Killing form on $\mathfrak{sp}(2n)$

I have the same question as this one from a long time ago. Is there an easy way to see that the Killing form on $\mathfrak{sp}(2n)$ is $\kappa(x,y) = (4n+2) \mathrm{tr}(xy)$?

For example, the Killing form for $\mathfrak{sl}(n)$ can be found from the easier to calculate case of $\mathfrak{gl}(n)$ using the decomposition $$\mathfrak{gl}(n) = \mathfrak{sl}(n) \oplus \mathbb{C}.$$ I don't know a similar argument for $\mathfrak{sp}(2n)$ though.

For the computation of the Killing form of the classical Lie algebras, the crucial thing is to note that a bilinear form $\rho: {\mathfrak g}\times {\mathfrak g}\to {\mathbb C}$ which is non-degenerate and associative - i.e. $\rho([X,Y],Z)=\rho(X,[Y,Z])$ - is unique up to scalar. The Killing form is associative in general and non-degenerate for semisimple Lie algebras, while for the trace form you can proceed directly.

With this at hand, it suffices to find any non-zero value of the forms under consideration to compute the scalar factor between them. For example, it is convenient to pick elements from a Cartan subalgebra because for them the Killing form is easy to compute.

Does this help?