Angular momentum (L) in z-axis of hydroxyl radical

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SUMMARY

The discussion focuses on calculating the expected value of angular momentum (Lz) for the hydroxyl radical (OH) in a rotational state. The relevant formula is Lz = mlħ, where ħ is the reduced Planck's constant (ħ = h/2π, with h = 6.62606876 × 10-34 J s). Participants express uncertainty about the role of the wavefunction in this calculation and the proper orbitals associated with the hydroxyl radical, specifically noting the electron configurations of hydrogen (1s1) and oxygen (1s2 2s2 2p4). The discussion also raises questions about the orientation of the angular momentum vector relative to the z-axis.

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Problem:
A hydroxyl radical in the gas phase is found to be in a rotational state with the wavefunction given, it's a long wavefunction so I'm not going to post unless needed.

a) Calculate the expected value of Lz for this radical in SI units.
I think Lz=mlħ is important for this but where does the given psi function come into play? I doubt it is extraneous information.

Information:
ħ = h/2\pi
h\ =\ 6.62606876(52)\ \times\ 10^{-34}\ J\ s
ml = -l, -l+1, -l+2, ... l-2, l-1, l
My problem: Unsure how to find the proper orbitals associated with OH radical. The hydrogen is 1s1 and the oxygen is 1s2 2s2 2p4.

b) Is the angular momentum vector parallel, perpendicular, or at an angle from the z-axis.

I'll attempt a solution if someone can point me in the right direction, thank you.
 
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I'm still struggling on this problem.

How does the lone electron effect the orbitals associated with the molecule?
(Lz)

Beginning attempt at solution: (possibly incorrect formulas)

The orbitals would give me the rotational quantum number (l), thus giving me spatial orientation quantum number (ml), with which I could solve for Lz=mlħ
 

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