AP EAPCET 20 May 2024 Shift 1 Solved Paper

In an octahedral complex, the $t_{2g}$ orbitals are lower in energy and the $e_g$ orbitals are higher in energy.For $[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$, Ni is in the $+2$ oxidation state, so the ion is $d^8$.$\mathrm{NH}_3$ is a weak-field ligand, so this is a high-spin octahedral complex.The $d^8$ configuration is $t_{2g}^{6}e_g^{2}$.Thus the two unpaired electrons occupy the $e_g$ orbitals, and there are no unpaired electrons in the $t_{2g}$ orbitals.The complex is paramagnetic, so the correct option is (3), $[\mathrm{Ni}(\mathrm{NH}_3)_6]^{2+}$.Checking the other options: $[\mathrm{Fe}(\mathrm{CN})_6]^{4-}$ is $\mathrm{Fe}^{2+}$ $d^6$ low-spin with $t_{2g}^{6}e_g^{0}$ and is diamagnetic.$[\mathrm{Fe}(\mathrm{CN})_6]^{3-}$ is $\mathrm{Fe}^{3+}$ $d^5$ low-spin with $t_{2g}^{5}e_g^{0}$; it is paramagnetic but has unpaired electrons in $t_{2g}$.$[\mathrm{Zn}(\mathrm{NH}_3)_6]^{2+}$ is $\mathrm{Zn}^{2+}$ $d^{10}$ and is diamagnetic.