Coordination Compounds

When reactant is cis isomer then following reaction takes place.

So, if reactant is cis - isomer then two isomers are produced.

When reactant is trans isomer then following reaction takes place.

So, if the reactant is trans isomer then only one isomer is produced.

Assume oxidation state of Cr in all the compounds = x (i)

In [Cr(H2O)6] Cl3 oxidation state of Cr is x + 0 $\times$ 6 + ($-$1 $\times$ ) = O

x + 0 $-$ 3 = O

x = + 3 (ii)

[Cr (C6 H6)2] oxidation state of Cr is x + 0 $\times$ 2 = 0

x = 0 (iii)

In K2 [ Cr(CN)2 (O)2 (O2) (NH3)] oxidation state of Cr is 1 $\times$ 2 + x + ($-$ 1 $\times$ 2) + ($-$2 $\times$ 2) + ($-$2) + 0 = 0

x = + 6

+ 3, 0 and + 6 is the correct answer.

Note : O2 molecule can have 0, $-$ 1, $-$ 2 oxidation state but in K2 [ Cr (CN)2 (O)2 (O2) NH3 ] if we choose zero as the oxidation state of O2 then for Cr oxidation state will be $+$ 4.

But + 4 oxidation state of Cr is unstable and $+$ 6 is most stable that is why we choose $-$ 2 oxidation state of O2.

The complexes with formula [Mabcd]n± can have three geometrical isomers.

As NO2− and SCN− are ambidentate ligands, therefore 4 × 3 = 12 geometrical isomers will be possible.

Here number of unpaired electrons = 3 $\therefore$ Spin only magnetic moment ($\mu$) =

B.M Note : Energy of t2g is less than eg.

As electrons always go to the lower energy level orbitals first, that is why electrons goes to the t2g orbital first.

Here number of unpaired electrons = 0 $\therefore$ Spin only magnetic moment ($\mu$) = 0 B.M Note : (1) As CN- is a strong field ligand so Energy gap between eg and t2g orbital is very high.

(2) [Fe(CN)6]4– is an octahedral complex.

And for octahedral complex Energy gap between eg and t2g orbital is called

0 or Crystal Field splitting Energy.

(3) Energy required to pair up the electron in same orbital is called Pairing Energy(P).

(4) For strong field ligand,

0 is very high and for weak field ligand,

0 is very low. (5) For strong field ligand,

0 > P, so when electron gets energy, pairing of electrons happens as for pairing of electrons very low energy requied.

(6) For weak field ligand,

0 < P, so when electron gets energy, pairing of electrons does not happens as the energy required to enter into the eg orbital is less than pairing energy.

That is why electron go to eg orbital first.

(7) As CN- is a strong field ligand so pairing of electron occurs.

Here number of unpaired electrons = 1 $\therefore$ Spin only magnetic moment ($\mu$) =

B.M Note : (1) In [Ru(NH3)6]3+ complex, NH3 is a strong field ligand so Energy gap between eg and t2g orbital is very high.

That is why pairing of electrons occurs.

Here number of unpaired electrons = 2 $\therefore$ Spin only magnetic moment ($\mu$) =

B.M $\therefore$ Correct order of the spin-only magnetic moment of metal ions V2+ > Cr2+ > Ru3+ > Fe2+

Compount is Fe(H2O)6]2 [Fe(CN)6] Cation is Fe(H2O)6]2+ Anion is [Fe(CN)6]4-

Degenerate orbitals means orbitals of equal energy.

Cr3+ forms an octahedral inner orbitals complex and its d orbital get splitted into two differnt energy level.

The two set of degenerate orbitals are (1) dxy, dyz and dxz (2) dx2 - y2 and dz2

To determine which statements are correct, let's analyze each one individually in the context of Valence Bond Theory (VBT) as it applies to transition metal complexes.

Statement (I): "Valence bond theory cannot explain the color exhibited by transition metal complexes."

Analysis: Color in Transition Metal Complexes: The colors of transition metal complexes arise from electronic transitions between different energy levels of the d-orbitals, specifically d-d transitions.

These transitions occur when an electron absorbs light energy and moves from a lower-energy d-orbital to a higher-energy d-orbital.

Valence Bond Theory Limitations: VBT focuses on the hybridization of atomic orbitals to form covalent bonds.

It does not account for the splitting of d-orbitals into different energy levels in the presence of ligands (known as crystal field splitting).

Therefore, VBT cannot explain the origin of color in these complexes because it doesn't address the electronic transitions responsible for color.

Conclusion: Statement (I) is correct.

Statement (II): "Valence bond theory can predict quantitatively the magnetic properties of transition metal complexes."

Analysis: Magnetic Properties: The magnetic behavior of a complex depends on the number of unpaired electrons in the metal ion.

Quantitative prediction requires calculating the magnetic moment, often using the formula: $\mu = \sqrt{n(n+2)} \ \text{Bohr Magnetons (BM)}$ where $n$ is the number of unpaired electrons.

Valence Bond Theory Capabilities: VBT can provide a qualitative idea about the magnetic properties by indicating whether a complex is paramagnetic (unpaired electrons present) or diamagnetic (no unpaired electrons).

However, VBT does not offer the tools to quantitatively predict the exact magnetic moment.

Accurate quantitative predictions require more advanced theories like Crystal Field Theory (CFT) or Ligand Field Theory (LFT).

Conclusion: Statement (II) is incorrect.

Statement (III): "Valence bond theory cannot distinguish ligands as weak and strong field ones."

Analysis: Weak and Strong Field Ligands: Ligands are classified based on their ability to split the d-orbitals of the metal ion, influencing the pairing of electrons.

Strong field ligands cause a large splitting, often leading to low-spin complexes.

Weak field ligands cause small splitting, leading to high-spin complexes.

Valence Bond Theory Limitations: VBT does not address the energy splitting of d-orbitals.

It assumes that all bonds are formed via overlap of orbitals without considering the effect of ligands on d-orbital energies.

Therefore, VBT cannot distinguish between weak and strong field ligands because it doesn't involve the spectrochemical series or orbital splitting concepts.

Conclusion: Statement (III) is correct.

Final Answer: Statements (I) and (III) are correct.

Statement (II) is incorrect.

Answer: Option D (I) and (III) only

As in a co-ordination compound, the strong field ligand causes higher splitting of the d-orbitals Also we know, strength of ligand $\propto$

Order of strength of ligand NH3 > H2O > Cl- Therefore decreasing order of wavelength absorbed is (I) > (II) > (III)

CN of [Fe(H2O)6]Cl2 = 6 For CN = 6, CFSE =

Here, n = number of electron in eg and n1 = number of electron in t2g Electronic configuration of Fe+2 according to crystal field theory =

$\therefore$ CFSE =

= - 0.4

CN of K2[NiCl4] = 4 For CN = 4, CFSE =

Here, n = number of electron in t2g and n1 = number of electron in eg Electronic configuration of Ni+2 according to crystal field theory =

$\therefore$ CFSE =

= - 0.8

Here in [Co(Cl)(en)2]Cl 'Cl' is monodentate so one coordinate linkage will be made with Co. 'en' is bidentate so two coordinate linkage will be made with Co.

There are two 'en ' present so 4 coordinate linkage will be made with Co.

$\therefore$ Total 5 coordinate linkage will be made with Co by Cl and en.

So C.N. of Co is 5.

Here in K3[Al(C2O4)3] C2O4-2 is bidentate so two coordinate linkage will be made with Al.

There are three C2O4-2' present so 6 coordinate linkage will be made with Al.

$\therefore$ So C.N. of Al is 6.