Which $d$ $d$ orbital is specified by $Y (θ, ϕ) = {(\frac{5}{8 π})}^{1 / 2} (3 {cos}^{2} θ - 1)$ $Y(theta,phi) = (5/(8pi))^(1//2) (3cos^2theta - 1)$ ?

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Which $d$ $d$ orbital is specified by $Y (θ, ϕ) = {(\frac{5}{8 π})}^{1 / 2} (3 {cos}^{2} θ - 1)$ $Y(theta,phi) = (5/(8pi))^(1//2) (3cos^2theta - 1)$ ?

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Answer 1 · 2016-04-25T06:57:03

Your angular wave function looks off. Every wave function $ψ_{n l m_{l}}$ $psi_(nlm_l)$ , which describes the state of a quantum mechanical system, must be at LEAST

orthogonal ( $\int_{allspace} ψ_{i}^{*} ψ_{j} d τ = 0$ $int_"allspace" psi_i^"*"psi_jd tau = 0$ where $i \neq j$ $i ne j$ )
normalizable ( $\int_{allspace} ψ_{i}^{*} ψ_{i} d τ = 1$ $int_"allspace" psi_i^"*"psi_id tau = 1$ )

and yours is not. The correct $Y_{l}^{m_{l}} (θ, ϕ)$ $Y_(l)^(m_l)(theta,phi)$ is:

$Y_{2}^{0} (θ, ϕ) = {(\frac{5}{16 π})}^{1/2} (3 {cos}^{2} θ - 1)$ $color(blue)(Y_(2)^(0)(theta,phi) = (5/(16pi))^"1/2" (3cos^2theta - 1))$

In this case, what we have here is ONLY the angular component of the wave function for the $d_{z^{2}}^{2}$ $d_(z^2)$ orbital using spherical coordinates. Actually, without knowing the radial component, we do not know the principal quantum number of your $d_{z^{2}}^{2}$ $d_(z^2)$ orbital.

The full wave function in spherical harmonics is written as the product of the radial and angular components:

$\mathbf(psi_(nlm_l)(vecr,theta,phi) = R_(nl)(vecr)Y_(l)^(m_l)(theta,phi))$

For instance, the full wave function for the $3d_(z^2)$ orbital is:

$psi_(3d_(z^2)) = psi_(320)(vecr,theta,phi) = R_(32)(vecr)Y_(2)^(0)(theta,phi)$

$= 1/(81sqrt(6pi))(Z/(a_0))^"3/2" ((Zr)/(a_0))^2e^(-Zr"/"3a_0) (3cos^2theta - 1),$

where $Z$ is the atomic number and $a_0$ is the Bohr radius ( $5.29177xx10^(-11) "m"$ ).

This, you would see is quite different and much more complicated.

You did not include $R_(nl)(vecr)$ , so the best I can do is say that you have accidentally specified all $d_(z^2)$ orbitals in existence. There exist the $3d_(z^2)$ , $4d_(z^2)$ , $5d_(z^2)$ , and $6d_(z^2)$ thus far that have been observed.

Generally they look like dumbbells along the z-axis with a torus (donut) on the xy-plane. The torus has the OPPOSITE sign to the lobes, and the lobes have the SAME sign as each other.

The $d_(z^2)$ have conic nodal surfaces. Specifically:

The $3d_(z^2)$ has $n - l - 1 = 3 - 2 - 1 = 0$ radial nodes and $n - 1 = 2$ total nodes. Thus, it has two angular nodes.

The $4d_(z^2)$ has $n - l - 1 = 4 - 2 - 1 = 1$ radial node and $n - 1 = 3$ total nodes. Thus, it has two angular nodes as before.

The $5d_(z^2)$ has $n - l - 1 = 5 - 2 - 1 = 2$ radial nodes and $n - 1 = 4$ total nodes. Thus, it has two angular nodes as before.

The $6d_(z^2)$ has $n - l - 1 = 6 - 2 - 1 = 3$ radial nodes and $n - 1 = 5$ total nodes. Thus, it has two angular nodes as before.

So you can see the pattern that the $d_(z^2)$ orbital simply gains radial nodes as the quantum number $n$ increases.

For instance, here is a picture of the $5d_(z^2)$ orbital:

![http://images.fineartamerica.com/](useruploads.socratic.org)

Now compare that to the $3d_(z^2)$ orbital.

![http://images.fineartamerica.com/](useruploads.socratic.org)

You should notice that the $5d_(z^2)$ is obviously more complex at the torus.

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Which $d$ $d$ orbital is specified by $Y (θ, ϕ) = {(\frac{5}{8 π})}^{1 / 2} (3 {cos}^{2} θ - 1)$ $Y(theta,phi) = (5/(8pi))^(1//2) (3cos^2theta - 1)$ ?

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Which ddd orbital is specified by Y(theta,phi) = (5/(8pi))^(1//2) (3cos^2theta - 1)Y(θ,ϕ)=(58π)1/2(3cos2θ−1)Y(theta,phi) = (5/(8pi))^(1//2) (3cos^2theta - 1)?

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Which $d$ $d$ orbital is specified by $Y (θ, ϕ) = {(\frac{5}{8 π})}^{1 / 2} (3 {cos}^{2} θ - 1)$ $Y(theta,phi) = (5/(8pi))^(1//2) (3cos^2theta - 1)$ ?