[Co(NH$_3$)$_6$]$^{3+}$
-Oxidation state of Co: $+3$
-Electronic configuration of Co$^{3+}$: $3d^6$
-NH$_3$ is a strong field ligand, causing pairing of electrons in the $t_{2g}$ orbitals.
-Distribution: $t_{2g}^6 e_g^0$
-Unpaired electrons: 0
[NiCl$_4$]$^{2-}$
-Oxidation state of Ni: $+2$
-Electronic configuration of Ni$^{2+}$: $3d^8$
-Cl$^-$ is a weak field ligand, causing no pairing of electrons.
-Distribution: $e_g^2 t_{2g}^6$
-Unpaired electrons: 2
Total unpaired electrons: $0 + 2 = 2$
Low spin tetrahedral complexes are not known.
Why is a solution of \(\text{Ni(H}_2\text{O})_6^{2+}\) green while a solution of \(\text{Ni(CN)}_4^{2-}\) is colourless? (At. No. of Ni = 28)
If the system of equations \[ x + 2y - 3z = 2, \quad 2x + \lambda y + 5z = 5, \quad 14x + 3y + \mu z = 33 \] has infinitely many solutions, then \( \lambda + \mu \) is equal to:}
The equilibrium constant for decomposition of $ H_2O $ (g) $ H_2O(g) \rightleftharpoons H_2(g) + \frac{1}{2} O_2(g) \quad (\Delta G^\circ = 92.34 \, \text{kJ mol}^{-1}) $ is $ 8.0 \times 10^{-3} $ at 2300 K and total pressure at equilibrium is 1 bar. Under this condition, the degree of dissociation ($ \alpha $) of water is _____ $\times 10^{-2}$ (nearest integer value). [Assume $ \alpha $ is negligible with respect to 1]