Shapes of Orbitals & Energies

Shapes of s-Orbitals

The boundary surface diagram of an orbital shows the region in space where the probability of finding the electron is about 90%.

s-Orbitals ($l = 0$)

All s-orbitals are spherically symmetric — the probability of finding the electron is the same in all directions at a given distance from the nucleus.

Key features:

• Shape: Sphere centred on the nucleus
• Size increases with $n$: $4s > 3s > 2s > 1s$
• Number of radial nodes = $n - 1$
• 1s: 0 nodes
• 2s: 1 node (a spherical surface where $|\psi|^2 = 0$)
• 3s: 2 nodes
• For 1s orbital: probability density is maximum at the nucleus and decreases as distance increases
• For 2s orbital: probability density first decreases to zero (node), then increases to a secondary maximum, then decreases again

What are Nodes?

A node is a region in space where the probability of finding the electron is zero ($|\psi|^2 = 0$).

Types of nodes:

• Radial nodes (spherical nodes): spherical surfaces where $\psi = 0$. Number = $n - l - 1$
• Angular nodes: regions where $\psi = 0$. Number = $l$
• Total nodes = $n - 1$

Orbital	$n$	$l$	Radial nodes ($n-l-1$)	Angular nodes ($l$)	Total ($n-1$)
1s	1	0	0	0	0
2s	2	0	1	0	1
2p	2	1	0	1	1
3s	3	0	2	0	2
3p	3	1	1	1	2
3d	3	2	0	2	2

Shapes of p-Orbitals

p-Orbitals ($l = 1$)

For $l = 1$, there are three values of $m_l$: $-1, 0, +1$, giving three p-orbitals designated as $p_x$, $p_y$, and $p_z$.

Key features:

• Shape: Dumbbell (two lobes on either side of the nucleus)
• Each p-orbital has a nodal plane through the nucleus:
• $p_x$: nodal plane is the $yz$-plane
• $p_y$: nodal plane is the $xz$-plane
• $p_z$: nodal plane is the $xy$-plane
• The two lobes have opposite signs of $\psi$ (positive and negative phase)
• All three p-orbitals are identical in shape and energy (degenerate) but differ in orientation
• Probability density is zero at the nucleus for all p-orbitals
• Size increases with $n$: $4p > 3p > 2p$

Shapes of d-Orbitals

d-Orbitals ($l = 2$)

For $l = 2$, there are five values of $m_l$: $-2, -1, 0, +1, +2$, giving five d-orbitals: $d_{xy}$, $d_{yz}$, $d_{xz}$, $d_{x^2-y^2}$, $d_{z^2}$

Key features:

• $d_{xy}$, $d_{yz}$, $d_{xz}$: each has four lobes lying between the respective axes
• $d_{x^2-y^2}$: four lobes lying along the $x$ and $y$ axes
• $d_{z^2}$: unique shape — two lobes along $z$-axis with a doughnut (torus) in the $xy$-plane
• All five d-orbitals are degenerate in the free atom (same energy)
• Each d-orbital has 2 angular nodes ($l = 2$)

[JEE Tip] The $d_{z^2}$ orbital looks different from the other four but is mathematically equivalent in energy. Don't assume it's special!

Energies of Orbitals

In Hydrogen Atom (One-electron System)

In hydrogen, the energy of an orbital depends only on the principal quantum number $n$: $$E_{1s} < E_{2s} = E_{2p} < E_{3s} = E_{3p} = E_{3d} < \ldots$$

Orbitals with the same $n$ but different $l$ are degenerate (same energy). This is because there is only one electron and no electron-electron repulsion.

In Multi-electron Atoms — The Key Difference!

In atoms with more than one electron, the energy depends on both $n$ and $l$: $$E_{1s} < E_{2s} < E_{2p} < E_{3s} < E_{3p} < E_{4s} < E_{3d} < E_{4p} < \ldots$$

Orbitals within the same subshell are still degenerate (e.g., all three 2p orbitals have the same energy), but different subshells within the same shell may have different energies.

Why Does This Happen? — Shielding and Penetration

Shielding effect: Inner electrons partially shield outer electrons from the full nuclear charge. The effective nuclear charge felt by an outer electron is $Z_{\text{eff}} = Z - \sigma$, where $\sigma$ is the shielding constant.

Penetration: Different subshells penetrate the inner electron cloud to different extents:

• s-orbitals penetrate the most (closest to nucleus, experience highest $Z_{\text{eff}}$)
• p-orbitals penetrate less than s
• d-orbitals penetrate even less

This is why, for the same $n$: $E_s < E_p < E_d < E_f$

And sometimes a lower-$l$ orbital in a higher shell can have lower energy than a higher-$l$ orbital in a lower shell. For example, $E_{4s} < E_{3d}$.

The $(n + l)$ Rule

The energy ordering of orbitals in multi-electron atoms follows the $(n + l)$ rule (also called the Madelung rule):

Orbitals are filled in order of increasing $(n + l)$ value. If two orbitals have the same $(n + l)$ value, the one with the lower $n$ fills first.

Orbital	$n$	$l$	$n + l$	Filling Order
1s	1	0	1	1st
2s	2	0	2	2nd
2p	2	1	3	3rd
3s	3	0	3	4th
3p	3	1	4	5th
4s	4	0	4	6th
3d	3	2	5	7th
4p	4	1	5	8th

The complete filling order: $$1s < 2s < 2p < 3s < 3p < 4s < 3d < 4p < 5s < 4d < 5p < 6s < 4f < 5d < 6p < 7s < 5f < 6d < 7p$$

[JEE Tip] The $(n+l)$ rule correctly predicts the filling order for most elements. But remember that after filling, 3d may become lower in energy than 4s — this explains why transition metals lose 4s electrons first during ionisation!

Key Point: In multi-electron atoms, orbital energy depends on both $n$ and $l$ due to shielding and penetration effects. The $(n+l)$ rule gives the correct filling order.

Solved Examples

Example 1: Number of Nodes

Calculate the number of radial nodes, angular nodes, and total nodes for a 3p orbital.

Solution: For 3p: $n = 3, l = 1$

• Radial nodes $= n - l - 1 = 3 - 1 - 1 = 1$
• Angular nodes $= l = 1$
• Total nodes $= n - 1 = 2$

Answer: 1 radial node, 1 angular node, 2 total nodes.

Example 2: Nodes for 4d Orbital

How many radial and angular nodes does a 4d orbital have?

Solution: For 4d: $n = 4, l = 2$

• Radial nodes $= 4 - 2 - 1 = 1$
• Angular nodes $= 2$
• Total nodes $= 3$

Answer: 1 radial node, 2 angular nodes.

Example 3: Identifying Orbital from Nodes

An orbital has 2 angular nodes and 1 radial node. Identify the orbital.

Solution: Angular nodes $= l = 2$ → d-orbital Radial nodes $= n - l - 1 = 1$, so $n - 2 - 1 = 1$, giving $n = 4$

Answer: 4d orbital.

Example 4: Comparing Orbital Energies

Arrange the following orbitals in order of increasing energy in a multi-electron atom: 4s, 3d, 4p, 3p.

Solution: Using $(n + l)$ rule:

• 3p: $n + l = 3 + 1 = 4$
• 4s: $n + l = 4 + 0 = 4$ (same as 3p, but higher $n$)
• 3d: $n + l = 3 + 2 = 5$
• 4p: $n + l = 4 + 1 = 5$ (same as 3d, but higher $n$)

For same $(n+l)$, lower $n$ has lower energy.

Answer: $3p < 4s < 3d < 4p$.

Example 5: Degeneracy in Hydrogen vs Multi-electron Atoms

Are 3s, 3p, and 3d orbitals degenerate in (a) hydrogen and (b) carbon?

Solution: (a) Hydrogen: Yes, they are degenerate. In hydrogen (one-electron system), energy depends only on $n$, so $E_{3s} = E_{3p} = E_{3d}$.

(b) Carbon: No, they are not degenerate. In multi-electron atoms, shielding causes $E_{3s} < E_{3p} < E_{3d}$.

Answer: Degenerate in hydrogen, not degenerate in carbon.

Example 6: Shape Identification

Describe the shape and number of nodes for a 2p orbital.

Solution:

• Shape: Dumbbell (two lobes on either side of the nucleus)
• Radial nodes $= n - l - 1 = 2 - 1 - 1 = 0$
• Angular nodes $= l = 1$ (one nodal plane through the nucleus)
• Total nodes $= 1$

Answer: Dumbbell shape with 1 angular node and 0 radial nodes.

Example 7: Using $(n+l)$ Rule

Which orbital fills first: 4s or 3d? Justify using the $(n+l)$ rule.

Solution:

• 4s: $n + l = 4 + 0 = 4$
• 3d: $n + l = 3 + 2 = 5$

Since $4 < 5$, 4s fills before 3d.

Answer: 4s fills first.

Example 8: Angular Nodes of d-Orbitals

How many angular nodes do d-orbitals have? Name the angular nodes for $d_{xy}$.

Solution: All d-orbitals have $l = 2$, so they have 2 angular nodes.

For $d_{xy}$, the lobes lie in the $xy$-plane between the axes. The nodal surfaces are the $xz$-plane and the $yz$-plane.

Answer: d-orbitals have 2 angular nodes. For $d_{xy}$, they are the $xz$ and $yz$ planes.

Example 9: Penetration and Relative Energy

Explain why in the same atom, a 2s electron is more tightly bound than a 2p electron.

Solution: The 2s orbital penetrates closer to the nucleus than the 2p orbital. Therefore, a 2s electron experiences a greater effective nuclear charge ($Z_{\text{eff}}$) and is held more strongly.

Because of this greater penetration: $$E_{2s} < E_{2p}$$

This is why, in multi-electron atoms, 2s lies lower in energy than 2p.

Answer: Greater penetration of the 2s orbital leads to higher effective nuclear charge and lower energy than 2p.

Example 10: Total Nodes for a Given Orbital

An orbital has $n = 5$ and $l = 3$. Calculate all nodes and identify the orbital.

Solution:

• $n = 5, l = 3$ → 5f orbital
• Radial nodes $= 5 - 3 - 1 = 1$
• Angular nodes $= 3$
• Total nodes $= 4$

Answer: 5f orbital with 1 radial node, 3 angular nodes, 4 total nodes.