A. I wonder if every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ is a square element in ${\rm Spin}_n^{\epsilon}(q)$?
B. When $Z(\Omega_{2m}^{\epsilon}(q))\cong C_2$, is the unique element of order two of $Z(\Omega_{2m}^{\epsilon}(q))$ a square element in $\Omega_{2m}^{\epsilon}(q)$？
Notes.
Here the ground field is a finite field $F_q$ with $q$ a power of some prime $p$.
We always set $n=2m\geq 6$ and $q^m\equiv \epsilon~({\rm mod}~4)$.
Some quoted results:
If $n=2m$ and $q^m\equiv \epsilon~({\rm mod}~4)$, then $\Omega_n^\epsilon(q)$ has a centre of order 2, and the spin group has the structure $4.{\rm P\Omega}_n^\epsilon (q)$ if $m$ is odd, and the structure $2^2.{\rm P\Omega}_n^\epsilon (q)$ (necessarily with $\epsilon=+$) if $m$ is even.
If $m\geq 3$, then $P\Omega^{\epsilon}(2m,q)$ is a finite simple group.
$P\Omega_6^+(q)=PSL_4(q)$.
$P\Omega_6^-(q)=PSU_4(q)$.
Analysis:
$\pi: \Omega_6^-(3)\rightarrow P\Omega_6^-(3)$
If $x^2=-1$, then $f(x)$ is an element of order 2, however $o(x)=4$, a contradiction by Richard Lyons's notes below.
The following websites may be useful to my question.