Which Metal Center Has the Highest Number of Unpaired Electrons Among VCl₄, Ni(CO)₄, [AuCl₄]⁻ and [CdBr₄]²⁻?

Correct Answer: Option (A) – VCl₄

Detailed Explanation of the Number of Unpaired Electrons

To identify the metal center with the highest number of unpaired electrons, each species must be analyzed separately. The correct approach is to determine the oxidation state of the central metal, calculate its d-electron configuration, and then determine how many electrons remain unpaired in the resulting species.

The four compounds contain the metal centers vanadium, nickel, gold, and cadmium. Their relevant electronic configurations are different, and this directly controls their magnetic behavior.

The final comparison is:

VCl₄: 1 unpaired electron

Ni(CO)₄: 0 unpaired electrons

[AuCl₄]⁻: 0 unpaired electrons

[CdBr₄]²⁻: 0 unpaired electrons

Therefore, VCl₄ has the highest number of unpaired electrons, and Option (A) is correct.

Step 1: Determine the Number of Unpaired Electrons in VCl₄

In VCl₄, each chloride ligand has an oxidation state of −1. Since the compound is electrically neutral, the oxidation state of vanadium can be calculated as follows:

x + 4(−1) = 0

x − 4 = 0

x = +4

Therefore, vanadium is present as V⁴⁺.

The atomic number of vanadium is 23, and the electronic configuration of neutral vanadium is:

V: [Ar] 3d³4s²

To form V⁴⁺, four electrons are removed. The two 4s electrons are removed first, followed by two electrons from the 3d subshell:

V⁴⁺: [Ar] 3d¹

A d¹ configuration contains exactly one unpaired electron. Therefore:

Number of unpaired electrons in VCl₄ = 1

Hence, VCl₄ is paramagnetic.

Why Does V⁴⁺ Have One Unpaired Electron?

The five d orbitals are available to accommodate d electrons. In a d¹ system, only one electron is present in the entire d subshell. Regardless of the detailed splitting pattern of the d orbitals, a single d electron cannot pair with another electron because no second d electron is present.

Therefore, the d¹ configuration of V⁴⁺ necessarily contains one unpaired electron. This makes VCl₄ the only species among the given options with an unpaired electron at the metal center.

Step 2: Determine the Number of Unpaired Electrons in Ni(CO)₄

In Ni(CO)₄, carbon monoxide is a neutral ligand. The complete complex is also neutral. Therefore, the oxidation state of nickel is:

Oxidation state of Ni = 0

Nickel has atomic number 28. The relevant electronic configuration of the nickel center in Ni(CO)₄ is a filled d shell:

Ni(0): 3d¹⁰

In a d¹⁰ configuration, all five d orbitals are completely filled, and every electron is paired.

Therefore:

Number of unpaired electrons in Ni(CO)₄ = 0

Hence, Ni(CO)₄ is diamagnetic.

Why Is Ni(CO)₄ Diamagnetic?

Carbon monoxide is a strong-field ligand and forms a stable zero-oxidation-state carbonyl complex with nickel. The nickel center has a completely filled d¹⁰ electronic configuration. Since all d electrons are paired, there is no unpaired electron available to produce paramagnetic behavior.

Ni(CO)₄ has a tetrahedral molecular geometry, but because the metal center is d¹⁰, all ten d electrons remain paired. Therefore, its number of unpaired electrons is zero.

Step 3: Determine the Number of Unpaired Electrons in [AuCl₄]⁻

For the complex ion [AuCl₄]⁻, each chloride ligand carries a charge of −1. Let the oxidation state of gold be x:

x + 4(−1) = −1

x − 4 = −1

x = +3

Therefore, the central metal ion is Au³⁺.

Gold has atomic number 79, and its ground-state electronic configuration is:

Au: [Xe] 4f¹⁴5d¹⁰6s¹

After the removal of three electrons:

Au³⁺: 5d⁸

Thus, Au³⁺ is a d⁸ metal ion.

Why Does [AuCl₄]⁻ Have Zero Unpaired Electrons?

The [AuCl₄]⁻ complex has a square planar geometry. Gold is a third-row transition metal, and 5d orbitals experience large ligand-field splitting. In the square planar arrangement of this d⁸ metal ion, the electrons occupy the lower-energy d orbitals in such a way that all eight d electrons are paired.

Therefore:

Number of unpaired electrons in [AuCl₄]⁻ = 0

Hence, [AuCl₄]⁻ is diamagnetic.

Step 4: Determine the Number of Unpaired Electrons in [CdBr₄]²⁻

In [CdBr₄]²⁻, each bromide ligand has a charge of −1. Let the oxidation state of cadmium be x:

x + 4(−1) = −2

x − 4 = −2

x = +2

Therefore, cadmium is present as Cd²⁺.

Cadmium has atomic number 48, and its ground-state electronic configuration is:

Cd: [Kr] 4d¹⁰5s²

When Cd forms Cd²⁺, the two 5s electrons are removed:

Cd²⁺: [Kr] 4d¹⁰

The d¹⁰ configuration contains five completely filled d orbitals. Therefore, all electrons are paired.

Number of unpaired electrons in [CdBr₄]²⁻ = 0

Hence, [CdBr₄]²⁻ is diamagnetic.

Comparison of All Four Metal Centers

The comparison clearly shows that only VCl₄ contains an unpaired electron at the metal center. All the other species contain metal centers with completely paired electrons.

Explanation of All Answer Options

Option (A): VCl₄

Option (A) is correct. In VCl₄, vanadium has an oxidation state of +4. The V⁴⁺ ion has a d¹ electronic configuration and therefore contains one unpaired electron. This is the highest number among all the species listed in the question.

Option (B): Ni(CO)₄

Option (B) is incorrect. Carbon monoxide is a neutral ligand, so nickel is in the zero oxidation state. The metal center has a d¹⁰ configuration in Ni(CO)₄. Since all d orbitals are completely filled, the number of unpaired electrons is zero.

Option (C): [AuCl₄]⁻

Option (C) is incorrect. Gold is present in the +3 oxidation state and therefore has a d⁸ configuration. The [AuCl₄]⁻ complex is square planar, and its d electrons are completely paired. Therefore, the metal center has zero unpaired electrons.

Option (D): [CdBr₄]²⁻

Option (D) is incorrect. Cadmium is present in the +2 oxidation state and has a d¹⁰ configuration. All ten d electrons are paired, so the number of unpaired electrons is zero.

Relationship Between Unpaired Electrons and Magnetic Behavior

The presence or absence of unpaired electrons determines whether a species is paramagnetic or diamagnetic. A species containing one or more unpaired electrons is attracted by an external magnetic field and is described as paramagnetic. A species in which all electrons are paired is weakly repelled by a magnetic field and is described as diamagnetic.

Among the four species in this question, VCl₄ is paramagnetic because the V⁴⁺ center contains one unpaired d electron. Ni(CO)₄, [AuCl₄]⁻, and [CdBr₄]²⁻ are diamagnetic because all electrons at their metal centers are paired.

Final Answer

The correct answer is Option (A): VCl₄. In VCl₄, vanadium is present as V⁴⁺, which has a d¹ electronic configuration and one unpaired electron. Ni(CO)₄ contains a d¹⁰ Ni(0) center, [AuCl₄]⁻ contains a square planar d⁸ Au³⁺ center, and [CdBr₄]²⁻ contains a d¹⁰ Cd²⁺ center; all three have zero unpaired electrons. Therefore, VCl₄ has the highest number of unpaired electrons.