No this is not possible.
When people talk about the distribution of eigenvalues they mean the expected eigenvalues when the unitary matrices are sampled with respect to some measure.
If you have a fixed unitary matrix $U$ then you can have absolutely any distribution of eigenvalues that lie on the unit circle in $\mathbb{C}$. For instance $\alpha I$ for any complex number $\alpha$ with $|\alpha|=1$ is a unitary matrix. It has $n$ eigenvalues all equal to $\alpha$.
If you don't like degeneracy then take any $n$ distinct $\alpha_i$ on the unit circle and then $\mathrm{Diag}(\alpha_1, \dots, \alpha_n)$, the diagonal matrix with $\alpha_1,\dots,\alpha_n$ on the diagonal is a unitary matrix with eigenvalues $\alpha_1, \dots, \alpha_n$.
If you don't like diagonal matrices then take your favorite unitary $V$ and then
$$
V \mathrm{Diag}(\alpha_1, \dots, \alpha_n) V^*
$$
is another unitary matrix with eigenvalues $\alpha_1,\dots, \alpha_n$.