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How do you determine this when given a list of elements? Please help!!!

2006-10-19 16:34:43 · 6 answers · asked by SuperCee 2 in Science & Mathematics Chemistry

6 answers

I think the preceding answers may mislead you, although they do provide some useful facts.

Here are the key concepts:
The fewer electrons in the outer shell, the more likely it is that the atom will lose its outer shell electrons. And those with only ONE electron in the outer shell are neatly lined up in the first column (Group I) of the periodic table. (Forget hydrogen, because it's a special case with only one electron IN TOTAL. But the other Group I elements are extremely reactive: Lithium, Sodium, Potassium, etc.)

And the atoms with only TWO electrons in the outer shell are lined up in the next column (Group II): Beryllium, Magnesium, Calcium, etc.

Now, the farther you go down the column, the more ACTIVE (likely to lose electrons) the atom is. Francium, at the bottom left, is the most reactive of all. That's because the single outer electron is farther from the nucleus and likely to be pulled off.

Final note:
The above are general concepts. What you really need to know for sure which element is more likely than which other element to give up an electron is a table ranking the atoms by how active they are. (I think the measure used here is called electronegativity.) Francium would be at one end of the table (likely to lose an electron), and Fluorine is at the other end (likely to gain an electron).

2006-10-19 17:00:06 · answer #1 · answered by actuator 5 · 0 0

It depends on the number of valence electrons. Those are the ones on the outer most energy level. The first energy level can hold 2 electrons, second can hold 8 electrons, and third can hold 8 electrons. Not sure about the levels past that. Look at the atomic number. That is how many electrons it has. If it has 6 valence electons then it is more likly to gain 2 electrons than give up 6. If it has 1 valence electron then it is more likly to lose that one. Hope this made sense.

2006-10-19 16:39:47 · answer #2 · answered by Luekas 4 · 0 0

I believe this is called "ionization affinity". The further away from the nucleus, the easier an electron is removed. This means you want to look to the bottom of the periodic table. Electrons experience a tighter pull moving from left to right across the periodic table (see the concept of effective nuclear charge).
Putting these together, you want the most lower left atom in a list of elements.

2006-10-19 16:40:01 · answer #3 · answered by davisoldham 5 · 0 0

Anions are adverse ions Cations are valuable ions So a and c are same b and d are same If atoms lose electrons, they're valuable. a million thank you to think of of that's they lack adverse so as that they are valuable vice versa in short 1st Q) Ans: the two b and d are spectacular 2nd Q) Ans: the two a and c are spectacular

2016-11-24 19:10:07 · answer #4 · answered by Anonymous · 0 0

the oxidising agent is most likely to lose electrons or just choose the element with the least valency(obviously it won't be a metal)

2006-10-19 16:42:26 · answer #5 · answered by Logan 1 · 0 0

mostly atoms that contain negative ions(such as sodium chloride)

2006-10-19 16:39:41 · answer #6 · answered by yo-yo 3 · 0 0

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