Question

Arrange the following orbitals in decreasing order of energy 
A. $n=3, l =0, m =0$ 
B. $n =4, l =0, m =0$ 
C. $n =3, l =1, m =0$ 
D. $n =3, l =2, m =1$ 
The correct option for the order is :

• A. $D > B > C > A$
• B. $B > D > C > A$
• C. $A > C > B > D$
• D. $D > B > A > C$

Hint:

When comparing orbitals, use Hund’s rule and the \( n + l \) value to determine energy levels.

Answer 1

The Correct Option is A

To determine the decreasing order of energy of the given orbitals, we need to use the n+l rule (also known as the Madelung rule), which helps us understand the order of filling of orbitals: 

1. n+l rule: The energy of the orbitals is determined based on the sum of the principal quantum number \(n\) and the azimuthal quantum number \(l\). The orbital with the lower sum of \(n+l\) has lower energy. If two orbitals have the same \(n+l\) value, the one with the lower \(n\) is lower in energy.

Now, let's determine the \(n+l\) values for each orbital:

1. Orbital A: \(n=3\), \(l=0\). Therefore, \(n+l = 3+0 = 3\).
2. Orbital B: \(n=4\), \(l=0\). Therefore, \(n+l = 4+0 = 4\).
3. Orbital C: \(n=3\), \(l=1\). Therefore, \(n+l = 3+1 = 4\).
4. Orbital D: \(n=3\), \(l=2\). Therefore, \(n+l = 3+2 = 5\).

Now, let's arrange these orbitals in decreasing order of energy based on the \(n+l\) values:

• Orbital D: \(n+l = 5\)
• Orbital B and C: \(n+l = 4\) (among these, Orbital B \((n=4)\) is higher in energy than Orbital C \((n=3)\))
• Orbital A: \(n+l = 3\)

Therefore, the correct order is: Orbital D > Orbital B > Orbital C > Orbital A.

The correct option is:

$D > B > C > A$