$\mathrm{Co}^{+3}=3 \mathrm{d}^{6} 4 \mathrm{s}^{0} 4 \mathrm{p}^{0}$
$\because$ in presence of strong field ligand, pairing of electrons occurs so in this complex no unpaired electron is present and
it is low spin complex.
(પરમાણુ ક્રમાંક $Co=27$)
$(I)$ $\left[\mathrm{Cr}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}\right] \mathrm{Br}_{2}$
$(II)$ $ \mathrm{Na}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]$
$(III)$ $\mathrm{Na}_{3}\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]\left(\Delta_{0}>\mathrm{P}\right)$
$(IV)$ $\left(\mathrm{Et}_{4} \mathrm{N}\right)_{2}\left[\mathrm{CoCl}_{4}\right]$