Due the fact that the fermions are the "block particles" and the bosons are the "carriers" I just came out with the question that, why the "block particle" have half-integer spin and the "carriers" have an integer spin?
[Physics] Why bosons have integer spin and fermions have half-integer ones
bosonsfermionsparticle-physicsspecial-relativityspin-statistics
Related Solutions
In contrast with the previous incorrect answers that I hadn't noticed, there isn't any ambiguity or confusion about the Bose-Einstein or Fermi-Dirac statistics for composite systems such as atoms.
A particle – elementary or composite particles – that contains an even number of elementary (or other) fermions is a boson; if it contains an odd number, it is a fermion. Bosons always have an integer spin; fermions always have a half-integer spin. By the spin-statistics theorem, the wave function of two bosons is invariant under their exchange but it is antisymmetric under the exchange of two identical fermions. These two rules are theorems for elementary particles and assuming this theorem, it's also trivial to prove these statements for composite particles.
In particular, for a neutral atom, the numbers of protons and electrons are equal so their total parity is even. That's why only neutrons matter. An isotope with an even number of neutrons is a boson (the whole atom: e.g. helium-4); an isotope with an odd number of neutrons is a fermion (the whole atom: e.g. helium-3).
That doesn't mean that composite bosons exhibit all the same physical phenomena like superfluidity that we may observe with some bosons.
The magnetic moment of a charged "elementary enough" particle scales like $1/m$ where $m$ is the mass of the particle. That's why the magnetic moment of the protons, neutrons, and nuclei are about 1,000-2,000 times smaller than the magnetic moments of the electrons. That's why the nuclear spins are largely negligible for the behavior of the atom in a magnetic field.
This is no contradiction because the whole atoms have a much larger magnetic moments than the nuclei separately – because of the neutrons: atoms are not "elementary enough" in this definition. Both the electrons' spins and their orbital angular momentum contribute to an atom's magnetic moment. Also, there exist a higher number of states because an atom is a typical example of the "addition of several angular momenta". The tensor product Hilbert space may be decomposed as a direct sum of Hilbert spaces with fixed values of the total angular momentum. The degeneracy and the magnetic moment of these components depend on the total angular momentum i.e. on the relative orientation of the nucleus-based and electron-related angular momenta.
In effect, the spectral lines of the whole atom exhibit the so-called hyperfine structure. Up to some approximation, the nuclear spin may be totally ignored. But when the considerations from the previous paragraph are properly account for, each spectral line is actually split to several nearby (3 or so orders of magnitude closer to each other) finer spectral lines, each of which corresponds to a different value of the total angular momentum of the whole atom (or, equivalently, a different value of $\vec J_{\rm electrons}\cdot \vec J_{\rm nucleus}$).
Spin-1/2 admits first order equations simply because $ (\mathbf{1/2,1/2})\otimes (\mathbf{0,1/2}) $ contains the representation $(\mathbf{1/2,0})$ so that a linear equation for free particles can be written (i.e. it contains a derivative acting on one field and returning one field). The first term in the product is the derivative that transforms as a Lorentz vector $(\mathbf{1/2,1/2})$, whereas $(\mathbf{0,1/2})$ and $(\mathbf{1/2,0})$ are left and right-handed spinors respectively. The clebsch-Gordan coefficients are nothing but that the gamma matrices.
Clearly the same is forbidden for the integer spins. For a scalar is trivial. For a spin-1 field $(\mathbf{1/2,1/2})$ one has that $( \mathbf{ 1/2, 1/2}) \otimes ( \mathbf{1/2,1/2}) $ does not contain $(\mathbf{1/2,1/2})$, and so on for higher integer spins. It is basically the group structure of Lorentz symmetry that forbids first order eq. for integer spins.
Best Answer
The fact that bosons have integer spin whereas fermions have half-integer is actually a result from the so-called spin-statistics theorem.
The definition of bosons and fermions is not in terms of spin, it is in terms of symmetry of the wave function under the exchange of particles. The spin-statistics theorem says that the wave function of an integer spin identical particles' system is symmetric under the exchange of particles and therefore those are bosons. On the other hand the wave function of an half integer identical particles' system is antisymmetric under the exchange of particles and thus they are fermions.
If you are interested in knowing why force carriers are bosons you can check this: Why are all force particles bosons?