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I'm having difficulty understanding why the base-centered orthorhombic crystal system is a unique crystal system. When I draw two base-centered orthorhombic unit cells next to each other there appears to be a primitive crystal that is contained within the union of the two lattices as shown in the image below.

Base-centered orthorhombic unit cells

This primitive crystal system appears to be monoclinic--though I wouldn't be surprised if I'm wrong on that front--if you take δ to be the typical β angle of a monoclinic crystal system (I've set δ ≠ 120° to avoid the rhombohedral scenario). Why does this not make the base-centered orthorhombic lattice redundant? Am I misguided in trying to understand these crystal systems in terms of their lattice points and unit cell lengths instead of their symmetry?

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All the centered Bravais lattices are redundant in the sense that it is possible to use a primitive cell of smaller volume instead. However, this often means that crystal symmetries are no longer aligned with the cell axes. For the orthorhombic system, you either have 2-fold rotation or screw axis parallel to the cell edges, or mirror planes or glide planes perpendicular to the cell edges. If you switch to a primitive lattice for the centered cases, these symmetry elements will no longer be aligned in this simple way (they might be on diagonals, making the symmetry operations harder to figure out).

So it is a convention, and one that is useful.

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  • $\begingroup$ Hexagonal until cells are often reduced to rhombic prisms consisting of only one-third of the hexagonal prism, which simplifies presentations. The sixfold or threefold axis is implied to pass through the obtuse-angle vertices of the rhombic bases, whose angles measure 120°. $\endgroup$ – Oscar Lanzi Feb 25 at 12:08
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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 orthorhombic, when those two axes are equal. Your primitive cell has what would be the two oblique axes of a monoclinic cell equal, therefore retaining the $D_\mathrm{2h}$ orthorhombic symmetry. Compare with this answer where a specific axis ratio turns an apparently tetragonal lattice into a cubic one.

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