Calculate the total spin and magnitic moment for atoms having atomic numbers `7`, `24`, `34` and `36`.

Explanation

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))`.