When creating term symbols, how do you know if the angular momentum $L$ is antisymmetric of symmetric? For example I'm trying to get the term symbol of $(1s)^{2}(2s)^{2}(2p)^2$ . 
In the answers they state the following:

The combination of angular momenta $L_1 = L_2 = 1$ gives $L = 2$
  (symmetric), $L = 1$ (antisymmetric) and $L = 0$ (symmetric). This must be combined with the spin wave function of opposite symmetry, thus $^1D_2,
^3P_{0, 1, 2}$  and $^1S_0.$

I totally understand this, except for how they assign symmetric and antisymmetric to the angular momenta. In the previous exercise I only had $L = 0$ and they said it was symmetric and antisymmetric. So how do I know if the angular momentum is symmetric or antisymmetric?
 A: For two particles with the same angular momentum $\ell_1=\ell_2=\ell$, the permutation symmetry follows immediately from the symmetries of the Clebsch-Gordan coefficients:
$$
C^{LM}_{\ell m_1;\ell m_2}=(-1)^{2\ell+L} C^{LM}_{\ell m_2;\ell m_1}
$$
so that (in accordance with the answer of @Superciocia), the symmetric states have $L$ even and the antisymmetric ones has $L$ odd.  This holds for any $\ell$.
The situation is much more complicated if you have $3$ or more particles. For instance, in the coupling of $3$ states with $\ell=1$, there are three sets of states with total $L=1$. One of them is symmetric but the other two have mixed symmetry.
One of the sets with mixed symmetry comes from coupling the first and second particles to $L_{12}=0$, then coupling this to the third to get $L=1$.  It’s easy to see the resulting states are neither symmetric nor antisymmetric (although they are antisymmetric under the exchange $1\leftrightarrow 2$.)
The symmetric states with $L=1$ are in fact linear combinations of $L_{12}=1$ and $L_{12}=2$ states.
A: The orbital angular momentum quantum number is just a number so it cannot be symmetric or antisymmetric.
Also, in this context you should specify that the "symmetry" is with respect to particle re-labelling/exchange. For Pauli exclusions etc.
The actual proof is (I think) quite mathematical with Slater determinants, but the general rule of thumb is that:

For odd numbers for the total angular momentum $(L= 1,3,5, ...)$ the
  spatial wavefunction is antisymmetric upon particle exchange.

You can test it though.  
I worked out the angular momentum states of your possible total $L$ systems.
In the following, on the LHS there will be the basis in the total (composite) angular momentum $|L, m_L\rangle$, while on the RHS there will be the state in the individual angular momentum basis $|\ell, m_\ell\rangle$. Each term on the RHS is a tensor product between particle $A$ and particle $B$, so $|1,0\rangle |1,-1\rangle$ means "$|1,0\rangle_A \otimes|1,-1\rangle_B$ ". In this specific case $\ell = 1$, always, as both electrons are the in $p$ orbital.
$L= 0$:
\begin{align}
\left|0,0\right> &= \frac1{\sqrt3} \left(
\big|1,1\big>\big|1,-1\big>
~~+~~
\big|1,-1\big>\big|1,1\big>
~~-~~
\big|1,0\big>\big|1,0\big>
\right)
\end{align}
If you swap labels $A\leftrightarrow B$, each term stays exactly the same, so $|0,0\rangle$ is symmetric.
$L =1$:
\begin{align}
\lvert1,1\rangle & = \frac{1}{\sqrt 2} \left( \lvert 1,0\rangle\lvert1,1\rangle - \lvert1,1\rangle\lvert1,0\rangle \right) \\
\lvert1,0\rangle & = \frac{1}{\sqrt 2} \left( \lvert1,-1\rangle\lvert1,1\rangle - \lvert1,1\rangle\lvert1,-1\rangle \right) \\
\lvert1,-1\rangle & = \frac{1}{\sqrt 2} \left( \lvert1,0\rangle\lvert1,-1\rangle - \lvert1,-1\rangle\lvert1,0\rangle\right)
\end{align}
If you swap labels $A\leftrightarrow B$, each line gets an overall minus sign, so $|1,*\rangle$ is antisymmetric.
$L=2$:
You can check every case yourself, but for instance:
\begin{align}
\left|2,0\right> &= \frac1{\sqrt6} \left(
\big|1,1\big>\big|1,-1\big>
~~+~~
\big|1,-1\big>\big|1,1\big>
~~+~~
\sqrt4\cdot
\big|1,0\big>\big|1,0\big>
\right)
\end{align}
If you swap labels $A\leftrightarrow B$, each term stays exactly the same, so $|2,0\rangle$ is symmetric.
