Q.

The allowable combinations of quantum numbers for each of the electron in 4s, 3p, 5d orbitals respectively is:

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a

n = 4, l = 0, ml = +l; n = 3, l = 2, ml = 1; n = 5, l = 3, ml = 0

b

n = 4, l = 0, ml = 0; n = 3, l = 1, ml = 0; n = 5, l = 2, ml = -1

c

n = 4, l = 0, ml = 0; n = 3, l = 2, ml = -1; n = 5, l = 3, ml = -2

d

n = 4, l = 0, ml = 0; n = 3,  l= 0, ml = 0; n = 5, l = 1, ml = 0

answer is C.

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Detailed Solution

  • For 4s n=4,l=0 and m=0
  • For 3p, n=3,l=1 and m=+1,0,-1
  • For 5d, n=5,l=2 and m=+2,+1,0,-1,-2
  • this thee we find in option 3
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