The d- and f- Block Elements Set-7

$4.90\ \mathrm{BM}$

$3.87\ \mathrm{BM}$

$2.83\ \mathrm{BM}$

$5.92\ \mathrm{BM}$

$[Ni(CN)_4]^{2-}:0\ \text{unpaired};\ [NiCl_4]^{2-}:2\ \text{unpaired}$

$[Ni(CN)_4]^{2-}:2\ \text{unpaired};\ [NiCl_4]^{2-}:0\ \text{unpaired}$

$[Ni(CN)_4]^{2-}:1\ \text{unpaired};\ [NiCl_4]^{2-}:1\ \text{unpaired}$

$[Ni(CN)_4]^{2-}:0\ \text{unpaired};\ [NiCl_4]^{2-}:0\ \text{unpaired}$

$2.08\times10^{2}\ \mathrm{kJ\ mol^{-1}}$

$3.33\times10^{2}\ \mathrm{kJ\ mol^{-1}}$

$1.56\times10^{2}\ \mathrm{kJ\ mol^{-1}}$

$2.49\times10^{2}\ \mathrm{kJ\ mol^{-1}}$

$Fe^{2+}$, because 3d orbitals are more contracted giving larger $\Delta_o$ than $Ru^{2+}$ which favours pairing.

Both are equally likely because $3d$ and $4d$ ions have nearly the same $\Delta_o$ and pairing energy $P$ for the same ligand.

$Ru^{2+}$, because 4d orbitals are more diffuse leading to larger $\Delta_o$ and generally lower pairing energy $P$, so electrons pair in $t_{2g}$.

$Ru^{2+}$, because 4d ions have higher pairing energy $P$ which forces electrons into $t_{2g}$ (low spin).

$La^{3+}$ and $Ce^{3+}$

$Yb^{3+}$ and $Lu^{3+}$

$Nd^{3+}$ and $Sm^{3+}$

$Gd^{3+}$ and $Tb^{3+}$

$5.92\ \text{BM}$

$1.73\ \text{BM}$

$3.87\ \text{BM}$

$4.90\ \text{BM}$

$[FeF_6]^{3-}$ — F$^-$ is a weak-field ligand, giving a high-spin $d^5$ configuration with electrons in antibonding $e_g$ orbitals, lengthening the bonds.

$[Fe(CN)_6]^{3-}$ — CN$^-$ is a strong-field ligand; the resulting low-spin arrangement puts more electron density into antibonding $e_g$ orbitals, lengthening the bonds.

Both have nearly the same Fe–ligand distance because bond lengths mainly depend on ionic radii, not ligand field strength.

$[FeF_6]^{3-}$ — F$^-$ gives weaker, less directional bonding, so steric/electronic factors make Fe–F bonds longer.

High-spin: $-1.6\,\Delta_o$, Low-spin: $-0.4\,\Delta_o$

High-spin: $-0.4\,\Delta_o$, Low-spin: $-1.6\,\Delta_o$

High-spin: $-0.4\,\Delta_o$, Low-spin: $-2.0\,\Delta_o$

High-spin: $-0.4\,\Delta_o$, Low-spin: $-2.4\,\Delta_o$

Low-spin (all electrons paired in $t_{2g}$)

High-spin (maximum unpaired electrons)

Intermediate-spin (neither fully high-spin nor low-spin)

Cannot be predicted without knowing ligand identity

A: low-spin; B: high-spin

A: high-spin; B: low-spin

Both A and B low-spin

Both A and B high-spin

$2.83\ \mu_B$

$3.87\ \mu_B$

$4.90\ \mu_B$

$5.92\ \mu_B$

$[Co(H_2O)_6]^{2+}$ has CFSE $-0.8\Delta_o$ and spin-only $\mu\approx3.87\ \mu_B$, whereas $[Fe(H_2O)_6]^{2+}$ has CFSE $-0.4\Delta_o$ and $\mu\approx4.90\ \mu_B$.

$[Fe(H_2O)_6]^{2+}$ has larger CFSE magnitude ($-0.8\Delta_o$) and a lower magnetic moment than the cobalt complex.

Both complexes have identical CFSE but differ in magnetic moments only due to different pairing energies.

$[Co(H_2O)_6]^{2+}$ has CFSE $-0.4\Delta_o$ and a higher magnetic moment than $[Fe(H_2O)_6]^{2+}$.

$[NiCl_4]^{2-}$ is square-planar and diamagnetic (0 unpaired); $[Ni(CN)_4]^{2-}$ is tetrahedral and paramagnetic (2 unpaired).

$[NiCl_4]^{2-}$ is square-planar and paramagnetic (1 unpaired); $[Ni(CN)_4]^{2-}$ is tetrahedral and paramagnetic (2 unpaired).

$[NiCl_4]^{2-}$ is tetrahedral and diamagnetic (0 unpaired); $[Ni(CN)_4]^{2-}$ is square-planar and paramagnetic (2 unpaired).

$[NiCl_4]^{2-}$ is tetrahedral and paramagnetic with 2 unpaired electrons; $[Ni(CN)_4]^{2-}$ is square-planar and diamagnetic (0 unpaired).

$1\ \mu_B$

$4\ \mu_B$

$10\ \mu_B$

$0\ \mu_B$

L1: high-spin, $\mu\approx4.90\ \mu_B$; L2: low-spin, $\mu\approx0\ \mu_B$; L3: low-spin, $\mu\approx0\ \mu_B$.

L1 and L2 low-spin ($\mu\approx0\ \mu_B$), L3 high-spin ($\mu\approx4.90\ \mu_B$).

All three are high-spin: L1 $4.90\ \mu_B$, L2 $4.90\ \mu_B$, L3 $4.90\ \mu_B$.

Only L3 is low-spin ($\mu\approx0\ \mu_B$); L1 and L2 are high-spin (both $4.90\ \mu_B$).