What are the nodal planes in $3 d_{x^{2} - y^{2}}^{2}$ $3d_(x^2-y^2)$ orbital?

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What are the nodal planes in $3 d_{x^{2} - y^{2}}^{2}$ $3d_(x^2-y^2)$ orbital?

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Answer 1 · 2018-02-22T21:21:29

This can be done rigorously or easily.

This is the easy way:

The $3 d_{x^{2} - y^{2}}^{2}$ $3d_(x^2-y^2)$ orbital has lobes that point along the $x$ $x$ and $y$ $y$ axes:

![https://chem.libretexts.org/](useruploads.socratic.org)

The lobes are alternating phase going around the orbital. Therefore, the two nodal planes point in between the lobes, i.e. bisecting the $x y$ $xy$ axes.

This is the hard way:

The angular wave function for the hydrogen atomic $3 d_{x^{2} - y^{2}}^{2}$ $3d_(x^2-y^2)$ orbital ( $(l, m_{l}) = (2, 2)$ $(l,m_l) = (2,2)$ ) is given by:

$Y_{2}^{2} (θ, ϕ) = \frac{1}{4} \sqrt{\frac{15}{2 π}} {sin}^{2} θ e^{2 i ϕ}$ $Y_(2)^(2)(theta,phi) = 1/4 sqrt(15/(2pi))sin^2thetae^(2iphi)$

The nodes occur when this function is equal to zero. Clearly, nonzero constants are not zero, and $e^{2 i ϕ} \neq 0$ $e^(2iphi) ne 0$ .

All we need to solve first then is:

$0 = {sin}^{2} θ$ $0 = sin^2theta$

If ${sin}^{2} θ = 0$ $sin^2theta = 0$ , then $θ = 0, π$ $theta = 0,pi$ . That gives rise to vertical nodal planes, but does not tell us which way they are oriented.

However, we can tell which phase the wave function is in based on the angle $ϕ$ $phi$ we are at on the $x y$ $xy$ plane.

$Y_{2}^{2} (θ, ϕ) \propto e^{2 i ϕ}$ $Y_(2)^(2)(theta,phi) prop e^(2iphi)$

We define that the function $e^{i ϕ}$ $e^(iphi)$ rotates the wave function by $ϕ$ $phi$ angle counterclockwise (rather than clockwise). We start from the front left orbital lobe as a reference point at $ϕ = 0$ $phi = 0$ . We can then find when the lobes are the opposite sign.

It is convenient to rewrite this as:

$e^{2 i ϕ} = cos (2 ϕ) + i sin (2 ϕ)$ $e^(2iphi) = cos(2phi) + isin(2phi)$

Here we use our intuition that lobes adjacent counterclockwise are opposite phases. So we use $ϕ = 0, \frac{π}{2}$ $phi = 0,pi/2$ to check:

$e^{0} ? = - e^{i π}$ $e^(0) stackrel(?" ")(=) -e^(ipi)$

Indeed,

$e^{i π} = cos (π) + i sin (π) = - 1$ $e^(ipi) = cos(pi) + isin(pi) = -1$ , so

$1 = - (- 1)$ $1 = -(-1)$

Therefore, the front left and front right lobes are opposite phases.

Nodal planes can only be in between lobes of opposite phases, so we have one nodal plane so far that is the one coming directly towards us.

Now we try the next two ( $ϕ = \frac{π}{2}, π$ $phi = pi/2,pi$ ) to find the other nodal plane:

$- e^{i π} ? = e^{2 i π}$ $-e^(ipi) stackrel(?" ")(=) e^(2ipi)$

Indeed:

$- (cos (π) + i sin (π)) = 1$ $-(cos(pi) + isin(pi)) = 1$

$cos (2 π) + i sin (2 π) = 1$ $cos(2pi) + isin(2pi) = 1$

Hence, the front right and rear right orbital lobes are opposite sign to each other.

Therefore, we find the second nodal plane horizontal to us, bisecting the $x y$ $xy$ axes.

That must also mean that the rear right and front left orbital lobes are the same sign. Let's check.

$e^{0} ? = e^{2 i π}$ $e^(0) stackrel(?" ")(=) e^(2ipi)$

$cos (0) + i sin (0) = cos (2 π) + i sin (2 π)$ $cos(0) + isin(0) = cos(2pi) + isin(2pi)$

$1 = 1$ $1 = 1$

Yep, we're good; the front left and rear right orbital lobes are same sign.

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