon, hydrogen and oxygen.
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Rhenium has the highest boiling point of the elements - Boiling point [/K]: 5869 [or 5596 °C (10105 °F)] (liquid range: 2410 K)
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H\n |    \nC=O:  (You need 2 lone pairs on Oxygen)\n |   \nH\n\nIt doesnt get much better than that :)
6677
Group I elements almost always have an oxidation number of +1, so the K (potassium) has a +1.  Unless they're in peroxide compounds, which is not the case here, oxygen usually has an oxidation number of -2.  The hardest one is the Mn.  We know that the whole compound has to have an oxidation number of 0, so let's do some math.  There is one +1, and there are 4 oxygens, each with -2.  \n+1 + ? + 4*(-2)=0\nThe answer is +7.  Therefore, Mn must have an oxidation number of +7. \nK=+1, Mn=+7, O=-2
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Boron, Carbon, Nitrogen, Oxygen
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There's nothing surprising about the normal Group oxidation state of +4.\n\nAll of the elements in the group have the outer electronic structure ns2npx1npy1, where n varies from 2 (for carbon) to 6 (for lead). The oxidation state of +4 is where all these outer electrons are directly involved in the bonding.\n\nAs you get closer to the bottom of the Group, there is an increasing tendency for the s2 pair not to be used in the bonding. This is often known as the inert pair effect - and is dominant in lead chemistry.\n\nHowever, just giving it a name like "inert pair effect" explains nothing. You need to look at two different explanations depending on whether you are talking about the formation of ionic or covalent bonds.
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MgBr2, SrBr2, CsBr\n\nYou can determine whether one has a greater ionic character than the other by looking at the electronegativity difference. The higher that is, the more ionic it is. In general, electronegativity is highest in the top right corner of the periodic table (excluding noble gases), and lowest in the bottom left. \n\nSo, to find the one with the highest ionic character, just choose the element that is farthest away from Br, on the table.
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a) [O 2-]\nb) [ I - ]\nc) [Na+]\nd) [Al 3+]\ne) [Ni 2+]\nf) [Mg 2+]
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Hydrogen, Oxygen, Nitrogen etc. are some elements\nCarbon tetrachloride, Methane, Ammonia etc are some such compounds.
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oxidation state for oxygen is usually -2 but it sumtimes have -1
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Ionization energy (1st ionization) increase as you go up and right on the periodic table, so fluorine has the highest.  Similar trends are noted in electronegativity.  Fluorine is the element with the highest electronegativity.\n\nThis trend in electronegativity is seen because going down the group you see the successive atoms have more electron levels which puts the valence electrons farther from the nucleus and easier to remove.  Going to the left on the table you are not adding any more electron levels, but the nuclei are becoming increasingly larger and therefore creating a stronger attraction for the valence electrons.