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In a multi - electron atom , which of t...

In a multi - electron atom , which of the following orbitals described by the three quantum numbers will have the same energy in the absence of magnetic and electric fields ?
(A) n=l, l=0,m=0 (B) n = 2, l =0 , m= 0
(C) n = 2 , l = 1 , m = 1 (D) n = 3, l =2 , m=1
(E) n=3,l=2,m=0.

A

(D) and (E)

B

(C) and (D)

C

(B) and (C)

D

(A) and (B)

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The correct Answer is:
To solve the question, we need to determine which of the given orbitals will have the same energy in a multi-electron atom in the absence of magnetic and electric fields. The key concept here is that orbitals with the same principal quantum number (n) and the same azimuthal quantum number (l) are degenerate, meaning they have the same energy. Let's analyze each option step by step: 1. **Option A: n = 1, l = 0, m = 0** - Here, n = 1 corresponds to the 1s orbital (l = 0). There is only one 1s orbital, so it does not have any degenerate orbitals. 2. **Option B: n = 2, l = 0, m = 0** - Here, n = 2 corresponds to the 2s orbital (l = 0). Similar to option A, there is only one 2s orbital, so it does not have any degenerate orbitals. 3. **Option C: n = 2, l = 1, m = 1** - Here, n = 2 and l = 1 corresponds to the 2p orbital. The 2p subshell has three degenerate orbitals (2px, 2py, 2pz). Therefore, this option has degenerate orbitals. 4. **Option D: n = 3, l = 2, m = 1** - Here, n = 3 and l = 2 corresponds to the 3d orbital. The 3d subshell has five degenerate orbitals (3dxy, 3dxz, 3dyz, 3dx2-y2, 3dz2). Therefore, this option also has degenerate orbitals. 5. **Option E: n = 3, l = 2, m = 0** - This is also a 3d orbital (specifically one of the five degenerate orbitals). Since it is part of the 3d subshell, it shares the same energy with the other 3d orbitals. ### Conclusion: The orbitals that will have the same energy in the absence of magnetic and electric fields are those in options C, D, and E. However, since the question asks for orbitals described by the three quantum numbers that will have the same energy, we can conclude that options C, D, and E are correct. ### Final Answer: The correct options are (C) n = 2, l = 1, m = 1; (D) n = 3, l = 2, m = 1; and (E) n = 3, l = 2, m = 0.

To solve the question, we need to determine which of the given orbitals will have the same energy in a multi-electron atom in the absence of magnetic and electric fields. The key concept here is that orbitals with the same principal quantum number (n) and the same azimuthal quantum number (l) are degenerate, meaning they have the same energy. Let's analyze each option step by step: 1. **Option A: n = 1, l = 0, m = 0** - Here, n = 1 corresponds to the 1s orbital (l = 0). There is only one 1s orbital, so it does not have any degenerate orbitals. 2. **Option B: n = 2, l = 0, m = 0** ...
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