${\left[\text{Ru}\right({\text{NH}}_3{)}_6]}^{3+}$ is an octahedral, ${\text{d}}^5$ low-spin complex, how many unpaired electrons does this complex have?

The correct option is B $1$
${\left[\text{Ru}\right({\text{NH}}_3{)}_6]}^{3+}:$ Octahedral

${}_{44}\text{Ru}:\left[\text{Kr}\right]4{\text{d}}^{7}5{\text{s}}^1$


${\text{Ru}}^{+}:\text{[Kr]}4{\text{d}}^{5}5{\text{s}}^0$


$\therefore$ Due to presence of strong field ligand electrons get paired, only $1$ unpaired electron is present.