Question

If the value of CFSE for Ni is $ {\vartriangle _0} $ then it is for Pd should be?
(A) 1.1 $ {\vartriangle _0} $
(B) 0.5 $ {\vartriangle _0} $
(C) 1.5 $ {\vartriangle _0} $
(D) None of the above

Answer 1

Hint : To know about the crystal field splitting energy (CFSE) it is important to know the crystal field theory of coordination compounds. According to crystal field theory bonding in coordination compounds is purely electrostatic which means that there is no overlapping, it is only the attraction between positively charged central metal atoms and negatively charged ligands.

Complete Step By Step Answer:
When the transition metal forms a coordination compound, the d-orbitals lose its degeneracy which means it splits into different energy levels. We know that there are five subshells in the d-orbitals, they are $ {d_{xy}} $ , $ {d_{yz}} $ , $ {d_{xz}} $ , $ {d_{{x^2} - {y^2}}} $ and $ {d_{{z^2}}} $ . The $ {d_{xy}} $ , $ {d_{yz}} $ , $ {d_{xz}} $ orbitals are together called as $ {t_{2g}} $ orbitals and $ {d_{{x^2} - {y^2}}} $ , $ {d_{{z^2}}} $ are together called as $ {e_g} $ orbitals. All $ {t_{2g}} $ orbitals are present between the axes and the $ {e_g} $ orbitals are present along the axes. The $ {t_{2g}} $ are the lower energy orbitals and $ {e_g} $ is the higher energy orbitals.
Hence the energy difference between these two orbitals is known as Crystal Field Splitting Energy (CFSE).
The electronic configuration of Ni is $ [Ar] $ $ 3{d^8} $ $ 4{s^2} $ .
The electronic configuration of Pd is $ [Kr] $ $ 4{d^{10}} $ .
So, Nickel(Ni) has 3d series and Palladium(Pd) has 4d series. On going from 3d to 4d series CFSE increases by half.
 Given, CFSE of Ni = $ {\vartriangle _0} $ .
CFSE of Pd = $ {\vartriangle _0} + \left( {\dfrac{1}{2}} \right){\vartriangle _0} = 1.5{\vartriangle _0} $ .
Hence option (C) is the right answer.

Note :
The splitting of orbitals is based on the ligands. Orbitals along the direction of the ligand experience more repulsion and go above the energy level and the orbital opposite to the direction of the ligand experiences less repulsion and lies below the energy level. The direction of the orbitals can be known from the geometry of the complex.