By page 80 of 'the finite simple groups' of Robert A. Wilson, we have the following the two results:
It is easy to find elements of the spin group which squre to $-1$, and hence the spin group is a proper double cover of the orthogonal group. We write ${\rm Spin}_n^{\epsilon}(q)$ for this group of shape $2.{\Omega_n^\epsilon}(q)$.
If $n$ is odd, or if $n=2m$ and $q^m\equiv -\epsilon~({\rm mod}~4)$, then $\Omega_n^\epsilon (q)$ is already simple, and the spin group has the stucture $2.\Omega_n^\epsilon (q)$. 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 $\epsilon=+$) is $m$ is even.
A. I wander if every generator of $Z({\rm Spin}_n^{\epsilon}(q))$ is a square element in ${\rm Spin}_n^{\epsilon}(q)$?
B. When $Z(\Omega_{2m}^{\epsilon}(q))\neq 1$, is the unique element of order two of $Z(\Omega_{2m}^{\epsilon}(q))$ a square element in $\Omega_{2m}^{\epsilon}(q)$？