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2.28    (iii) Which of the following orbitals are possible?
                        1p, 2s, 2p and 3f

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1p is NOT possible because for n=1, the value of l is zero. (for \ p,l=1)

2s is possible because, when n=2,l=0,1(for \ s,l=0).

2p is possible because when n=2,l=0,1(for \ p,l=1).

3f is NOT possible because for n=3, the value of l=0,1,2(for \ f,l=3)

 

 

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Divya Prakash Singh

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