To explain bonding in complexes valence bond theory was proposed by Linus Pauling. The main postulates of VB theory are as follows:

(i) The central atom loses a requisite number of electrons to form the cation. The number of electrons lost is equal to the valence of the resulting cation.

(ii) The central cation makes available a number of vacant orbitals equal to its coordination number for the formation of dative bonds with the ligands.

(iii) The cation orbitals hybridize to form new set of equivalent hybrid orbitals with definite directional characteristics.

(iv) The non-bonding metal electrons occupy the inner d-orbitals and do not participate in the hybridization.

(v) In the presence of strong ligands such as CN^-, NO, CO, the d-electrons are rearranged vacating some d-orbitals (when the number of d-electrons are more than 3 only) which can participate in hybridization.

(vi) In the presence of weak ligands such as F^-, Cl^-, H₂O, etc., the d-electrons are not rearranged.

(vii) The d-orbitals involved in the hybridization may be either (n - 1)d orbitals or outer d-orbitals.

The complexes formed by the involvement of (n - 1)d orbitals in hybridization are called inner orbital complexes or low spin complexes. The complexes formed by the involvement of d-orbitals of outer orbit are called outer orbital complexes or high spin complexes.

(viii) Each ligand contains a lone pair of electrons. A dative bond is formed by the overlap of a vacant hybrid orbital of metal ion and a filled orbital of ligand.

(ix) The complex will be paramagnetic, if any unpaired electrons present, otherwise diamagnetic.

(x) The number of unpaired electrons in a complex gives out the geometry of the complexes or vice versa.

The atomic number of Cr and Fe are 24 and 26, respectively. Which of the following is paramagnetic?

\([Cr(NH_3)_6]^{3+}\)

The correct answer is option 4. \([Cr(NH_3)_6]^{3+}\).

To determine which of the given complexes is paramagnetic, we need to analyze the electronic configuration and the presence of unpaired electrons in each complex.

1. \([Cr(CO)_6]\)

Oxidation State: Chromium is in the zero oxidation state \((Cr^0)\) in this complex.

Electronic Configuration of \(Cr^0\):\( \text{Cr}: [Ar] 3d^5 4s^1 \)

Configuration in the Complex: The carbon monoxide \((CO)\) is a strong field ligand, which causes pairing of electrons in the d-orbitals. Hence, the 3d electrons will be paired.

Magnetism: Since all the d-electrons are paired, \([Cr(CO)_6]\) is diamagnetic.

2. \(Fe(CO)_5\)

Oxidation State: Iron is in the zero oxidation state \((Fe^0)\) in this complex.

Electronic Configuration of \(Fe^0\): \( \text{Fe}: [Ar] 3d^6 4s^2 \)

Configuration in the Complex: Carbon monoxide \((CO)\) is a strong field ligand, leading to the pairing of electrons in the \(3d\) orbitals.

Magnetism: As with \([Cr(CO)_6]\), the CO ligand causes pairing of all d-electrons, making \(Fe(CO)_5\) diamagnetic.

3. \([Fe(CN)_6]^{4−}\)

Oxidation State: Iron is in the +2 oxidation state (Fe^2+) in this complex.

Electronic Configuration of \(Fe^{2+}\):\( \text{Fe}^{2+}: [Ar] 3d^6 \)

Configuration in the Complex: Cyanide \(CN^−)\) is a strong field ligand, which causes pairing of electrons in the \(3d\) orbitals.

Magnetism: In the presence of strong field ligands like \(CN^−\), all the d-electrons are paired, making \([Fe(CN)_6]^{4−}\) diamagnetic.

4. \([Cr(NH_3)_6]^{3+}\)

Oxidation State: Chromium is in the +3 oxidation state \(Cr^{3+})\) in this complex.

Electronic Configuration of \(Cr^{3+}\): \( \text{Cr}^{3+}: [Ar] 3d^3 \)

Configuration in the Complex: Ammonia \((NH_3)\) is a weak field ligand, meaning it does not cause significant pairing of d-electrons. As a result, \(Cr^{3+}\) retains its unpaired electronsagnetism: With three unpaired electrons in the 3d orbitals, \([Cr(NH_3)_6]^{3+}\) is paramagnetic.

Summary:

The complex that is paramagnetic due to the presence of unpaired electrons is: \([Cr(NH_3)_6]^{3+}\)