A) 3, 2, 1, \[\frac{1}{2}\]
B) 4, 2, - 1, \[\frac{1}{2}\]
C) 4, 1, 0, \[-\frac{1}{2}\]
D) 5, 0, 0, \[\frac{1}{2}\]
Correct Answer: B
Solution :
Key Idea: First find the orbitals from the given values of n and \[l\]. Then find .their energies according to \[n+l\] rule. 3, 2, 1, 1/2 means 3 d (\[\because l=2\] means d) 4, 2, - 1,1/2 means 4 d 4, 1, 0, -1/2 means 4 p (\[\because l=1\] means p) 5, 0, 0, 1/2 means 5 s (\[\because l=0\] means s) \[n+l\] for \[3d=3+2=5\] \[n+l\] for \[4d=4+2=6\] \[n+l\] for \[4p=4+1=5\] \[n+l\] for \[5s=5+0=5\] \[\because \] Greater the value of \[n+l\], more will be energy. \[\therefore \] 4d orbital will have highest energy. \[4d>5s>4p>3d\]
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