Let $(q)_n=(1-q)(1-q^2)\cdots(1-q^n)$ with $(q)_0:=1$. Define a $q$-exponential by
$$e_q(z)=\sum_{n\geq0}\frac{z^n}{(q)_n}.$$
There is a notion of $q$-Eulerian polynomials of type $A$, see the reference. I like to introduce **$q$-Eulerian polynomial of type B** via the generating function
$$\sum_{n\geq1}B_n(t,q)\frac{z^n}{(q)_n}
=\frac{(e_q(z)-e_q(tz))\cdot(e_q(tz)+te_q(z))}{e_q(2tz)-te_q(2z)}.$$
Now, expand $B_n(t,q)$ as a polynomial
$$B_n(t,q)=\sum_{k=0}^nB_{n,k}(q)t^k$$
and call $B_{n,k}(q)$ **$q$-Eulerian numbers type B**. The first few terms are:
\begin{align} B_1(t,q)&=1+t, \\
B_2(t,q)&=1+(2q+4)t+t^2, \\
B_3(t,q)&=1+(7q^2+7q+9)t+(7q^2+7q+9)t^2+t^3.
\end{align}
Here is an earlier MO problem. This time, I'm interested in a specialized aspect of it. For instance, $B_n(t,1)$ become the ordinary Eulerian polynomials of type $B$ whose coefficients are listed at OEIS. On the other hand, the polynomials $B_n(t,-1)$ do not appear anywhere. You may look at the first few of these: $B_1(t,-1)=1, B_2(t,-1)=1+t, B_3(t,-1)=1+2t+t^2$,
$$B_4(t,-1)=1+9t+9t^2+t^3 \qquad\text{and} \qquad
B_5(t,-1)=1+12t+22t^2+12t^3+t^4.$$

QUESTION 1.Is there some interpretation of the coefficients in $B_n(t,-1)$?

QUESTION 2.Can you provide a proof for the unimodality of $B_n(t,-1)$?