Optical isomes $= 0$
If the complex $MA_2B_2$ is $dsp^2$ hybridised then the shape of this complex is square planar
Both isomers are optically inactive due to the presence of plane of symmetry.
Optical isomers $= 0$
${\left[\mathrm{Fe}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Co}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{3+},}$
${\left[\mathrm{Cu}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+},\left[\mathrm{Cr}\left(\mathrm{H}_2 \mathrm{O}\right)_6\right]^{2+}}$
($en=$ એથીલીનડાયએમાઈન )