= Solution
Let $U$ be a four-dimensional <symplectic vector space>. The <symplectic contraction of an exterior square> splits $\Lambda^2U$ into its five-dimensional <primitive exterior square> $W$ and the invariant line spanned by the inverse symplectic bivector $\Omega$. Choose $\operatorname{vol}=\Omega\wedge\Omega/2$. Exterior multiplication defines a <symmetric bilinear form> on $\Lambda^2U$ by $\xi\wedge\eta=B(\xi,\eta)\operatorname{vol}$; it is symmetric because both degrees are two. In a symplectic basis, $\Omega\wedge\Omega=2\operatorname{vol}\ne0$, and $W=\Omega^\perp$, so the restriction to $W$ is <nondegenerate>.
The <symplectic Lie algebra> $\mathfrak{sp}_4$ preserves this form, giving a homomorphism $\mathfrak{sp}_4\to\mathfrak{so}(W)$. It is injective: an element acting trivially on $W$ also acts trivially on $\Omega$, hence on all of $\Lambda^2U$. In a four-vector basis, the identities $Ae_i\wedge e_j+e_i\wedge Ae_j=0$ first force every off-diagonal coefficient of $A$ to vanish, then force $A_{ii}+A_{jj}=0$ for every pair. Over $\mathbb C$ these equalities imply $A=0$.
Both Lie algebras have <dimension> ten: $\dim\mathfrak{sp}_4=4\cdot5/2$ and $\dim\mathfrak{so}_5=5\cdot4/2$. The injective map is therefore an <isomorphism>, proving the <exceptional isomorphism between sp4 and so5>:
$$
\boxed{\mathfrak{sp}_4(\mathbb C)\cong\mathfrak{so}_5(\mathbb C).}
$$
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