= Solution
Use the <basis> <matrices> $M_{ab}=E_{ab}-E_{ba}$ from <4/a/solution>. Define the <Pauli matrices> and the two-by-two identity by
$$
\sigma_1=\begin{pmatrix}0&1\\1&0\end{pmatrix},\qquad
\sigma_2=\begin{pmatrix}0&-i\\i&0\end{pmatrix},\qquad
\sigma_3=\begin{pmatrix}1&0\\0&-1\end{pmatrix},\qquad I=I_2.
$$
The following five <Hermitian> four-by-four <matrices>, written as <Kronecker products>, realize a Euclidean <Clifford algebra>:
$$
\gamma_1=\sigma_1\otimes I,\quad
\gamma_2=\sigma_2\otimes I,\quad
\gamma_3=\sigma_3\otimes\sigma_1,\quad
\gamma_4=\sigma_3\otimes\sigma_2,\quad
\gamma_5=\sigma_3\otimes\sigma_3,
\qquad \gamma_a\gamma_b+\gamma_b\gamma_a=2\delta_{ab}I_4.
$$
Define the <Lie algebra representation> on $\mathbb C^4$ by
$$
\boxed{\rho(M_{ab})=\frac14[\gamma_a,\gamma_b]=\frac12\gamma_a\gamma_b\qquad(a<b).}
$$
These are <anti-Hermitian matrices>, appropriate to the real compact <Lie algebra>. For complete explicitness, all ten <representation> <matrices> are
$$
\boxed{\begin{array}{c|c||c|c}
M_{12}&\tfrac i2\sigma_3\otimes I&M_{13}&-\tfrac i2\sigma_2\otimes\sigma_1\\
M_{14}&-\tfrac i2\sigma_2\otimes\sigma_2&M_{15}&-\tfrac i2\sigma_2\otimes\sigma_3\\
M_{23}&\tfrac i2\sigma_1\otimes\sigma_1&M_{24}&\tfrac i2\sigma_1\otimes\sigma_2\\
M_{25}&\tfrac i2\sigma_1\otimes\sigma_3&M_{34}&\tfrac i2 I\otimes\sigma_3\\
M_{35}&-\tfrac i2 I\otimes\sigma_2&M_{45}&\tfrac i2 I\otimes\sigma_1
\end{array}}
$$
Each table entry is the image under $\rho$ of the <basis> <matrix> immediately to its left. Since the two-by-two factors are specified above, these <Kronecker products> specify every entry of each four-by-four <matrix>.
To verify the <representation> law, the <Clifford algebra> relations imply
$$
[\rho(M_{ab}),\gamma_c]=\delta_{bc}\gamma_a-\delta_{ac}\gamma_b.
$$
Apply this identity to $\rho(M_{cd})=\tfrac14[\gamma_c,\gamma_d]$ and use the <commutator product rule>. The result is
$$
[\rho(M_{ab}),\rho(M_{cd})]
=\delta_{bc}\rho(M_{ad})-\delta_{ac}\rho(M_{bd})
-\delta_{bd}\rho(M_{ac})+\delta_{ad}\rho(M_{bc}),
$$
exactly the <Lie bracket> relation in <4/a/solution>. Thus $\rho$ is a <Lie algebra representation>.
It remains to identify its <highest weight>, rather than just its <dimension>. The commuting <Hermitian> <Cartan subalgebra> operators are
$$
-i\rho(M_{12})=\tfrac12\sigma_3\otimes I,\qquad
-i\rho(M_{34})=\tfrac12I\otimes\sigma_3.
$$
Their simultaneous <eigenvalues> are $(x,y)=(\pm\tfrac12,\pm\tfrac12)$. Converting with $(a,b)=(x-y,2y)$ gives exactly the four <weights> of $V(0,1)$ in <4/b/solution>. A state with $(x,y)=(\tfrac12,\tfrac12)$ has <highest weight> $(0,1)$, so this <Clifford realization of the so5 spin representation> is the requested $r$.
Finally, $\rho(M_{12})^2=-I_4/4$, so
$$
\boxed{e^{2\pi\rho(M_{12})}=-I_4,\qquad e^{2\pi M_{12}}=I_5.}
$$
This explicitly exhibits the distinction between the <spinor representation> of $\mathfrak{so}(5)$, or of $\operatorname{Spin}(5)$, and a single-valued <group representation> of $SO(5)$.
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