InterviewSolution
Saved Bookmarks
| 1. |
Calculate the total spin and magnitic moment for atoms having atomic numbers `7`, `24`, `34` and `36`. |
|
Answer» The electronic configruation are : ` ._7 N : 1s^2 , 2 s^2 2 p^3` unpaired electron `=3` ` ._(24) Ce : 1 s ^2 , 2s^ 2 2p^6 , 3 p^7 3d^5 . 4s^1` unpaired electron `= 6` ` ._( 34) Se : 1 s^2 , 2a^(2)2p^6 , 2p^2 3p^(6)3d^5 , 4s^1` unpaired electron =6 ` ._(24) : 1a^2 , 2a^(2) 2p^6 , 3s^2 3p^6 3d^(10) , 4s^2 4p^4` unpaired electron `=2` ` ._(36) Kr : 1s^2 , 2s^2 2p^6 , 2s^2 sp^6 3d^(10) , 4s^2 4p^6` unpaired electron `= ` ` :.` Total spin for an atom ` = += 1//2 ` no. of unpairred electron For`._7N`, it is `= +- 3 //2 ` For` ._(24)Cr`, it is `= +- 3` For` ._(34)Se`, it is `= +- 1` For `._(36)` Kr, it is `=0` Also magnetic moment ` = sqrt ([n (n+2)])` Bohr magnetion For `._7N`, it is `= sqrt (( 15))` For` ._(24)` Cr, it is` = sqrt (( 48))` For `._(34 )` Se it is `= sqrt 8` For `._(36)` Kr , it is `= sqrt ((0))`. |
|