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:
The topic is Atomic Structure.
We need to arrange a given set of atomic orbitals in the order of their increasing energy for multi-electron atoms. This is governed by the Aufbau principle, specifically the \((n+l)\) rule.
Step 2: Key Formula or Approach:
The energy of an orbital in a multi-electron atom is determined by the \((n+l)\) rule:
1. An orbital with a lower value of \((n+l)\) has lower energy.
2. If two orbitals have the same value of \((n+l)\), the orbital with the lower value of \(n\) has lower energy.
Step 3: Detailed Explanation:
Let's calculate the \((n+l)\) value for each orbital. Note that the magnetic quantum number \(m\) does not affect the energy of the orbital itself.
A. \(n=3, l=2\) \(\implies\) \((n+l) = 3+2 = 5\) (This is a 3d orbital).
B. \(n=4, l=0\) \(\implies\) \((n+l) = 4+0 = 4\) (This is a 4s orbital).
C. \(n=6, l=1\) \(\implies\) \((n+l) = 6+1 = 7\) (This is a 6p orbital).
D. \(n=5, l=1\) \(\implies\) \((n+l) = 5+1 = 6\) (This is a 5p orbital).
E. \(n=2, l=1\) \(\implies\) \((n+l) = 2+1 = 3\) (This is a 2p orbital).
Now, we arrange the orbitals in increasing order of their \((n+l)\) values:
\((n+l)=3\) (E) < \((n+l)=4\) (B) < \((n+l)=5\) (A) < \((n+l)=6\) (D) < \((n+l)=7\) (C).
Since there are no ties in the \((n+l)\) values, the order is determined directly.
The final order of increasing energy is E < B < A < D < C.
Step 4: Final Answer:
The increasing order of energy is \(E<B<A<D<C\).