We were speaking about this in class, but I can't understand it quite well. What would happen if (hypothetically of course) the allowed $m_s$ values were $-1$, $0$ and $1$? What impact does this have on our current Periodic Table? What impact does it have on the other quantum numbers?
I don't see any relation between $m_s$ and the other quantum numbers... which is why I'm asking so I can really understand it.
There are several issues here which should be carefully disentangled.
It is independent. The reason is because $n, l, m_l$ are quantum numbers which deal with the spatial coordinates of the electron, i.e. its position and momentum in 3D space as given by (for example) $(x, y, z)$-coordinates. On the other hand, spin is a separate property, often called an "intrinsic angular momentum" which cannot be visualised in 3D space.*
In nonrelativistic quantum mechanics, the Schrödinger equation actually only describes the spatial part (i.e. $n, l, m_l$); the spin part is just tacked on to that at the end. The spin and spatial parts of a wavefunction are completely separate from one another and don't directly interact. (Indirectly they still do, via things like the Pauli exclusion principle.)
As mentioned in the comments, spin-1 particles (and more generally, particles with integer spin, i.e. bosons) do not obey the Pauli exclusion principle. In the context of an atom, that means that two "spin-1 electrons" can share the same four quantum numbers $(n, l, m_l, m_s)$.†
However, the Periodic Table, and chemistry generally, crucially depends on the Pauli exclusion principle. Consider the case of lithium: we say it has an electron configuration of $\mathrm{1s^2 ~2s^1}$. Why does the third electron go into the $\mathrm{2s}$ orbital? Well, it's because the first two went into the $\mathrm{1s}$ orbital, and thus have quantum numbers $(1, 0, 0, +1/2)$ and $(1, 0, 0, -1/2)$. Since the third electron can't share all four quantum numbers with either of its two predecessors, it's forced into the next-lowest orbital, i.e. $(2, 0, 0, \pm 1/2)$ (the sign is irrelevant).
With spin-1 electrons there is no such constraint, so every spin-1 electron would simply sit in its lowest possible energy state, namely the $\mathrm{1s}$ orbital.‡ The true electronic configuration of the $n$-th element would then simply be $\mathrm{1s}^n$. So, chemistry as we know it would not exist.
This is the easy part: each orbital would just hold three electrons instead of two. So the 1s orbital would hold up to 3 electrons. The 2p subshell would hold 9 electrons (3 orbitals of 3 electrons each). The 3d subshell would hold 15 electrons.
Consider the 2–8–8–18... pattern that we see in the (normal, spin-1/2) Periodic Table:
In hypothetical spin-1-electron land with hypothetical Pauli exclusion principle, this would be 3–12–12–27:
It's not hard to see that these numbers are just the previous numbers multiplied by $3/2$.
Footnotes
* You may have seen oversimplified depictions of spin as the electron rotating about its own axis, like how the Earth rotates to cause the day/night cycle. These depictions are all incorrect and physically unreasonable.
† Note that the Pauli exclusion principle refers to particles that are identical, or indistinguishable. Two electrons on the same atom are indistinguishable, and thus cannot share the same four quantum numbers. Conversely, if we were talking about two separate hydrogen atoms (which are distinguishable), then it is perfectly OK for each of the electrons to have the same four quantum numbers.
‡ In fact, this is not entirely true. If you raise the temperature high enough, then some of the electrons will start to occupy higher-energy states. This concept is similar to a Boltzmann distribution, but is more properly defined by Bose–Einstein statistics. However, the energy gaps between successive orbitals are very large, so at "realistic" temperatures this is not likely to happen.
