1.

The quantum number of sixelectrons are givenbelow,Arrangethemin orderof increasingenergies . If anyofthesecombinationhashave thesameenergylists:(1) n-4,l=2 , m_(1)= 2m_(1) = (1)/(2) (2)n=3, l = 2, m_(1)= 1 ,m_(s) = (1)/(2) (3 ) n=4, l = 1 , m_(1) = =0 m_(s ) = (1)/(2) (4 )n=3: l= 2 m_(1)= - 2 m_(s ) = - (1)/(2) (5 )n=3, l = 1,m_(1)= - 1m_(s ) = (1)/(2) (6)n=4, l = 1, m_(1)= 0m_(s ) = (1)/(2)

Answer»

SOLUTION :(1)L =2so dand n= 4 so4dsubshell
(2)l=2so dand n=3 so3dsubshell
(3 )l =1so p andn=4so 4p
(5 )l =1 sop andn= 3 so 3p
(6)l =1so p andn=4so 4p
This4d3d , 3d4parrange as perenergythan
(5) `3p(4 ) 3d = 2 (3d)lt (6) 4p= (3) 4p lt (1)4D (5 )lt (4) = (2)lt (6)= (3) lt (1) to`energyincreases(2) ,(3) (4) and(6) has sameenergy .


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