It’s an s orbital with no angular dependence, so it is spherically symmetric.
The wavefunction does not depend on θ or φ → probability is the same in all directions.
Probability density (probability per unit volume).
P(r)=4πr2∣R(r)∣2
Because the area/volume of a spherical shell increases with r.
A boundary enclosing ~90% probability of finding the electron.
Because E=0 is defined for a free electron at infinite separation; bound electrons have E<0 since energy must be added to remove them.
The energy level and average size of the orbital.
Orbital shape (ℓ=0 s, ℓ=1 p, ℓ=2 d, ℓ=3 f).
Orbital orientation in space, values from −ℓ to +ℓ.
n−1
1s: 0, 2s: 1, 3s: 2.
A region where Ψ = 0 so ∣Ψ∣ 2 =0 → zero probability of finding the electron.
Orbitals are standing wave solutions; only wavefunctions that satisfy boundary conditions are allowed.
ℓ= 1
−1,0,+1
3 (because 2ℓ+1=3)
Because ℓ can only be 0 to n−1; to have
ℓ=1, n must be ≥ 2.
A nodal plane through the nucleus where Ψ = 0
The phase/sign of the wavefunction Ψ, not electrical charge.
Three p orbitals oriented along the x, y, and z axes (same energy/shape, different orientation).
Ψ can be +/– (phase); Ψ² is always positive and gives probability density.
Wavefunction phase/sign, not charge.
1 (since angular nodes = ℓ and for p, ℓ=1).
0 (since n−ℓ−1=0).
A nodal plane through the nucleus splits the wavefunction into opposite phases → two lobes.
Same-phase overlap is constructive → bonding; opposite-phase overlap is destructive → antibonding.
1 (because ℓ=1).
1 (because n−ℓ−1=3−1−1=1).
Because it has a radial node that splits the orbital into inner and outer regions.
Ψ=0 and the wavefunction changes sign (phase flips).
No — they represent the phase of Ψ.
Higher n orbitals are more spread out → more probability at larger r.
A nodal plane through the nucleus splits Ψ into two opposite phases → two lobes.
Phase (sign) of Ψ, not electrical charge.
Bond strength depends on how well orbitals overlap between nuclei (constructive overlap).
Overlap of orbitals with the same phase → builds electron density between nuclei → bonding.
It affects penetration and shielding, which control how strongly the nucleus stabilizes the electron.
Only certain wave patterns fit boundary conditions, producing stable wavefunctions with nodes.
ℓ = 2
5, because m ℓ =−2,−1, 0, +1, +2 (so 2ℓ+1=5).
d xy, d xz, dyz, dx2-y2, dz2
dx2-y2
d z2
between the x and y axes in the xy plane
wavefunction phase/sign not charge
in the xy plane between the x and y axes
the xz plane and yz plane
4-2-1= 1
n-1= 3
ℓ =3
7 (since 2ℓ+1=7)
−3,−2,−1,0,+1,+2,+3
n=4 (4f is the first).
They’re core-like and shielded → poor overlap with ligands → mostly ionic bonding.
Many unpaired f electrons → high magnetic moments.
The probability of finding an electron at a distance r from the nucleus.
Higher n means higher energy and larger average radius.
Only n. So orbitals with same n are degenerate (3s = 3p = 3d).
Both n and ℓ, due to shielding and penetration.
Penetration is how much electron density is close to the nucleus; more penetration → higher Zeff → lower energy.
3s<3p<3d
0
1 nodal plane through nucleus
2
2ℓ+1.
s = 1, p = 3, d = 5.
dx2-y2
: Electron–electron repulsion causes shielding/penetration → orbital energy depends on n and ℓ, not just n.
Electrons fill the lowest energy orbitals first (ground state configuration).
Max 2 electrons per orbital with opposite spins.
In degenerate orbitals, electrons occupy singly first with parallel spins, then pair.
Yes (degenerate, energy depends only on n).
No. Usually 3s<3p<3d due to penetration/shielding.
s:1, p:3, d:5, f:7
The net positive charge an electron feels after shielding by other electrons
Zeff= Z - S
Shielding/screening by other electrons (depends on electron configuration/orbitals).
Because electron–electron repulsion shields the nucleus.
Z increases while shielding changes only slightly (electrons added to same shell)
its 2s valence electron is shielded by 1s^2 --> low Zeff--> easy to remove
In multi-electron atoms, electrons repel each other and shield nuclear charge → electrons feel less than +Z.
Zeff = Z-S
H is a 2body problem; he is a 3 body problem with e- - e- repulsion
H: 1s^1, He: 1s^2
Even with pairing repulsion, 1s is far lower energy than 2s.
Energy cost of placing two electrons in the same orbital (mainly from e⁻–e⁻ repulsion).
2s
