Atomic term symbol

atomic term symbol

1. SAJAD AHMAD SHEER GUGRI Roll no. 16062120033 kashmir university [phone] DEPTT . OF CHEMISTRY UNIVERSITY OF KASHMIR Topic: atomic term symbol

2. TERM SYMBOL : - Term symbol of an electron arrangement reflects the total orbital angular momentum( L ), total spin angular momentum( S ) and total angular momentum( J ). In other words it represents energy of a particular electron arrangement.

3. MICROSTATES: The different ways in which the electrons can be rearranged in the orbitals of a subshell are called microstates of the confrigation. . No. of microstates= ( 3 l+2)! /q!(4l+2-q)! Where l =value of orbital q =no. of electrons

4. Term symbol for a particular atomic state term is written as; 2s+1 χ j L = 0 1 2 3 4…. X = S P D F G….. Where S total spin quantum no.2s+1 is spin multiplicity. L total orbital quantum number. J Total angular quantum number.

5. F0r d 1 system Total no. microstates = 10!/1!9! = 10 It means , the single electron can be arrange in this five sub orbitals in different 10 ways and representation will have different combination ML & MS value. +2 +1 0 -1 -2 ml = +2, +1, 0,-1 ,-2[for 1 e-] S = + ½ Hence 2S+1= 2 L =2 i;e X =D 2 D J =|L+S|,….|L-S| J =L-S {if orbital is lesser than half filled} J =L+S {if orbitals are more than half filled}

6. Determination of total wave function is micro states table for d 2 confrigation . example of free metal ion ( V 3+ ) Total no microstates = 45 First we determine M L VALUE M L = Ʃml i =ml 1 + ml 2 ml 1 = +2, +1, 0 ,-1, -2 ml 2 = +2 , +1, 0 -1, -2 +4………………….-4 Middle values also possible +2 +1 0 -1 -2

7. The range of total orbital quantum no. =+4, +3,+2,+1, 0 ,-1, -2,-3,-4 Next we determine total M S value M S =Ʃm si =m s1 +m s2 m s1 = +½ ,-½ m s2 = +½ ,-½ Total M S possible = +1 , 0 ,-1 By using these total spin quantum no. &total orbital quantum number. we can construct micro states table .

8. M L /M S +1 -1 4 (2 + ,2 - ) 3 (2 + ,1 + ) (2 + ,1 - ) (2 - ,1 + ) (2 - ,1 - ) 2 (2 + ,0 + ) (2 + ,0 - ) (1 + ,1 - )(2 - ,0 + )(1 - ,1 + ) (2 - ,0 - ) 1 (1 + ,0 + ) (2 + ,-1 + ) (2 + -1 - ) (1 + ,0 - ) (1-,0 + )(2-,-1 + ) (1 - ,0 - ) (2 - ,-1 - ) (1 + ,-1 + ) (2 + ,-2 + ) (2 + ,-2 - ) (1 + ,-1 - )(0 + ,0 - )(1 + ,1 - )(2 + ,-2 - ) (1 - ,-1 - )(2 - ,-2 - ) -1 (-1 + ,0 + )(-2 + ,1 + ) (-2 + ,1 - ) (-1 + ,0 - )(-1 - ,0 + ) (-2 - ,1 + ) (-1 - ,0 - ) (-2 - ,1 - ) -2 (-2 + ,0 + ) (-2 + ,0 - ) (-1 + ,-1 - )(-2 - ,0 + ) (-2 - ,0 - ) -3 (-2 + ,-1 + ) (-2 + ,-1 - ) (-2 - ,-1 + ) (-2 - ,-1 - ) -4 (-2 + ,-2 - ) Construction table of Pauli's allowed microstates (a + , a + ) and (a - ,a - ) are disallowed according to Pauli's exclusion principle

9. M L =±4 {1 times} M L =±3 {4 times} M L =±2 {5 times} M L =±1 {8 times} M L = 0 {9 times} AND M S = ±1 {10 times} M S = 0 {25 times}

10. a) first possibility L =4 &S =0 ml Term will be 1 G Total orbital degeneracy (2L+1)(2S+1) 2×4+1 = 9 J =|L+S|…|L-S| J =|4+0|…|4-0| =4 So there is no splitting of 1 G Hence complete term symbol = 1 G 4 b) L =3 & S = 1 Term will be 3 F +2 +1 -1 -2

11. Degeneracy (2L+1)(2S+1) = (2×3+1)(3) =21 J =|3+1| …|3-1| F term further splitting into three energy levels having J value 4, 3, 2 complete thee terms are: 3 F 4 , 3 F 3 , 3 F 2 c) L =2 & S =0 term will be 1 D J =|2+0|…..|2-0| = 2 1 D 2 there is no splitting degeneracy= (2×2+1)(1) =5 d ) L=1 & S=1 term will be 3 P J=|1+1|…..|1-1| = 2, 1 ,0 3 P 2 , 3 P 1 , 3 P degeneracy=( 2×1+1)(3) =9

12. e ) L =0 & S=0 term will be 1 S j = 0 complete term symbol 1 S degeneracy only one The terms derived for an electronic configuration have different energies. for example the d 2 configuation give rise five terms that are , 1 G , 3 F, 1 D , 3 P, 1 S these five terms have different energies. These terms represents different inter electronic repulsion .the ground state term (lowest energy) can be determined by using Hund’s rule:

13. 1) for a given configuation ;the ground state term(term that has lowest energy) is that which has highest spin multiplicity. The ground state; therefore , have highest number of unpaired electrons and they give maximum repulsion and high exchange energy. for example the ground state term for d 2 conf. is 3 F or 3 P. 2) if the two states have same multiplicity the state will be highest value of L will be the ground state. for d 2 ; 3 F & 3 P both same multiplicity but 3 F higher value of L .3F has lower energy. Hence 3F is the ground state 3) for a given electronic configuation , if multiplicity are same & vales of L same for the atomic states, then the state having lower j value will be lowest energy if the subshells is lesser than half filled , state having highest value of j will be lower energy if subshell is more than half filled.

14. Splitting of d 2 configuration