Question

(i) How many geometrical isomers are possible in each of the following complexes ?

(I) \( \mathrm{[Cr(C_2O_4)_3]^{3-}} \)

(II) \( \mathrm{[Co(NH_3)_3Cl_3]} \)

(ii) \( \mathrm{[Co(NH_3)_6]^{3+}} \) is an inner orbital complex whereas \( \mathrm{[Ni(NH_3)_6]^{2+}} \) is an outer orbital complex. Why ?

[Atomic number : Co = 27, Ni = 28]

(iii) Write the formula of Wilkinson catalyst and its use.

Hint:

M(AA)3 has no geometrical isomer; MA3B3 has fac and mer. Co(III) d6 with strong field NH3 is d2sp3; Ni(II) d8 uses 4d, so sp3d2. Wilkinson catalyst is [RhCl(PPh3)3].

Answer 1

Step 1: Part (i)(I), \( \mathrm{[Cr(C_2O_4)_3]^{3-}} \):
Oxalate is a symmetrical bidentate ligand and the complex is octahedral of the type \( \mathrm{M(AA)_3} \).
All three ligands are alike and all positions are equivalent, so cis-trans isomerism cannot arise.
The number of geometrical isomers is 1, that is, no geometrical isomerism. The complex shows only optical isomerism.

Step 2: Part (i)(II), \( \mathrm{[Co(NH_3)_3Cl_3]} \):
This is an octahedral complex of the type \( \mathrm{MA_3B_3} \).
In the facial (fac) isomer the three \( \mathrm{NH_3} \) groups sit on one triangular face of the octahedron.
In the meridional (mer) isomer the three \( \mathrm{NH_3} \) groups lie in one plane that cuts the middle of the octahedron.
So 2 geometrical isomers are possible.

Step 3: Part (ii), the inner orbital complex:
In \( \mathrm{[Co(NH_3)_6]^{3+}} \) cobalt is \( \mathrm{Co^{3+}} \) with configuration \( 3d^6 \).
\( \mathrm{NH_3} \) is a strong field ligand and forces the electrons to pair up in the 3d orbitals.
Two 3d orbitals become empty. They mix with one 4s and three 4p orbitals to give \( d^2sp^3 \) hybrids.
These hybrids use inner 3d orbitals, so the complex is an inner orbital complex. It is diamagnetic.

Step 4: Part (ii), the outer orbital complex:
In \( \mathrm{[Ni(NH_3)_6]^{2+}} \) nickel is \( \mathrm{Ni^{2+}} \) with configuration \( 3d^8 \).
Even after pairing, eight electrons fill four of the five 3d orbitals. Only one 3d orbital is empty, but two are needed for \( d^2sp^3 \).
So the metal uses outer 4d orbitals with 4s and 4p to give \( sp^3d^2 \) hybrids.
This is an outer orbital complex and it is paramagnetic with two unpaired electrons.

Step 5: Part (iii), Wilkinson catalyst:
The formula is \( \mathrm{[RhCl(PPh_3)_3]} \), tris(triphenylphosphine)chloridorhodium(I).
It is used as a homogeneous catalyst for the hydrogenation of alkenes.

Final Answer:
(i)(I) none (1 form only); (i)(II) 2 (fac and mer). (ii) \( \mathrm{Co^{3+}} \) gives \( d^2sp^3 \) inner orbital; \( \mathrm{Ni^{2+}} \) gives \( sp^3d^2 \) outer orbital. (iii) \( \mathrm{[RhCl(PPh_3)_3]} \), used for hydrogenation of alkenes.