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The two $\ce{CHR}$ groups at opposite corners of the ring are co-planar, and both isomers are symmetric about this plane. Yes, seeing a molecule in three dimensions when you have a paper drawing is di …

When you have atoms bonded all in one plane, there will be $p$ orbitals oriented perpendicular to the plane which may not interact significantly with adjacent atoms. Such orbitals would then be called …

In the structure as drawn here, the axis is perpenducular to the plane of the paper passing through the sulfur atom.

Monoclinic lattices do not have their two oblique axes equal; or in terms of point group symmetry, the $C_\mathrm{2h}$ symmetry characteristic of monoclinic lattices is promoted to $D_\mathrm{2h}$, therefore … Your primitive cell has what would be the two oblique axes of a monoclinic cell equal, therefore retaining the $D_\mathrm{2h}$ orthorhombic symmetry. …

In a benzene molecule draw the mirror line from one carbon atom, called C-1, through the opposite carbon. For each degenerate energy level you can construct one orbital that's symmetric under reflect …

In your drawings you are correct that 5 and 7 are enantiomers. But 1 and 2 are inconsistent with tetrahedral bonding at the carbon atom (when two bonds are in the plane of the paper the others must b …

So to have both the base-centered translational symmetry and the cubic point-group symmetry, you have to allow translation of the unit cell corners to the centers of all the faces, not just one opposing … " into "face-centered" like what happened with cubic symmetry. …

The comments are right. The molecule shown is not chiral and will not show enantiomerism. But still, remember that an alkane ring, unlike an aromatic ring, is generally not planar and take that into …

An article by Bradley et al. [1] documents $\ce{[Cr(N(SiMe3)2)2(THF)2]}$ as a square planar complex. However, the $\ce{[Cr(N(SiMe3)2)3NO]}$ complex is pseudotetrahedral $(C_\mathrm{3v}).$ Figures be …