Question

Arrange the following atomic orbitals of multi electron atoms in order of increasing energy.
A. \(n=3, l=2, m=+1\)
B. \(n=4, l=0, m=0\)
C. \(n=6, l=1, m=0\)
D. \(n=5, l=1, m=+1\)
E. \(n=2, l=1, m=+1\)

• A. \(C < D < B < A < E\)
• B. \(B < A < E < C < D\)
• C. \(E < C < D < B < A\)
• D. \(E < B < A < D < C\)

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
In multi-electron atoms, the energy of an orbital depends on both the principal quantum number (\(n\)) and the azimuthal quantum number (\(l\)). The sequence is determined by the Bohr-Bury rule (\(n+l\) rule). 
Step 2: Key Formula or Approach: 
1. Orbitals with lower \((n+l)\) values have lower energy. 
2. If two orbitals have the same \((n+l)\) value, the one with the lower \(n\) value has lower energy. 
Step 3: Detailed Explanation: 
Let's calculate \((n+l)\) for each orbital: 
A. \(n=3, l=2 \implies (n+l) = 3+2 = 5\) (3d orbital) 
B. \(n=4, l=0 \implies (n+l) = 4+0 = 4\) (4s orbital) 
C. \(n=6, l=1 \implies (n+l) = 6+1 = 7\) (6p orbital) 
D. \(n=5, l=1 \implies (n+l) = 5+1 = 6\) (5p orbital) 
E. \(n=2, l=1 \implies (n+l) = 2+1 = 3\) (2p orbital) 
Now, sort the orbitals by their \((n+l)\) values: 
E (3) \(<\) B (4) \(<\) A (5) \(<\) D (6) \(<\) C (7). 
There are no ties in \((n+l)\) value in this set, so the order is straightforward. 
Step 4: Final Answer: 
The increasing order of energy is \(E < B < A < D < C\).