1.

Calculate 'Q' for last electron of Ga. where Q=n+l+ maximum possible value of 'm'.

Answer»

SOLUTION :`Ga(31) 1s^(2)2s^(2)2p^(6)3S^(2)3p^(6)4s^(2)3D^(10)4P^(1)`
n=4
l=1 n+1+m=6
m=1


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