f-orbital

An f-orbital, defined by the quantum number $l=3$ , is a region of space around the nucleus where an electron is most likely to be found. The letter ”f” is of spectroscopic origin, standing for ‘fundamental’. f-block elements have either hexagonal close pack, cubic close pack or body centred cubic structures. For non-cubic symmetry systems, we describe them using a general set of orbitals, which are derived in the same way as the d-orbitals. For cubic symmetry systems, they are better described by another set of f-orbitals known as the cubic set, which are derived by taking a different set of linear combinations. Both sets are eigenfunctions of the Schrödinger equation.

One-electron f-orbital wavefunctions can be expressed using the total wavefunction of a hydrogenic atom:

$\psi=R_{n,l}(r)Y_l^{\vert m_l\vert}(\theta,\phi)$

where

$R_{n,l}(r)=N_{n,l}\frac{1}{r}\biggr$\frac{r}{na_0}\biggr$^{l+1}e^{-\frac{r}{na_0}}L_{n-l-1}^{2l+1}\biggr$\frac{2r}{na_0}\biggr$$ , the radial wavefunction, is the radial component of $\psi$ .
$Y_l^{\vert m_l\vert}(\theta,\phi)=N_l^{\vert m_l\vert}P_l^{\vert m_l\vert}(cos\theta)e^{im_l\phi}$ , the spherical harmonics, is the angular component of $\psi$ .
$L_{n-l-1}^{2l+1}=\sum_{k=0}^{n-l-1}(-1)^k\frac{(n+l)!}{k!(k+2l+1)!(n-l-1-k)!}\biggr$\frac{2r}{na_0}\biggr$^k$ are the associated Laguerre polynomials.
$P_l^{\vert m_l\vert}(cos\theta)=\frac{1}{2^ll!}(1-cos^2\theta)^{\frac{\vert m_l\vert}{2}}\frac{d^{l+\vert m_l\vert}}{dcos\theta^{l+\vert m_l\vert}}(cos^2\theta-1)^l$ are the associated Legendre polynomials.
$N_{n,l}=\sqrt{\frac{(n-l-1)!}{2n(n+l)!}\frac{2^{2l+3}}{na_0}}$ is the normalisation constant of the radial wavefunction.
$N_l^{\vert m_l\vert}=\frac{1}{\sqrt{2\pi}}\biggr\[\frac{2l+1}{2}\frac{(l-\vert m_l\vert)!}{(l+\vert m_l\vert)!}\biggr\]^{\frac{1}{2}}$ is the normalisation constant of the spherical harmonics.
$n$ is the principal quantum number, where $n=1,2,\cdots$
$l$ is the orbital angular momentum quantum number (also known as azimuthal quantum number), where $l=0,1,2,\cdots$ and $l\leq n-1$ .
$m_l$ is the magnetic quantum number, where $-l\leq m_l\leq l$ .

General set

The principal quantum number, $n$ , is also called a shell. Since $-3\leq m_l\leq 3$ when $l=3$ , there are seven f-orbitals that are characterised by the set of quantum numbers $\{n,l,m_l\}$ of $\{n,3,3\}$ , $\{n,3,2\}$ , $\{n,3,1\}$ , $\{n,3,0\}$ , $\{n,3,-1\}$ , $\{n,3,-2\}$ and $\{n,3,-3\}$ in each shell for $n\geq 4$ . Each of the seven f-orbitals has three angular nodes. Consider the set $\{4,3,0\}$ . Substituting these values into the explicit formula for $R_{n,l}(r)Y_l^{\vert m_l\vert}(\theta,\phi)$ yields:

$R_{4,3}(r)Y_3^{0}(\theta,\phi)=r^3(5cos^3\theta-3cos\theta)f(r)$

where $f(r)=\frac{1}{3072\sqrt{5\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}r^2e^{-\frac{r}{4a_0}}$ .

Converting from spherical coordinates to Cartesian coordinates, where $z=rcos\theta$ , gives

$\psi_{4f_{z^3}}=z^35f(r)-3r^2f(r)\;\;\;\;\;\;\;\;482$

Eq482 is known as the $4f_{z^3}$ orbital. When $z=0$ or $z=\sqrt{\frac{3r^2}{5}}$ or $z=-\sqrt{\frac{3r^2}{5}}$ , $\psi_{4f_{z^3}}=0$ . $z=0$ represents the $xy$ –nodal plane, while $z=\pm\sqrt{\frac{3r^2}{5}}$ implies that the angular nodes occur at two conical surfaces with their apices at the origin, extending at $\theta=arccos\biggr$\pm\frac{1}{\sqrt{3}}\biggr$$ along the $z$ -axis.

For the sets $\{4,3,1\}$ and $\{4,3,-1\}$ , we have ${\psi_{f_1}=\frac{1}{2048\sqrt{15\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}r^3e^{-\frac{r}{4a_0}}sin\theta(5cos^2\theta-1) e^{i\phi}}$ and ${\psi_{f_{-1}}=\frac{1}{2048\sqrt{15\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}r^3e^{-\frac{r}{4a_0}}sin\theta(5cos^2\theta-1) e^{-i\phi}}$ , respectively. These two wavefunctions include the factor $e^{\pm i\phi}$ , which may complicate calculations when they undergo symmetry operations. Therefore, we take linear combinations of these wavefunctions to form simpler wavefunctions. The first linear combination is ${\psi_{f_1}+\psi_{f_{-1}}=\frac{1}{1024\sqrt{15\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}r^3e^{-\frac{r}{4a_0}}sin\theta\cos\phi(5cos^2\theta-1)}$ , which normalises to ${\frac{1}{1024\sqrt{30\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}r^3e^{-\frac{r}{4a_0}}sin\theta\cos\phi(5cos^2\theta-1)}$ , or equivalently in Cartesian coordinates:

$\psi_{4f_{xz^2}}=(5z^2-r^2)xf(r)\;\;\;\;\;\;\;\;483$

where $f(r)=\frac{1}{1024\sqrt{30\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}e^{-\frac{r}{4a_0}}$ , $x=rsin\theta cos\phi$ and $z=rcos\theta$ .

One angular node occurs at $x=0$ or when $sin\theta cos\phi=0$ , which represents the $yz$ -nodal-plane. The second and third angular nodes correspond to $5cos^2\theta-1=0$ . These nodes are described by two conical surfaces with their apices at the origin, extending at $\theta=arccos\biggr$\pm\sqrt{\frac{1}{5}}\biggr$$ along the $z$ -axis.

The second normalised linear combination is ${\psi_{f_1}-\psi_{f_{-1}}=\frac{1}{1024\sqrt{30\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}r^3e^{-\frac{r}{4a_0}}(5cos^2\theta-1)sin\theta\sin\phi}$ , or equivalently,

$\psi_{4f_{yz^2}}=(5z^2-r^2)yf(r)\;\;\;\;\;\;\;\;484$

where $y=rsin\theta sin\phi$ .

One angular node occurs at $y=0$ , which represents the $xz$ -nodal plane. The second and third angular nodes correspond to $5cos^2\theta-1=0$ . These nodes are described by two conical surfaces with their apices at the origin, extending at $\theta=arccos\biggr$\pm\sqrt{\frac{1}{5}}\biggr$$ along the $z$ -axis.

Question

If ${\psi_{f_1}}$ and ${\psi_{f_{-1}}}$ are already normalised, why do we need to normalise a linear combination of them? How do we normalise a linear combination of ${\psi_{f_1}}$ and ${\psi_{f_{-1}}}$ ?

Answer

When forming a linear combination of normalised wavefunctions, the result is not necessarily normalised. Consider a general linear combination $\varphi=a\phi_1+b\phi_2$ , where $a$ and $b$ are coefficients. The normalisation condition for $\varphi$ requires that $\int\vert\varphi\vert^2d\tau=1$ , where $d\tau$ represents the volume element in spherical coordinates. Expanding $\int\vert a\phi_1+b\phi_2\vert^2d\tau=1$ and noting that $\phi_1$ and $\phi_2$ are orthonormal, we have $\vert a\vert^2+\vert b\vert^2=1$ . If $\vert a\vert^2+\vert b\vert^2\neq 1$ , which is the case for the two linear combinations of ${\psi_{f_1}}$ and ${\psi_{f_{-1}}}$ , $\varphi$ will not be normalised.

To normalise ${\psi_{f_1}-\psi_{f_{-1}}}$ , we have ${N^2\int\vert\psi_{f_1}-\psi_{f_{-1}}\vert^2d\tau=1}$ or

$-N^2\frac{1}{15728640\pi}\biggr$\frac{1}{a_0}\biggr$^9\int_0^{\infty}\int_0^{\pi}\int_0^{2\pi}r^6e^{-\frac{r}{2a_0}}(5cos^2\theta-1)^2 sin^2\theta sin^2\phi r^2drsin\theta d\theta d\phi=1$

Using ${\int_0^{\infty}r^ne^{-ar}dr=\frac{n!}{a^{n+1}$ (see this article for proof), ${\int_0^{2\pi}sin^2\phi d\phi=\pi}$ and setting $u=cos\theta$ and $du=-sin\theta d\theta$ gives $N=\frac{1}{i\sqrt{2}}$ .

For the remaining sets $\{4,3,2\}$ , $\{4,3,-2\}$ , $\{4,3,3\}$ and $\{4,3,-3\}$ , we apply the same logic to give:

Wavefunction	$\boldsymbol{\mathit{f(r)}}$	Angular nodes
$\psi_{4f_{z(x^2-y^2)}}=z(x^2-y^2)f(r)$	$\frac{1}{1024\sqrt{3\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}e^{-\frac{r}{4a_0}}$	$xy$ -plane $x=\pm y$ -planes
$\psi_{4f_{xyz}}=xyz2f(r)$	$\frac{1}{1024\sqrt{3\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}e^{-\frac{r}{4a_0}}$	$xy$ -plane $yz$ -plane $xz$ -plane
$\psi_{4f_{x(x^2-3y^2)}}=x(x^2-3y^2)f(r)$	$\frac{1}{3072\sqrt{2\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}e^{-\frac{r}{4a_0}}$	$yz$ -plane $x=\pm\sqrt{3}y$ -planes
$\psi_{4f_{y(3x^2-y^2)}}=y(3x^2-y^2)f(r)$	$\frac{1}{3072\sqrt{2\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}e^{-\frac{r}{4a_0}}$	$xz$ -plane $x=\pm\frac{1}{\sqrt{3}}y$ -planes

Cubic set

Out of the seven cubic f-orbitals, three of them, $\psi_{4f_{z^3}}$ , $\psi_{4f_{z(x^2-y^2)}}$ and $\psi_{4f_{xyz}}$ are the same as the general set of f-orbitals. The other four derived by taking different linear combinations of general set of f -orbitals. They are:

Linear combination	Normalised wavefunction	$\boldsymbol{\mathit{f(r)}}$	Angular nodes
$-\frac{\sqrt{3}}{2}(\psi_{f_1}-\psi_{f_{-1}})-\frac{\sqrt{5}}{2}(\psi_{f_3}-\psi_{f_{-3}})$	$\psi_{4f_{y^3}}=y^3f(r)-\frac{3}{5}r^2yf(r)$	$\frac{\sqrt{10}}{3072\sqrt{\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}e^{-\frac{r}{4a_0}}$	$xz$ -plane Conical surfaces with apices at the origin, extending $\theta=arcsin\biggr$\pm\sqrt{\frac{3}{5}}\biggr$$ along the $y$ -axis
$-\frac{\sqrt{3}}{2}(\psi_{f_1}+\psi_{f_{-1}})+\frac{\sqrt{5}}{2}(\psi_{f_3}+\psi_{f_{-3}})$	$\psi_{4f_{x^3}}=x^3f(r)-\frac{3}{5}r^2xf(r)$	$\frac{\sqrt{10}}{3072\sqrt{\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}e^{-\frac{r}{4a_0}}$	$yz$ -plane Conical surfaces with apices at the origin, extending $\theta=arcsin\biggr$\pm\sqrt{\frac{3}{5}}\biggr$$ along the $x$ -axis
$\frac{\sqrt{5}}{2}(\psi_{f_1}+\psi_{f_{-1}})+\frac{\sqrt{3}}{2}(\psi_{f_3}+\psi_{f_{-3}})$	$\psi_{4f_{x(z^2-y^2)}}=x(z^2-y^2)f(r)$	$\frac{\sqrt{6}}{3072\sqrt{\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}e^{-\frac{r}{4a_0}}$	$yz$ -plane $z=\pm y$ -planes
$\frac{\sqrt{5}}{2}(\psi_{f_1}-\psi_{f_{-1}})-\frac{\sqrt{3}}{2}(\psi_{f_3}-\psi_{f_{-3}})$	$\psi_{4f_{y(z^2-x^2)}}=y(z^2-x^2)f(r)$	$\frac{\sqrt{6}}{3072\sqrt{\pi}}\sqrt{\biggr$\frac{1}{a_0}\biggr$^9}e^{-\frac{r}{4a_0}}$	$xz$ -plane $z=\pm x$ -planes

Question

1. How do we know which set of linear combinations result in the cubic set of wavefunctions?
2. How do we determine the angular nodes of $\psi_{4f_{y^3}}$ ?

Answer

1. The linear combinations that give the cubic set of wavefunctions must transform according to the $O_h$ point group.
2. The angular nodes occur when $\psi_{4f_{y^3}}=y(y^2-\frac{3}{5}r^2)f(r)=0$ , which implies that $y=0$ or $y^2-\frac{3}{5}r^2=0$ . Obviously, $y=0$ corresponds to the $xz$ -nodal-plane. $y^2=\frac{3}{5}r^2$ , which states that $y$ increases as $r$ increases in spherical coordinates, represents two conical surfaces with their apices at the origin. Substituting $y=rsin\theta sin\phi$ gives $sin^2\theta sin^2\phi=\frac{3}{5}$ . When $\phi=\frac{\pi}{2}$ (along the $y$ -axis), the angles that the conical surfaces make with the $y$ -axis are $\theta=arcsin\biggr$\pm\sqrt{\frac{3}{5}}\biggr$$ .

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