we know from our experiments that, the element attained octet ( Noble gas configuration) is always stable. but why 8 is the stability no....why not any thing else..what is the physical/ electric reason for that.

Most of these answers seem to be effectively saying, "8 electrons form a stable shell because the valence shell has 8 slots and a full shell is the most stable configuration," but I think the real question being asked is, "why does the valence shell have 8 slots in the first place?" Although the presence of 8 slots is often taught as if it were a purely empirical observation (probably because, historically, it was for a long time), it can actually be derived from the axioms of quantum mechanics and the geometry of 3-dimensional space. Below is an overview of some of the relevant features of the derivation. If you solve Schroedinger's equation for an electron in the potential well of a nucleus, you can find every possible bound state that the electron could possibly have. Finding the solution involves making sure that the electron's wavefunction repeats itself at intervals that allow it to "fit" around the nucleus. By analogy, consider a wave on a length of string. It has peaks and troughs at a certain period. The goal is to find which periods (wavelengths) will "fit" on the string, such that if you bring the two ends of the string together to form a loop around the nucleus, there will be no discontinuities in the wave. The resulting waves are called harmonics: A loop is sufficient to wrap around a point in a 2 dimensional plane, but electrons and atoms exist in 3 dimensions. Hence, we have to make sure the electron's wavefunction "fits" in both of the two angular degrees of freedom around a point in 3 dimensions. The waves that "fit" around a central point in 3 dimensions are called spherical harmonics: Since there are two angular degrees of freedom around a point in 3 dimensions, it takes two numbers to identify a given solution that fits. As a result, there are a total of three numbers needed to fully identify a wavefunction solution: n: quantization in energy l (lowercase L): quantization in first angular degree of freedom m: quantization in second angular degree of freedom Purely from the math of the solution and the rules of geometry, one finds that the following rules must be satisfied in order for the solution to fit: n must be a positive integer (1, 2, 3, ...) for a given n, l can be any integer from 0 to n-1 inclusive for a given l, m can be any integer from -l to l inclusive In addition, the properties of the electron dictate that: for a given n, l, and m, there are two possible electron states: one for spin up and one for spin down Putting all these rules together, we can count the number of "slots" available for electrons to fill up a given shell. For the lowest energy shell (n=1), l can be any integer from 0 to 0. That is, it can only be 0. In turn, for l=0, m can be only -0 to 0, meaning it can also only be 0. This means there is only one allowed combination with n=1, namely the one where n=1, l=0, and m=0. Since electrons can be spin up or spin down, this means there are 2 total slots available. This is why hydrogen and helium are stable with 2 electrons -- because those 2 electrons fill the n=1 shell. For the next energy shell (n=2), l can be 0 to 1. For l=0, m can be only 0. For l=1, m can be -1, 0, or 1. This forms the following set of allowed values: n=2, l=0, m=0 n=2, l=1, m=-1 n=2, l=1, m=0 n=2, l=1, m=1 Since there are 4 allowed values, the shell has 8 slots for electrons (accounting for spin). read less

No its not necessary that every element needs eight electron in their outermost shell to be stable,actually it follows by almost all s- and p-block elements and both the orbitals can have maximum number of eight electron(s+p=8 electrons) so that they follow it. This octet rule does not work for d- and f-block elements as they show various stable oxidation states. read less