[gitesh]

Teachers always say that atoms with octate or duplate completed are stable. But what is the reason behind it..but why only eight or two outermost electronic configuration is stable?

[mjc123]

Haven't any of your teachers explained it? Are they doing their job? Or does each assume one of the others has explained it?
What do you think? Any obvious reasons why that may be more stable than, say, with one electron more or less?

[Arkcon]

Fir starters, why are some 8, and some 2?  Is there a difference between the two?  That won't completely answer your question, but its a start.  And it prevents you from considering them equivalent.

[AdiDex]

Its mine question too..
I think that it is just for minimize their energy.
They have fully filled orbitals with all paired electron.
So they do not have electrons for reaction .(you can say that for reaction initiation they need a high ammount of energy )

But i think it is not the whole story as their (Noble gases) energy is also low . So how this happens
.as electrons are just getting filled into the atoms.
So it should increase their energy.

[mjc123]

OK, let's think about this. Why do atoms (main group atoms, anyway - transition metals and d/f orbitals are another kettle of fish) often (not always!) adopt in their compounds a quasi-noble-gas configuration of s²d⁶? (Or 1s² in the case of H^-, He, Li⁺ etc.)
Let's think about ionic compounds, and compare, say, NaF and MgO. Here are the first 3 ionisation energies of Na and Mg (kJ/mol):
Na: 496, 4562, 6912
Mg: 738, 1451, 7733
And some electron affinities:
O: -142, +844
F: -328
The 3s electrons (one for Na, two for Mg) are ionised relatively easily, but there is a big jump to the 2p electrons, which are much more strongly held.
Now suppose you have ionic compounds Na⁺F^- and Mg⁺O^-. And it occurs to you that if you transferred an electron from the cation to the anion, giving divalent ions, you would get four times the lattice energy, considerably stabilising your compound. You would have to expend some energy removing the second electron from the cation and forcing it onto the anion (note that EA2 of O is positive, meaning you have to put in energy to add the second electron). For Mg and O, that cost is ca. 2300 kJ/mol. The lattice energy of Mg²⁺O^2- is 3795 kJ/mol, so if we assume it would be ca. 900 kJ/mol for Mg⁺O^-, you can see that the extra lattice energy more than compensates for the energy used in making the divalent ions. For NaF, however, (LE 910 kJ/mol), the energy cost is over 4500 kJ/mol (I can find no EA2 for F, presumably because F^2- doesn't occur, because it can't be stabilised in a compound), so even increasing the LE by a factor of 4 won't recover this energy. The same argument will show that formation of Mg³⁺O^3- would be energetically unfavourable.
(In addition, ions with a 3s electron (Mg⁺, F^2-, O^3-) would be considerably larger than those with a neon configuration, so the lattice energy of Na²⁺F^2- would be less than 4 times Na⁺F^-, and Mg⁺O^- less than a quarter, and Mg³⁺O^3- less than 9/4, that of Mg²⁺O^2-.)
So we see that energetic considerations dictate that the noble gas configuration is the logical place to stop. Removing 2p electrons from a cation, or adding 3s electrons to an anion, doesn't make sense energetically, because of the large energy gap between 2p and 3s. Note that this applies to the overall energetics of compound formation. In isolation, a noble gas configuration is not necessarily particularly stable. We have seen that EA2 of O is positive - an isolated O^2- would like to get rid of an electron. And an isolated Na atom is stable relative to Na⁺ and a free electron.
Similar arguments could be made for covalent bonding, e.g. a 3-valent carbon could gain extra bonding energy by forming a 4th bond, but to form a 5th bond would require utilising high-energy atomic orbitals, and is unfavourable. (See e.g. this thread http://www.chemicalforums.com/index.php?topic=77716.0.)

[AdiDex]

Bro thanks for your reply.
But my question is for atom not for molecule.
Can you elaborate for atoms

[mjc123]

I assume you're talking about the noble gases themselves. As I explained, for non-noble-gas atoms in isolation, a noble-gas configuration is not necessarily the most stable. There is nothing magical about an octet. (Apart from Schubert's.)
We can extend the arguments I made before. The IE of Ne (ca. 2000 kJ/mol) is much higher than that of Na because you are removing a 2p electron, not 3s. If we assume Ne⁺F^- would have a similar lattice energy to NaF, the energetics are against it. Ne^- doesn't form for the same reasons as F^2-.
For covalent bonding, again the argument is that to have unpaired electrons available for bonding, you need to promote electrons to much higher energy orbitals. The higher noble gases, such as Xe, where the energy gap is less, can do this, and form some compounds. As to why, say, Ar is reluctant to form expanded-octet compounds when P, S and Cl do, I would say the promotion energy increases with nuclear charge, and the process becomes less favourable in the order P-S-Cl-Ar.
In short, the s²p⁶ configuration marks the transition point from one energy level (or set of levels) and another significantly higher in energy. It is, as I said, a logical stopping point in varying the electron configurations of atoms in bonding. (And for noble gases of course, that means not bonding.)