C

过$P$作$PC\bot y$轴于$C$，过${P\’}$作$QD\bot y$轴于$D$.
$\because \angle POQ=90^{\circ}$，$\angle COD=90^{\circ}$，
$\therefore \angle POC+\angle QOC=90^{\circ}$，$\angle QOD+\angle QOC=90^{\circ}$，
$\therefore \angle POC=\angle QOD$，
$\because \angle PCO=\angle QDO=90^{\circ}$，$OP=OQ$，
在$\triangle PCO$和$\triangle QDO$中，
$\left\{\begin{array}{l}{∠PCO=∠QDO}\\{∠POC=∠QOD}\\{PO=QO}\end{array}\right.$，
$\therefore \triangle PCO$≌$\triangle QDO\left(AAS\right)$，
$\therefore QD=PC=2$，$OD=CO=3$，
$\therefore Q$的坐标是$\left(3,2\right)$.
故选：$C$.