no. of electrons of sodium with n+l value 3 are....

Dear Student,

Sodium = 11 electron and shell configuration = 2,8,1 and in terms of n+1 is
n = 0 +1 = 1 and electron in shell = 2
n = 1+1 = 2 and electron in shell = 2
n = 2+1 = 3 and electron in shell = 6 & 1 = 6+1 =7

so total in n+1=3 is 7 electrons.

Regards.
 

  • 0
Na(11) = 1s2 , 2s2 , 2p6 , 3s1
Here,
n                  l              subshell     (n+l)        no of electons
1                 0                  s                1                    2

2                 0                 s                  2                     2
                   1                 p                  3                     6


3                  0               s                    3                     1
 
Therefore there are 7e- in Na having (n+l) = 3

HOPE IT HELPS)
  • -1
7 electrons
  • 1
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