Coordination Chemistry

Understanding metal-carbon back bonding in metal carbonyls:

Back bonding occurs when:

1. Metal donates electron density from filled d-orbitals
2. Into empty π* antibonding orbitals of CO
3. This strengthens M-C bond and weakens C-O bond

Factors favoring strong back bonding:

• High electron density on metal (negative charge, low oxidation state)
• Good orbital overlap between metal d-orbitals and CO π* orbitals

Analyzing each complex:

(A) [Ti(CO)₆]²⁻

• Ti in oxidation state: -2 (very electron-rich)
• Configuration: Ti²⁻ has d⁶ configuration
• Very high electron density on metal
• Strong back donation expected
• Highest back bonding

(B) [V(CO)₆]⁻

• V in oxidation state: -1
• Configuration: V⁻ has d⁶ configuration
• High electron density, but less than Ti²⁻
• Strong back bonding, but less than (A)

(C) Cr(CO)₆

• Cr in oxidation state: 0
• Configuration: Cr⁰ has d⁶ configuration
• Neutral metal, moderate electron density
• Good back bonding, but less than charged species

(D) [Mn(CO)₆]⁺

• Mn in oxidation state: +1
• Configuration: Mn⁺ has d⁶ configuration
• Positive charge reduces electron density
• Weakest back bonding among these

Trend: As metal becomes more electron-rich (more negative charge), back bonding increases.

Order: [Ti(CO)₆]²⁻ > [V(CO)₆]⁻ > Cr(CO)₆ > [Mn(CO)₆]⁺

Answer: (A) [Ti(CO)₆]²⁻

1. Isomers of ( type)

The complex has the general form , where is the bidentate ligand ethylenediamine ( ) and is . This octahedral complex exhibits two types of stereoisomerism: geometrical and optical.

Geometrical Isomers: Two are possible:

isomer: The two chloride ligands ( ) are adjacent to each other.

isomer: The two chloride ligands ( ) are opposite to each other.

Optical Isomers (Enantiomers):

The isomer has a plane of symmetry and is achiral (no optical isomers).

The isomer lacks a plane of symmetry and is chiral, existing as a pair of non-superimposable mirror images (a pair of enantiomers).

2. Isomers of ( type)

The complex has the general form . This octahedral complex has only one possible geometrical arrangement (no isomerism), but it is inherently chiral due to the arrangement of the three bidentate ligands.

Geometrical Isomers: Only one geometrical form is possible.

Optical Isomers (Enantiomers): This form lacks a plane of symmetry and is chiral, existing as a pair of non-superimposable mirror images.

3. Total Number of Isomers

The total number of all possible isomers (geometrical and optical) for both complexes together is the sum of the isomers for each complex:

$ $

Answer: 5

Step 1: Analysis of Requirements

The complex(es) must satisfy three criteria:

Zero Magnetic Moment ( ): Requires zero unpaired electrons ( ), meaning the complex is diamagnetic. This typically implies a low-spin configuration for , , , or a configuration.

Zero Dipole Moment ( ): Requires the complex to have a symmetrical geometry so that all bond dipoles cancel out. Common examples are (e.g., isomers may have a dipole, usually do not) and ( isomers often have ).

CFSE : Crystal Field Stabilization Energy of is calculated for a low-spin octahedral complex.

$ \text{d}^6 \mathbf{(t_{2g})^6(e_g)^0} \mathbf{\text{d}^6} \mathbf{[\text{Mn}(\text{CO})_5(\text{CH}_3)]} \mathbf{d} \text{CO} \text{CH}_3 \text{Mn} +1 \text{CH}_3 -1 \text{Mn}(I) = \mathbf{\text{d}^6} \text{Mn} 5\text{CO} 1\text{CH}_3 \text{CO} \text{d}^6 \mu=0 \text{CFSE} = -2.4\Delta_0 \text{Mn}(\text{CO})_5(\text{CH}_3) \mathbf{[\text{trans}-\text{Ni}(\text{ethylenediamine})_2\text{Cl}_2]} \mathbf{d} \text{en} \text{Cl} -1 \text{Ni} +2 \text{Ni}(II) = \mathbf{\text{d}^8} \text{d}^8 \text{CFSE} = (6 \times -0.4\Delta_0) + (2 \times 0.6\Delta_0) = -2.4\Delta_0 + 1.2\Delta_0 = -1.2\Delta_0 \text{d}^8 t_{2g}^6 e_g^2 \mu \neq 0 \mu=0 \text{trans-M}(\text{AA})_2\text{X}_2 \mathbf{[\text{trans}-\text{Co}(\text{CN})_4(\text{H}_2\text{O})_2]^{-}} \mathbf{d} \text{CN} -1 \text{H}_2\text{O} \text{Co} +3 \text{Co}(III) = \mathbf{\text{d}^6} \text{CN} \text{d}^6 \mu=0 \text{CFSE} = -2.4\Delta_0 \text{trans}-\text{Co}(\text{CN})_4(\text{H}_2\text{O})_2^{-} \text{CN} \text{H}_2\text{O} \text{CN} \text{H}_2\text{O} \text{CN} \text{H}_2\text{O} \text{trans-M}(\text{A})_4\text{B}_2 \text{A} \text{B} \text{CN} \sigma/\pi \text{H}_2\text{O} \sigma \text{trans} \mathbf{[\text{trans}-\text{Fe}(\text{CN})_4\text{Cl}_2]^{4-}} \mathbf{d} \text{CN} -1 \text{Cl} -1 \text{Fe} +2 \text{Fe}(II) = \mathbf{\text{d}^6} \text{CN} \text{d}^6 \mu=0 \text{CFSE} = -2.4\Delta_0 \text{trans}-\text{Fe}(\text{CN})_4\text{Cl}_2^{4-} \text{trans-M}(\text{A})_4\text{B}_2 \text{CN} \text{Cl} \text{d}^6 \mu=0 \text{CFSE} = -2.4\Delta_0 \text{trans-M}(\text{A})_4\text{B}_2 \mu_d=0$).

Therefore, the correct options are (C) and (D).

Structure of

The complex anion has an ion at the center coordinated by four groups.

Nature of the Ligand: The tetrahydroborate anion ( ) acts as a bidentate ligand in this complex, forming a unique type of bond known as a three-center, two-electron ( ) bond through its hydrogen atoms.

Coordination: Each of the four groups uses two of its hydrogen atoms to bridge to the central ion. * Each group contributes two hydrogen atoms to the coordination sphere.

1. Coordination Number of

The coordination number is the total number of bonds formed.

Number of ligands

Number of bridging hydrogen atoms per ligand

Coordination Number (number of bonds)

2. Number of Bridging Hydrogen Atoms

The bridging hydrogen atoms are those that form bridges.

Since each of the four ligands contributes two bridging hydrogen atoms:

Total Bridging Hydrogen Atoms

The coordination number and the number of bridging hydrogen atoms are 8 and 8, respectively.

To determine the order of the M–C bond strength in the given species , , , and , we need to consider the metal-to-carbon monoxide (CO) back-bonding capability, which is influenced by the oxidation state and electronic configuration of the central metal.

1. In organometallic complexes such as these, the strength of the M–C bond is largely dependent on the back-bonding from the metal to the antibonding orbitals of CO (π-backbonding). The greater the extent of back-bonding, the stronger the M–C bond.
2. The back-bonding strength generally increases with the electron density on the metal. Thus, metals with a higher oxidation state will have weaker back-bonding due to reduced electron density.
3. Let's evaluate each complex:
   • : Chromium in is in the zero oxidation state. Chromium has the electronic configuration . Here, ligands use the orbitals for back-bonding.
   • : Manganese in this complex is in the +1 oxidation state. The loss of one electron reduces the back-bonding ability compared to .
   • : Titanium here is in the -2 oxidation state, which facilitates high electron density and strong back-bonding to CO. Titanium's configuration is .
   • : Cobalt in the +1 oxidation state has less electron density for back-bonding, similar to manganese.
4. Based on the above consideration, the complex with titanium having the lowest oxidation number and therefore the greatest electron density for back-bonding is expected to have the strongest M–C bond, followed by cobalt and chromium. Manganese, due to higher oxidation, will have the weakest bond.

Thus, the order of M–C bond strength is: > > > .

Therefore, the correct option is: III > IV > I > II.