$\def\eq{\leftrightarrow}\def\sset{\subseteq}$An example of such a formula is
$$\phi(p,q)=(\Box p\to p)\land(p\to(q\eq\neg\Box q)).$$

**Lemma.** The formula $\phi$ satisfies condition 1.

**Proof:**
Let $(F,R)$ be a finite frame. For all $i\in\omega$, let $W_i=\{w\in F:w\vDash\Box^i\bot\}$, where $\Box^i=\underbrace{\Box\cdots\Box}_i$. Since $W_0\sset W_1\sset W_2\sset\cdots$ and $F$ is finite, there is an $n$ such that $W_n=\bigcup_{i\in\omega}W_i$. Note that $W_n$ is the largest converse well-founded generated subframe of $F$. By well-founded recursion, there exists a unique $X\sset W_n$ such that
$$x\in X\iff\exists y\,(x\mathrel Ry\land y\notin X)$$
for all $x\in W_n$. Then $F\vDash\phi$ under the valuation
$$\begin{align*}w\vDash p&\iff w\in W_n,\\w\vDash q&\iff w\in X.\end{align*}$$

$W_n$ is defined in $F$ by the ground formula $\Box^n\bot$. To see that $X$ is also definable by a ground formula, put $\xi_0=\bot$ and $\xi_{i+1}=\neg\Box\xi_i$ for each $i$ (that is, $\xi_{2i}=(\Diamond\Box)^i\bot$ and $\xi_{2i+1}=(\Diamond\Box)^i\Diamond\top$). By induction on $i$, we can show
$$\forall x\in W_i\:(x\in X\iff F,x\vDash\xi_i),$$
hence $X$ is definable by the formula $\Box^n\bot\land\xi_n$. Thus, the substitution $\sigma(p)=\Box^n\bot$, $\sigma(q)=\xi_n$ satisfies $F\vDash\sigma(\phi)$. **QED**

To see that $\phi$ satisfies 2, note that condition 2 is equivalent to “$\phi$ is not unifiable”. Since Löb’s rule $\Box\alpha\to\alpha\mathrel/\alpha$ is admissible in **K**, any unifier of $\phi$ also unifies $p$, hence $q\eq\neg\Box q$; but the latter is not unifiable, as it is not satisfiable in nonempty reflexive frames. Let me spell out a more explicit argument for completeness:

**Lemma.** The formula $\phi$ satisfies condition 2.

**Proof:**
Let $(F,R)=(\mathbb N,\{(n+1,n):n\in\mathbb N\})$ be the infinite descending irreflexive intransitive chain, and assume for contradiction that $F\vDash\sigma(\phi)$ for a ground substitution $\sigma$. Then $F\models\sigma(\Box p\to p)$ implies $n\models\sigma(p)$ for all $n\in\mathbb N$ by induction on $n$, hence also $F\models(\sigma(q)\eq\neg\Box\sigma(q))$. Thus,
$$n+1\models\sigma(q)\iff n+1\nvDash\Box\sigma(q)\iff n\nvDash\sigma(q).$$
On the other hand, for every ground formula $\alpha$, we can prove by induction on the complexity of $\alpha$ that $\{n\in\mathbb N:n\models\alpha\}$ or its completement is finite, thus for all sufficiently large $n\in\mathbb N$, we have
$$n+1\models\sigma(q)\iff n\models\sigma(q).$$
This is a contradiction. **QED**

The argument actually works for all logics between $\mathbf K$ and $\mathbf{Alt_1=K}\oplus\Diamond p\to\Box p$.