Q.The arrangement of orbitals on the basis of energy is based upon their (n+l) value. Lower the value of (n+l), lower is the energy. For orbitals having same values of (n+l), the orbital with lower value of n will have lower energy.
I. Based upon the above information, arrange the following orbitals in the increasing order of energy.

Concept: Effective nuclear charge and the $(n+\ell)$ rule — orbital energy increases with $(n+\ell)$; for equal $(n+\ell)$, the orbital with lower $n$ has lower energy.

Reasoning:

1. Compute $(n+\ell)$ for each orbital:

   • $1s$: $1+0=1$
   • $2s$: $2+0=2$
   • $2p$: $2+1=3$
   • $3s$: $3+0=3$
   • $3p$: $3+1=4$
   • $4s$: $4+0=4$
   • $4d$: $4+2=6$
   • $4f$: $4+3=7$
   • $5s$: $5+0=5$
   • $5p$: $5+1=6$
   • $5d$: $5+2=7$
   • $5f$: $5+3=8$
   • $6s$: $6+0=6$
   • $6p$: $6+1=7$
   • $6d$: $6+2=8$
   • $7s$: $7+0=7$
   • $7p$: $7+1=8$

2. Arrange by increasing $(n+\ell)$; for ties, lower $n$ comes first.

I.

1. $1s < 2s < 2p < 3s$
2. $3s < 3p < 4s < 4d$
3. $4d < 5p < 6s < 4f < 5d < 5f$
4. $7s < 5f < 6d < 7p$ II. …

Why the $(n+\ell)$ Rule Works

The energy of an orbital in a multi-electron atom depends not only on the principal quantum number $n$ but also on the azimuthal quantum number $\ell$. This is because electrons in orbitals with higher $\ell$ experience more shielding from inner electrons, reducing the effective nuclear charge they feel. The $(n+\ell)$ rule (also called the n+l rule or Aufbau principle) captures this: orbitals with a smaller $(n+\ell)$ value are lower in energy. When two orbitals have the same $(n+\ell)$, the one with the smaller $n$ is lower in energy because it is closer to the nucleus on average.

Let’s apply this step by step.

I. Arranging orbitals in increasing order of energy

1. (a) 1s, 2s, 3s, 2p

Compute $(n+\ell)$ for each:

• 1s: $n=1,\ \ell=0 \implies 1+0 = 1$
• 2s: $n=2,\ \ell=0 \implies 2+0 = 2$
• 3s: $n=3,\ \ell=0 \implies 3+0 = 3$
• 2p: $n=2,\ \ell=1 \implies 2+1 = 3$

Order by $(n+\ell)$: 1s (1) < 2s (2) < 3s (3) and 2p (3). Since 3s and 2p both have $(n+\ell)=3$, compare $n$: 2p has $n=2$, 3s has $n=3$, so 2p is lower.

Increasing order: 1s < 2s < 2p < 3s

2. (b) 4s, 3s, 3p, 4d

Compute $(n+\ell)$:

• 4s: $4+0 = 4$
• 3s: $3+0 = 3$
• 3p: $3+1 = 4$
• 4d: $4+2 = 6$

Order by $(n+\ell)$: 3s (3) < 4s (4) and 3p (4) < 4d (6). For 4s and 3p, both have $(n+\ell)=4$: 3p has $n=3$, 4s has $n=4$, so 3p is lower.

Increasing order: 3s < 3p < 4s < 4d

3. (c) 5p, 4d, 5d, 4f, 6s

Compute $(n+\ell)$:

• 5p: $5+1 = 6$
• 4d: $4+2 = 6$
• 5d: $5+2 = 7$
• 4f: $4+3 = 7$
• 6s: $6+0 = 6$

Group by $(n+\ell)$:

• $(n+\ell)=6$: 5p, 4d, 6s
• $(n+\ell)=7$: 5d, 4f

Within $(n+\ell)=6$, compare $n$: 4d ($n=4$) < 5p ($n=5$) < 6s ($n=6$).

Within $(n+\ell)=7$, compare $n$: 4f ($n=4$) < 5d ($n=5$).

Increasing order: 4d < 5p < 6s < 4f < 5d

4. (d) 5f, 6d, 7s, 7p

Compute $(n+\ell)$:

• 5f: $5+3 = 8$
• 6d: $6+2 = 8$
• 7s: $7+0 = 7$
• 7p: $7+1 = 8$ …