SHELL, SUB-SHELL AND ORBITAL 1. Shell is a group of orbitals with - TopicsExpress



          

SHELL, SUB-SHELL AND ORBITAL 1. Shell is a group of orbitals with the same value of n. The size of shell is determined by its P.Q.No. of n. The maximum number of electrons can accommodated by a shell is given by the formula 2n2. 2. Sub – Shell is a group of orbitals with the same value of (l ) (Azimuthal Q.No.) is called sub shell. All the orbitals in the sub shell have same energy i.e. they are degenerated. The number of orbitals in any sub shell tells us the value of its magnetic Q.No. One orbital can accommodate maximum 2 . RULES FOR DISTRIBUTION OF ELECTRONS IN ORBITALS The method of distribution of electrons in sub shells and orbitals is called electronic configuration. There are three rules for electronic configuration. They are (1) Auf – Bau Principle:- It states that electrons enter the orbital in the order of increasing energy. The energy of the electron (orbital) is given by the value (n + l ). If two sub shells have the same value of (n + ) value then the sub shell with lower n value would have lower energy e.g. for S = 0, P = 1, d = 2, f = 3 Energy inversing order is 1S < 2S < 2P < 3S < 3P < 4S < 3d < 4P. 2. Pauli Exclusion Principle:- According to this principle (1) An orbital can hold a maximum of two electrons. (2) When there are two electrons in an orbital, they must have opposite spins. OR No two electrons in an orbital can have the same set of four Q.No. OR No two in an orbital have the same spins e.g. two electrons in an orbital 2p has quantum No. For 1st = n = 2, = 1, m = + 1 and ms = + ½ For 2nd = n = 2, = 1, m = + 1 but ms = – ½ Since three Q.No. are alike but spin Q. No. is different. 3. Hund’s Rule:- According to this rule if orbitals of the same energy are available. Electrons will distribute in them in such away to give maximum number of unpaired electrons. These electrons with the same energy (degenerate) have the same spins. e.g. E.C of C6 = 1S2, 2S2, 2Px2 Py Pz (Wrong) 1S2, 2S2, 2Px1 Py1 Pz (Correct) For N7 = 1S = 2S2, 2Px2, 2Py1, 2P2 (Wrong) 1S2, 2S2, 2Px1, 2Py1, 2Pz1 (Correct)
Posted on: Thu, 11 Dec 2014 06:59:33 +0000

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