Question

What is the maximum spin multiplicity for 4d-orbital?
(a).4
(b).6
(c).5
(d).0

Answer 1

Hint:Spin multiplicity is the number of probable orientations of the spin angular momentum corresponding to a given total spin quantum number (S), for the same spatial electronic wave-function’. The number possible orientation is measured as 2S+1.

Complete step by step solution:
> The formula used for calculating spin multiplicity is 2S+1, Where, S= 2xmaximum number of unpaired electrons in 4d orbital x\[\dfrac{1}{2}\].
4d orbital can have 5 unpaired electrons. Therefore
S= 5x\[\dfrac{1}{2}\]
Spin multiplicity 2S+1=(2x5x\[\dfrac{1}{2}\]) +1= 6
Hence the maximum oxidation spin multiplicity of 4d-orbital is 6. The correct option is (b).
> For a multi-electron atom, there are a number of configurations of spin (S) and orbital (L) angular momentum quantum numbers which give the same value of total angular momentum (J). This is known as spin-orbit coupling, L-S coupling, or Russell-Saunders coupling and is denoted by the term symbol\[{}^{2s+1}{{L}_{J}}\]. Provided L can be greater than equal to S, there are 2S+1 values of spin possible for corresponding values of L that result in the same value of J. So there are (2S+1)(2L+1) degenerate states for that atom. The factor 2S+1 is called the spin multiplicity for that configuration. For example if S=3/2 and L=2, there is a spin multiplicity factor of 4, so J can be in one of 20 possible microstates, after including all possible spin and orbital values.

Note: The maximum number of unpaired electrons in a d orbital will be always 5 electrons. Because each electron gets occupied in \[{{d}_{xy}},{{d}_{yz}},{{d}_{xz}},{{d}_{{{x}^{2}}-{{y}^{2}}}}\]$ and$ \[{{d}_{{{z}^{2}}}}\]. Then only we get maximum spin multiplicity.