Question

(i) (I) Explain why \( \mathrm{[Fe(CN)_6]^{3-}} \) is an inner orbital complex, whereas \( \mathrm{[FeF_6]^{3-}} \) is an outer orbital complex.

[Atomic number : Fe = 26]

(II) Write the formula of the given complex :

Hexaamminenickel (II) chloride

(ii) A solution of \( \mathrm{[Ni(H_2O)_6]^{2+}} \) is green in colour while a solution of \( \mathrm{[Ni(CN)_4]^{2-}} \) is colourless. Explain.

[Atomic number : Ni = 28]

Hint:

Fe(III) is 3d5. Strong field CN pairs electrons and gives d2sp3; weak field F gives sp3d2. Colour needs unpaired d electrons for a d-d transition.

Answer 1

Step 1: Part (i)(I): Electronic state of the metal ion:
Iron has atomic number 26, so the Fe atom is \( [\mathrm{Ar}]\,3d^6 4s^2 \).
In both complexes iron is in the +3 state, so Fe(III) is \( 3d^5 \).
Five d electrons occupy the five 3d orbitals, one in each, when no pairing takes place.

Step 2: Why \( \mathrm{[Fe(CN)_6]^{3-}} \) is inner orbital:
Cyanide is a strong field ligand. It forces the five d electrons to pair up in three 3d orbitals.
This leaves two 3d orbitals empty. They mix with one 4s and three 4p orbitals.
The hybridisation is \( d^2sp^3 \). Because the inner (n-1)d orbitals are used, it is an inner orbital complex.
The complex has one unpaired electron, so it is low spin and paramagnetic.

Step 3: Why \( \mathrm{[FeF_6]^{3-}} \) is outer orbital:
Fluoride is a weak field ligand. It cannot pair the electrons, so all five d electrons stay unpaired.
The inner 3d orbitals are not free. So the metal uses one 4s, three 4p and two 4d orbitals.
The hybridisation is \( sp^3d^2 \). Because the outer nd orbitals are used, it is an outer orbital complex.
It has five unpaired electrons, so it is high spin.

Step 4: Part (i)(II): Formula of the complex:
Hexaammine means six \( \mathrm{NH_3} \) ligands. Nickel is in the +2 state.
The complex ion is \( \mathrm{[Ni(NH_3)_6]^{2+}} \), which needs two chloride ions outside the bracket.
\[ \mathrm{[Ni(NH_3)_6]Cl_2} \]

Step 5: Part (ii): Configuration of Ni(II):
Ni has atomic number 28, so Ni(II) is \( 3d^8 \).

Step 6: Why \( \mathrm{[Ni(H_2O)_6]^{2+}} \) is green:
Water is a weak field ligand. The \( 3d^8 \) ion keeps two unpaired electrons in an octahedral field.
The complex absorbs light in the red region and moves an electron from a lower d level to a higher d level (d-d transition).
The light that is not absorbed reaches the eye, so the solution looks green.

Step 7: Why \( \mathrm{[Ni(CN)_4]^{2-}} \) is colourless:
Cyanide is a strong field ligand. It pairs the two unpaired electrons of \( 3d^8 \).
One 3d orbital becomes empty, and the ion is \( dsp^2 \) hybridised and square planar.
No unpaired electron is left, so no d-d transition is possible in visible light. The solution shows no colour.

Final Answer:
(i)(I) Strong field CN pairs the electrons and uses inner 3d orbitals; weak field F does not and uses outer 4d orbitals. (II) \( \mathrm{[Ni(NH_3)_6]Cl_2} \). (ii) The aqua complex has unpaired electrons and shows d-d absorption; the cyano complex has none.\[ \boxed{\mathrm{[Ni(NH_3)_6]Cl_2}} \]