= Solution
Let $V=\mathbb C^4$ with its <nondegenerate> <alternating bilinear form>. Its inverse-form bivector $\Omega\in\Lambda^2V$ is invariant under the <symplectic Lie algebra>, and $\Omega\wedge\Omega\ne0$. Choose $\operatorname{vol}=\Omega\wedge\Omega/2$. The <exterior product> defines a <symmetric bilinear form> on $\Lambda^2V$ by
$$
\xi\wedge\eta=B(\xi,\eta)\operatorname{vol}.
$$
It is symmetric because two-forms commute under the wedge product. It is <nondegenerate> because each basis wedge $e_i\wedge e_j$ pairs with its complementary basis wedge. Since $B(\Omega,\Omega)=2$, the five-dimensional <orthogonal complement>
$$
W=\Omega^\perp=\Lambda_0^2V
$$
also has a <nondegenerate> <symmetric bilinear form>. This is the <primitive exterior square>, equivalently the <kernel> of <symplectic contraction of an exterior square>.
Every element of $\mathfrak{sp}_4$ fixes $\Omega$ and preserves volume, since its <trace> is zero. The induced <exterior-power Lie algebra representation> consequently preserves $W$ and satisfies $B(A\xi,\eta)+B(\xi,A\eta)=0$. Thus it gives a <Lie algebra homomorphism>
$$
\varphi:\mathfrak{sp}_4\longrightarrow\mathfrak{so}(W,B).
$$
We check <injectivity> explicitly. If $A$ acts as zero on $W$, it also kills $\Omega$, and therefore acts as zero on all of $\Lambda^2V=W\oplus\mathbb C\Omega$. For three distinct indices $i,j,k$, the coefficient of $e_k\wedge e_j$ in $A(e_i\wedge e_j)$ is $A_{ki}$, so all off-diagonal entries of $A$ vanish. Its action on $e_i\wedge e_j$ is then $(A_{ii}+A_{jj})e_i\wedge e_j$. All these sums vanish, which in characteristic zero forces every diagonal entry to vanish. Hence $A=0$.
Finally,
$$
\dim\mathfrak{sp}_4=4+3+3=10=\frac{5\cdot4}{2}=\dim\mathfrak{so}(W,B).
$$
The injective <Lie algebra homomorphism> is therefore surjective. Over $\mathbb C$, every <nondegenerate> <symmetric bilinear form> admits a basis in which its matrix is the identity, so this constructs the <exceptional isomorphism between sp4 and so5>:
$$
\boxed{\mathfrak{sp}_4(\mathbb C)\cong\mathfrak{so}_5(\mathbb C).}
$$
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