Coordination Chemistry

The magnetic moment of a metal complex depends on the number of unpaired electrons in the d-orbitals of the central metal ion. The formula for magnetic moment is given by: Where is the number of unpaired electrons. The more unpaired electrons there are, the higher the magnetic moment. Let’s analyze the complexes:
- (1) : For Ni²⁺, which has an electronic configuration of , in the presence of chloride ions (Cl⁻), a weak field ligand, it results in two unpaired electrons. The magnetic moment will be moderate.
- (2) : has an electronic configuration of , and is a strong field ligand, so it causes pairing of the electrons. In this complex, no unpaired electrons are present, making the magnetic moment zero (since is diamagnetic).
- (3) : Cr³⁺ has an electronic configuration of , and ammonia (NH₃) is a weak field ligand, so there will be three unpaired electrons, making this complex paramagnetic with a relatively high magnetic moment.
- (4) : Co³⁺ has an electronic configuration of , and fluoride ions (F⁻) are weak field ligands. This results in four unpaired electrons, so the complex will have a high magnetic moment.
Thus, the highest magnetic moment will be found in and , with Cr³⁺ having a relatively higher number of unpaired electrons compared to the other complexes. The correct answer is Option (2).

Let's analyze the hybridization of the metal ion in each of these complexes:
- : The complex consists of Ni²⁺ surrounded by six water molecules. In the case of Ni²⁺ (which has a d⁸ electron configuration), the central metal ion typically undergoes octahedral geometry, which corresponds to an hybridization. This hybridization is correct for this complex.
- : The complex consists of Co³⁺ surrounded by six fluoride ions. The Co³⁺ ion, with a d⁶ electron configuration, also adopts octahedral geometry.
Therefore, the hybridization of Co³⁺ in this complex is . This is correct. - : The complex consists of Cu²⁺ surrounded by four ammonia molecules. The Cu²⁺ ion (with a d⁹ electron configuration) typically adopts a square planar geometry (not octahedral), which corresponds to hybridization. However, Cu²⁺ generally forms a tetrahedral geometry in the presence of NH₃, which should be . Hence, this match is incorrect.
- : The complex consists of Mn²⁺ surrounded by four chloride ions. The Mn²⁺ ion (with a d⁵ electron configuration) adopts a tetrahedral geometry, and the hybridization for a tetrahedral geometry is . Hence, the hybridization is incorrect.

Thus, the incorrect match is for and , but only (1) has the correct hybridization.
Therefore, the correct answer is Option (1).

Let’s examine the isomerism exhibited by :
- Geometrical Isomerism (1): This compound exhibits geometrical isomerism (cis-trans isomerism). In the complex , the two chloride ions and the two ammonia molecules can arrange themselves in different spatial configurations, either cis or trans. The cis-isomer has both chloride ions adjacent to each other, while the trans-isomer has the chloride ions opposite each other. - Cis: The chloride ions are adjacent to each other. - Trans: The chloride ions are opposite to each other.
- Linkage Isomerism (2): Linkage isomerism occurs when a ligand can bind through two different atoms. For example, NO₂⁻ can bind to the metal center through the nitrogen atom or the oxygen atom. In the given complex , there is no possibility of linkage isomerism since chloride and ammonia ligands do not have two binding sites.
- Ionisation Isomerism (3): Ionisation isomerism occurs when a coordination compound has the same molecular formula but gives different ions in solution. In this case, the complex does not exhibit ionisation isomerism, as no other ions are produced in solution.
- Coordination Isomerism (4): Coordination isomerism occurs when the metal ions in the coordination sphere can switch places. However, this complex does not exhibit coordination isomerism, as there is only one metal ion in the coordination sphere.
Thus, the correct answer is Option (1), as the complex exhibits geometrical isomerism.

We are asked to find the total number of ions produced when the complex dissociates in aqueous solution.
The given complex is . This complex consists of a central metal ion coordinated with six ammonia molecules (denoted as ) and three chloride ions as counterions.
Step 1: Dissociation in water
When the complex is placed in water, it dissociates into its components. The dissociation reaction is as follows: Here:
- is the complex ion.
- The 3 ions are the counterions.
Step 2: Counting the ions produced
The dissociation produces:
- 1 ion of
- 3 ions of Thus, the total number of ions produced is: Conclusion:
The total number of ions produced from the dissociation of is 6.

In the complex , the ions contribute -2 charge each.
Since there are three ions per complex, the total negative charge is .
Let the oxidation number of Co be .
Since the complex is neutral, the sum of charges must equal zero.
For each Co complex, we have: Thus, the oxidation number of Co is +3.