Francium, right on the bottom of the group one. If you remember size of the atoms increases as you go down a column and it decreases as you go across the table. The larger the atom, the easier it is to get rid of the valance electrons and easier it is to react with other atoms. Now if you look at the periodic table, francium has the largest radius therefore it is the most reactive element.
2006-10-29 03:14:41
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answer #1
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answered by smarties 6
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Most Reactive Element
2016-10-04 11:16:18
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answer #2
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answered by wilfrid 4
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This Site Might Help You.
RE:
what is the most reactive metal in the periodic table?
thx
2015-08-06 23:48:03
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answer #3
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answered by Reuven 1
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Alkali metals and metals in general are very reactive, some more than others, but most form compounds with other elements quite easily. Sodium (Na) and potassium (K), Lithium (Li) are some of the most reactive metals. Francium and Cesium are the most reactive elements in this group.
2006-10-29 03:14:20
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answer #4
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answered by BLANK 4
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While the answer, Francium, is probably theoretically correct, there has never been enough Francium created to be able to experimentally determine that.
So the most reactive metal that is available for testing is Cesium.
2006-10-29 03:37:11
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answer #5
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answered by Alan Turing 5
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Carson is the most reactive metal whereas flourine is the most reactive non metal
2015-02-13 16:07:00
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answer #6
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answered by Neeraj 1
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Francium. It is the most electropositive of all metals (or elements for that matter). This means that it has the largest ability to lose electrons
2006-10-29 03:13:47
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answer #7
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answered by Dr. J. 6
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Sodium(Na) and Potassium(K) are highly reactive.Na is stored in kerosene because it can react with air in room temperature and blast.The reactivity series can be given as:
Potassium
Sodium
Calcium
Magnesium
Aluminium
Zinc
Iron
Tin
Lead
Hydrogen
Copper
Mercury
Silver
Gold
2006-10-29 03:23:20
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answer #8
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answered by Mayank Sharma 2
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francium (group 1)
2006-10-29 10:01:02
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answer #9
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answered by Baby #1 born August 2009 6
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Fr
2006-10-29 04:14:13
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answer #10
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answered by Aaron F 1
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