What are the nodal planes in #3d_(x^2-y^2)# orbital?
1 Answer
This can be done rigorously or easily.
This is the easy way:
The
The lobes are alternating phase going around the orbital. Therefore, the two nodal planes point in between the lobes, i.e. bisecting the
This is the hard way:
The angular wave function for the hydrogen atomic
#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
All we need to solve first then is:
#0 = sin^2theta#
If
However, we can tell which phase the wave function is in based on the angle
#Y_(2)^(2)(theta,phi) prop e^(2iphi)#
We define that the function
It is convenient to rewrite this as:
#e^(2iphi) = cos(2phi) + isin(2phi)#
Here we use our intuition that lobes adjacent counterclockwise are opposite phases. So we use
#e^(0) stackrel(?" ")(=) -e^(ipi)#
Indeed,
#e^(ipi) = cos(pi) + isin(pi) = -1# , so
#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 (
#-e^(ipi) stackrel(?" ")(=) e^(2ipi)#
Indeed:
#-(cos(pi) + isin(pi)) = 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
That must also mean that the rear right and front left orbital lobes are the same sign. Let's check.
#e^(0) stackrel(?" ")(=) e^(2ipi)#
#cos(0) + isin(0) = cos(2pi) + isin(2pi)#
#1 = 1#
Yep, we're good; the front left and rear right orbital lobes are same sign.