Question

Draw orbital-energy splitting diagrams and use the spectrochemical series to show the orbital occupancy for each of the following (assuming thatis a weak-field ligand):

(a)${\mathbf{\left[}{\mathbf{\text{MoCl}}}_{\mathbf{6}}\mathbf{\right]}}^{\mathbf{3}\mathbf{-}}$(b) ${\left[\mathbf{Ni}{\left({\mathbf{H}}_{\mathbf{2}}\mathbf{O}\right)}_{\mathbf{6}}\right]}^{\mathbf{2}\mathbf{+}}$(c)${\left[\mathbf{Ni}\mathbf{(}\mathbf{CN}{\mathbf{)}}_{\mathbf{4}}\right]}^{\mathbf{2}\mathbf{-}}$

Short answer

The final answer is

(a) 3 d electrons (b) 8 d electrons (c) 8 d electrons

Step-by-step solution

Given compounds

Given: The compounds to draw the orbital- energy spitting diagrams:

(a)${\left[{\text{MoCl}}_6\right]}^{3-}$(b)${\left[\text{Ni}{\left({\text{H}}_2\text{O}\right)}_6\right]}^{2+}$(c)${\left[\mathrm{Ni}(\mathrm{CN}{)}_4\right]}^{2-}$

Crystal field splitting energy  

Basic Concept: The crystal field splitting energy represents the difference in energy between the two higher energy d orbitals and the three lower energy d orbitals; there are five orbitals. Because of the electrostatic connection between the metal cation and the ligands' negative charge, they are separated into two higher energy orbitals and three lower energy orbitals.

Spectrochemical series (from weak-field to strong-field ligands):$1^{-}<{\text{Cl}}^{-}<{\text{F}}^{-}<{\text{H}}_2\text{O}<{\text{SCN}}^{-}<{\text{NH}}_3<\text{en}<{\text{NO}}_2^{-}<{\text{CN}}^{-}<\text{CO}$

- The lower the splitting energy is $\left(△\right)$, the weaker the field.

- the stronger the field is, the higher the splitting energy is$\left(∆\right)$

Oxidation state

a) ${\left[{\text{MoCl}}_6\right]}^{3-}$- the oxidation state of Mo in this complex is , so the number of electrons is 3. (Electron configuration of Mo: [Ae] $4{\mathrm{d}}^{5}5{\mathrm{s}}^1$) is a weak-field ligand.

Oxidation state 

b) ${\left[\mathrm{Ni}\left({\mathrm{H}}_2\mathrm{O}\right){\text{l}}_6\right]}^{3-}$- the oxidation state of in this complex is +2, so the number of electrons is 8. (Electron configuration of $\text{Ni}:\left[\text{Ar}\right]3{\text{d}}^{8}4{\text{s}}^2$) is a weak-field ligand, so this is high-spin octahedral complex.

Oxidation state

c) ${\left[\text{Ni}{\left(\text{CN}\right)}_4\right]}^{2-}$- the oxidation state of Ni in this square planar complex is , so the number of d electrons is 8. Here we have different positions of orbitals. (Electron configuration of $\text{Ni}:\left[\text{Ar}\right]3{\text{d}}^8{\text{4s}}^2$) is a strong-field ligand, so this is low-spin diamagnetic complex.