This couldn't be achieved even by $\Sigma_3$ sentences. First note that $L_{\omega_1^{CK}}$ (as any other model of $\mathsf{KP}\omega+L=V$) satisfies the scheme of $\Sigma_3$-reflection:
$$\varphi(\vec{p})\to \exists a \;(\mathsf{Trans}(a)\land \vec{p}\in a\land (\varphi(\vec{p}))^a),\text{ where $\varphi$ is $\Sigma_3$}.$$
And note that there is a $\Pi_2$ sentence $F$ such that for each transitive set $a$, the sentence $F$ is true in $a$ iff $a$ is of the form $L_{\omega(1+\alpha)}$.
Henceforth for any $\Sigma_3$ sentence $\varphi$, if $L_{\omega_1^{CK}}\models\varphi$ then there is a transitive set $a\in L_{\omega_1^{CK}}$ such that $a\models \varphi\land F$ which means that $a$ is of the form $L_{\alpha}$, where $\alpha<\omega_1^{CK}$ and $L_{\alpha}\models \varphi$.

Note that $\omega_1^{CK}$ is the least $\alpha$ such that $L_{\alpha}$ is a model of $\mathsf{KP}\omega-\mathsf{Foundation}$. And the theory $\mathsf{KP}\omega-\mathsf{Foundation}$ could be axiomatized by a single $\Pi_3$ sentence (the only axiom that isn't $\Pi_2$ is the axiom of $\Sigma_1$-collection). Note that everywhere in this answer the classes $\Pi_n$ were understood as consisting of formulas that start with an unbounded quantifier prefex $\vec{\forall}\vec{x}_1\ldots \vec{Q}\vec{x}_n$ followed by a $\Delta_0$ formula. However if we switch to the classes $\hat\Pi_n$ defined in terms of alternation depth of unbounded quantifiers (bounded quantifiers could appear anywhere) the answer changes. The axiom of $\Sigma_1$-collection is a $\hat\Pi_2$-sentence and hence $\mathsf{KP}\omega-\mathsf{Foundation}$ is $\hat\Pi_2$-axiomatizable.

However I don't know whether there is a $\hat\Sigma_2$ sentence that "captures" $L_{\omega_1^{CK}}$. Non-existence of $\hat\Pi_1$ sentence is trivial due to downward-absoluteness. And non-existence of $\hat\Sigma_1$ sentence follows from the fact that $\mathsf{KP}\omega$ proves $\Sigma$-reflection (the class $\Sigma$ is exactly $\hat\Sigma_1$).