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

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

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

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

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