Question

element 24 ground - and structer state for Configuration 1st this element is what about the elements 7,80 °F love, "No, mg,

Answer 1

structure for a element is written as 2s+1_LJ

where s=spin quantum number, it is either non-negative integer or zero only if number of electrons is even or half integer if number of electron is odd. spin multiplicityy=2S+1

L=total orbital angular momentum quantum number(a non-negative integer)

L= 0,1,2,3 respectively for S,P,D ,F and so on

J=Total angular momentum quantum number (Non -negative integer or half integer)=L+S,L+S-1,...........L-S

if orbital is less than half-filled then J=L-S

If more than half filled J=L+S

IF exactly halffilled or ful-filled then J=S

Structure is determined by valance orbital of element.

YFN = 15² 28² 283 hit L=1+0 1 = 0 so s S = 3/=3* Y2 C3 on paired 25 + 1 = 2x3 + 1 = 4 J: L+S 0 + 3/2 = 3/2 structure fore IN is 45 + $0= 15? 25? ap4 TL 1-2x1 + 1x01 1x1 52+0-11 so p 5 = 2*4- Ita unpacreede 25+ 1 = 2x1 + 1 = 3 J=12+5). 11-5] & 11 + 1 - 11-1 F2, 1,0 Due to Porebital is more than half filled soompound state team symbol, J = 'hts! so streecture for 8 is p +90 = 152 282 2p5 til t=ax! + 0*2+1x-1 - 2 1 - 1 So L=1 IS = 1/2 (Due to one unpaired el in 2 + 1 = 2 r 1 at :2 present) JE L&S: 1 + ² = 3% (More than half felled) " Streucture 120 l My Note Book Scanned with

Date:_//_Page nos yote = 13² 282 apo LLL L = 2x1 + 0x2+2x- = 2-2 = 0 so s S = = as t 1 = 1 J: Its : 0+0 EO Structeerce: As, Y No 18² 28² 286 35' 1=0 so s S = %= 25 + 1 = 2x1 + 1 = 2 I J = 0+1/2 = 1/2 i structure: 2 - 12 M = 15² 25² 2p 35² 12 - L=0 So S S=0 = as t1 = 2x0+1 = 1 JFO = Lts - Structure to > 18 Ar & 16² 26² 206 35 36 AL 14 2 . = 2 x 1+02+2x-1 O so s SEO (No unpaired eo is present. 25 + 1 = 1 J= 0 to ²0 . Structure list - 19 K = 15² 262 266 35² 3po 45' Do L L = 0 = S - S=1/2 28 t 1 = 2x1 + 1 = 2 J = 1/2 = Its so structure S. Ny Note Bock with CamScanner

Date: +2002 = 152 252 2p6 352 3p6 43 11 10 TS S=0 25+ 1 = 2 x 0 + 1 = 1 JE 0 streecturce for this is o -