AP EAPCET 06-Jul-2022 Shift 2 Paper

I. $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ Atomic number of $\mathrm{Fe}=26$Ground state configuration of $\mathrm{Fe}=[\mathrm{Ar}]\,3d^6 4s^2$$\mathrm{Fe}^{2+}=[\mathrm{Ar}]\,3d^6$ Since, cyanide is a powerful ligand, therefore, all the electrons will be paired.Number of unpaired electrons in $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}=0$II. $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}$Atomic number of $\mathrm{Fe}=26$Ground state configuration of $\mathrm{Fe}=[\mathrm{Ar}]\,3d^6 4s^2$$\mathrm{Fe}^{2+}=[\mathrm{Ar}]\,3d^6$ Number of unpaired electrons in $[\mathrm{Fe}(\mathrm{H}_2\mathrm{O})_6]^{2+}=4$. III. $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}$Atomic number of $\mathrm{Co}=27$Ground state configuration of $\mathrm{Fe}=[\mathrm{Ar}]\,3d^7 4s^2$$\mathrm{Co}^{2+}=[\mathrm{Ar}]\,3d^7$Number of unpaired electrons in $[\mathrm{Co}(\mathrm{H}_2\mathrm{O})_6]^{2+}=3$So, the correct order is II > III > I.