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Considering the sign of orbital and assuming the $z$-axis as a principal axis, for me, it looks like that it has two perpendicular $C_2$ axes that penetrate the lobes, so I think it is $C_\mathrm{2v}$. But the book (Inorganic chemistry by Miessler and Tarr) says it is $D_\mathrm{2h}$.

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Note that $C_\mathrm{2v}$ is a subgroup of $D_\mathrm{2h}$. This means that you're just overlooking some of the symmetry elements. In addition to the $C_2$ axes penetrating the lobes "end to end", you also have a $C_2$ axis perpendicular to the orbital plane through its center.

Your $C_\mathrm{2v}$ guess is also inconsistent because that point group only has one $C_2$ axis, but you said yourself that you already found two.

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  • $\begingroup$ Thank you for your comment. I misunderstood the criteria for C and D groups. Now it makes sense! $\endgroup$ – skwon Jan 26 '20 at 23:13
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Imagine that the $x$ axis and $y$ axis lobes have a different color (phase). Locate the principal axis, that of highest rotational symmetry, if there is more than one just choose one. A $C_2$ or 180 degree rotation will make the orbital indistinguishable from its starting position if this points out of the plane of the orbital (i.e. I choose this to be along $z$ if orbital is in $xy$-plane). Any 'D' point group has a 2-fold, $C_2$ or 180 degree rotation axis, perpendicular to the principal axis so the $\mathrm{d}_{xy}$ belongs to the $D_\mathrm{2h}$ point group. You can check other symmetry elements by looking at the point group table. A $C_\mathrm{2v}$ molecule, i.e. one with symmetry of water molecule, has no 2-fold axis perpendicular to the principal axis. If you do not distinguish $x$ and $y$ with 'color' the orbital will have $D_\mathrm{4h}$ symmetry.

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  • $\begingroup$ OMG I just misunderstood that if there is a perpendicular C2 axis, then it goes to C group. Thank you for your kind explanation & correction. $\endgroup$ – skwon Jan 26 '20 at 23:11

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