Question

If the crystal field stabilisation energy in octahedral field for Ni is $\Delta_0$, then under same condition, the 18. value of CFSE for Pd will be

• A. 1.5 $\Delta_0$
• B. $\Delta_0$
• C. 0.9 $\Delta_0$
• D. 0.5 $\Delta_0$

Answer 1

Answer: (A)

1.5 $\Delta_0$

Answer 2

Crystal Field Stabilization Energy ($\Delta_0$) generally increases down a group in the transition series for the same ligands and oxidation state. This is because the d-orbitals become more diffuse and experience greater repulsion from the ligands, leading to a larger splitting.

For transition metals, the $\Delta_0$ values typically increase by about 50% (1.5 times) when moving from the 3d series to the 4d series, and by another 10-20% (1.1-1.2 times) when moving from the 4d series to the 5d series.

Nickel (Ni) is a 3d transition metal, and Palladium (Pd) is a 4d transition metal, both belonging to Group 10.

If the CFSE for Ni (3d series) is $\Delta_0$, then for Pd (4d series) under the same conditions, it will be approximately 1.5 times that value.

CFSE (Pd) $\approx$ 1.5 $\times$ CFSE (Ni)

CFSE (Pd) $\approx$ 1.5 $\Delta_0$