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