= Solution
Let $r=V(0,1)$ be the four-dimensional <spinor representation>. The <tensor product representation> splits into its <symmetric square> and <exterior square>, with <dimensions> ten and six. If $v$ is a <highest-weight vector> of $r$, then $v\otimes v$ is a <highest-weight vector> of <weight> $(0,2)$ in the <symmetric square>. The <irreducible representation> with that <highest weight> has <dimension> ten by <4/b/solution>, exhausting the <symmetric square>:
$$
S^2r\cong V(0,2)=\mathbf{10}.
$$
For the <exterior square>, add each pair of distinct <weights> from <4/b/solution>. The resulting <multiset> of <weights> is
$$
(1,0),\quad(-1,2),\quad(1,-2),\quad(-1,0),\quad(0,0)^{[2]}.
$$
Its maximal <weight> $(1,0)$ is the <highest weight> of a five-dimensional <irreducible representation>. It accounts for the four nonzero <weights> and one zero <weight>, leaving a single zero <weight>. By <complete reducibility of semisimple Lie algebra representations>, that remaining one-dimensional summand is the <trivial representation>. Thus $\Lambda^2r\cong\mathbf5\oplus\mathbf1$, and the <spinor tensor square for so5> is
$$
\boxed{r\otimes r\cong V(0,0)\oplus V(1,0)\oplus V(0,2)
=\mathbf1\oplus\mathbf5\oplus\mathbf{10}.}
$$
The <dimension> check is $16=1+5+10$. The <trivial representation> lies in the <exterior square>, reflecting the invariant nondegenerate alternating <bilinear form> of this <pseudoreal representation>. Although $r$ itself does not descend to $SO(5)$, its <tensor square> does: the central element $-1\in\operatorname{Spin}(5)$ acts as $(-I_4)\otimes(-I_4)=I_{16}$.
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