Explanation

(b)
Maximum oxidation state of transition metals =number of electrons in ($n - 1)d$ orbitals + number of electrons in $ns$ orbital.
The electronic configuration of
Os=[xe]$4f^{14},5d^{6}.6s^{2}$
$\therefore\$Maximum oxidation state 6+2=8
$\therefore$The highest oxidation state exhibited by transition metal is +8 $e.g.,\ {OsO}_{4}$.