All questions of Coordination Compounds for NEET Exam

The iron complex may be treated as cationic part, and C5H5- is a bidentate ligand therefore name can be assigned as follows “dicyclopentadienyl Iron (II) cation”.

Explanation:

In the formation of a complex entity, a central atom/ion acts as a Lewis acid. This can be explained as follows:

Lewis Acid and Lewis Base:
According to Lewis acid-base theory, a Lewis acid is a species that accepts a pair of electrons to form a coordinate covalent bond, while a Lewis base is a species that donates a pair of electrons to form a coordinate covalent bond.

Formation of Complex Entity:
A complex entity is formed by the coordination of a central atom/ion with one or more ligands. Ligands are molecules or ions that donate a pair of electrons to the central atom/ion to form a coordinate covalent bond.

Role of Central Atom/Ion:
In the formation of a complex entity, the central atom/ion acts as a Lewis acid because it accepts a pair of electrons from the ligands to form a coordinate covalent bond. The central atom/ion has an incomplete outer shell, which makes it electron deficient and thus able to accept electrons from other species.

Examples:
Some examples of complex entities and their central atom/ion are as follows:

- In [Fe(CN)6]4-, Fe2+ acts as the central ion, which accepts electrons from the CN- ligands.
- In [Cu(NH3)4]2+, Cu2+ acts as the central ion, which accepts electrons from the NH3 ligands.
- In [Ag(NH3)2]+, Ag+ acts as the central ion, which accepts electrons from the NH3 ligands.

Conclusion:
Thus, we can conclude that in the formation of a complex entity, the central atom/ion acts as a Lewis acid because it accepts a pair of electrons from the ligands to form a coordinate covalent bond.

Explanation:

The complex ion dichlorido bis(ethylene diamine) cobalt (III) has the formula [Co(en)2Cl2]+. Let's analyze the structure and isomers of this complex ion to determine the correct statement.

Structure:
The central cobalt (III) ion is coordinated by two ethylenediamine (en) ligands and two chloride (Cl) ligands. The ethylenediamine ligand is a bidentate ligand, meaning that it can form two coordination bonds with the central metal ion. The chloride ligand is a monodentate ligand, meaning that it can form only one coordination bond with the central metal ion.

Isomers:
Isomers are compounds that have the same chemical formula but different structural arrangements. In this case, there are three possible isomers for the complex ion [Co(en)2Cl2]+.

1. Geometrical isomerism:
Geometrical isomerism occurs when there is restricted rotation around a bond, resulting in different spatial arrangements. In the case of [Co(en)2Cl2]+, there are two possible geometrical isomers:

- cis-[Co(en)2Cl2]+: In this isomer, the two ethylenediamine ligands are adjacent to each other, and the two chloride ligands are also adjacent to each other.
- trans-[Co(en)2Cl2]+: In this isomer, the two ethylenediamine ligands are opposite to each other, and the two chloride ligands are also opposite to each other.

2. Optical isomerism:
Optical isomerism occurs when a compound is chiral, meaning it does not possess a plane of symmetry. Chiral compounds exist in two forms known as enantiomers, which are mirror images of each other. In the case of [Co(en)2Cl2]+, there is only one chiral center, which is the cobalt ion.

- Each geometrical isomer can exist as either an R or S enantiomer, resulting in two optically active isomers.
- The trans-[Co(en)2Cl2]+ isomer is optically inactive because its mirror image can be superimposed on it.

Conclusion:
Based on the analysis, the correct statement is option A. The complex ion dichlorido bis(ethylene diamine) cobalt (III) has three isomers, two of them are optically active (cis-[Co(en)2Cl2]+ R and S enantiomers) and one is optically inactive (trans-[Co(en)2Cl2]+).

Introduction:
In coordination chemistry, the oxidation state of a metal ion refers to the charge that the metal ion would have if all the ligands were removed along with the electron pairs that were shared with the ligands. The oxidation state of a metal ion can range from positive to negative depending on the number of electrons it has gained or lost.

Explanation:
To determine the oxidation state of nickel in each complex, we need to consider the oxidation states of the other elements and the overall charge of the complex.

a) Ni(CO)4:
In this complex, each carbon monoxide (CO) ligand is considered neutral because carbon has an oxidation state of +2 and oxygen has an oxidation state of -2. Therefore, the overall charge of the complex is 0. Since there are no other ligands present, the oxidation state of nickel must be 0.

b) [Cr(NH3)6]2[NiF6]3:
In this complex, the oxidation state of chromium is +3 since each ammonia (NH3) ligand is neutral and the overall charge of the complex is 2+. The oxidation state of fluorine is -1. Therefore, to balance the charges, the oxidation state of nickel must be +3.

c) [Ni(NH3)6](BF4)2:
In this complex, the oxidation state of boron in the tetrafluoroborate (BF4) ion is +3, and the oxidation state of fluorine is -1. Since the overall charge of the complex is 0, the oxidation state of nickel must be +2 to balance the charges.

d) K4[Ni(CN)6]:
In this complex, the oxidation state of potassium is +1. The cyanide (CN) ligand is considered neutral, with carbon having an oxidation state of +2 and nitrogen having an oxidation state of -3. Therefore, to balance the charges, the oxidation state of nickel must be +2.

Conclusion:
Among the given complexes, the complex [Cr(NH3)6]2[NiF6]3 has the highest oxidation state of nickel, which is +3.

The hybridization of an atom refers to the mixing of atomic orbitals to form new hybrid orbitals, which have different shapes and energies than the original atomic orbitals. In the case of tetracarbonyl nickel, the nickel atom is bonded to four carbon monoxide (CO) ligands.

The central nickel atom in tetracarbonyl nickel is surrounded by four ligands, which means it has a coordination number of 4. In order to determine the hybridization of the nickel atom, we can use the concept of the Valence Shell Electron Pair Repulsion (VSEPR) theory.

According to VSEPR theory, the electron pairs around the central atom will arrange themselves in a way that maximizes the distance between them, resulting in a geometry that minimizes electron-electron repulsion. In the case of tetracarbonyl nickel, the four CO ligands are arranged in a symmetrical tetrahedral geometry around the central nickel atom.

To determine the hybridization of the nickel atom, we can count the number of electron pairs around it. In tetracarbonyl nickel, there are four sigma bonds between the nickel atom and the four CO ligands, as well as four lone pairs on the nickel atom. This gives a total of eight electron pairs around the nickel atom.

The hybridization of an atom is determined by the number of sigma bonds and lone pairs around it. In this case, since there are four sigma bonds and four lone pairs, the nickel atom in tetracarbonyl nickel is sp3 hybridized.

The sp3 hybrid orbitals are formed by the mixing of one s orbital and three p orbitals on the nickel atom. These hybrid orbitals are oriented in a tetrahedral arrangement, pointing towards the four CO ligands.

In summary, the hybridization of the nickel atom in tetracarbonyl nickel is sp3, as there are four sigma bonds and four lone pairs around the central atom.

Trioxalato aluminate (III) and tetrafluorido-borate (III) ions

- Trioxalato aluminate (III) ion: [Al(C2O4)3]^3-
- Tetrafluorido-borate (III) ion: [BF4]^-

Correct combination

The correct combination of these ions is option 'C', which is [Al(C2O4)3]^3- and [BF4]^-. This means that the trioxalato aluminate (III) ion has a 3- charge and the tetrafluorido-borate (III) ion has a 1- charge.

Explanation

- The trioxalato aluminate (III) ion consists of one aluminum ion (Al^3+) complexed with three oxalate ions (C2O4^2-) to balance the charge.
- The tetrafluorido-borate (III) ion consists of one boron ion (B^3+) complexed with four fluoride ions (F-) to balance the charge.
- Therefore, the correct notation for these ions is [Al(C2O4)3]^3- and [BF4]^-.

Explanation:
The given complex is [Ir(OCN)2(H2O)3]. It has two isomers, i.e., linkage and polymerization isomers.

Linkage Isomerism:
In linkage isomerism, the ligand can coordinate through a different atom of the same ligand to form a different compound. Here, the OCN^- ion can coordinate via nitrogen (N) or carbon (C).

- If OCN^- binds through N atom, it forms a linkage isomer.
- If OCN^- binds through C atom, it forms a normal isomer.

Hence, [Ir(OCN-N)2(H2O)3] and [Ir(OCN-C)2(H2O)3] are linkage isomers.

Polymerization Isomerism:
In polymerization isomerism, the ligand can have different bridging modes between two metal centers. Here, the OCN^- ion can act as a bridging ligand to form a polymer chain.

- If OCN^- bridges between two Ir centers, it forms a polymerization isomer.
- If OCN^- binds to only one Ir center, it forms a normal isomer.

Hence, [Ir(OCN)2(H2O)3] and [Ir(OCN)2(μ-OCN)(H2O)2] are polymerization isomers.

Conclusion:
Thus, the given complex exhibits both linkage and polymerization isomerism.

The answer to the question is option 'D' - K4[Ni(CN)4].

Explanation:

In order to determine the oxidation state of the transition metal in each complex, we need to analyze the ligands and their charge contributions.

a) [Co(NH3)6Cl2]:
- In this complex, we have six ammine (NH3) ligands and two chloride (Cl) ligands.
- Ammine is a neutral ligand, so it does not contribute any charge.
- Chloride is a negatively charged ligand, contributing a charge of -1 each. Since there are two chloride ligands, the total charge contribution is -2.
- Since the overall charge of the complex is 0, the oxidation state of the central transition metal (Co) must be +2 to balance the charge.
- Therefore, the transition metal in this complex is in the +2 oxidation state.

b) [Fe(H2O)6]SO4:
- In this complex, we have six water (H2O) ligands and a sulfate (SO4) ligand.
- Water is a neutral ligand, so it does not contribute any charge.
- Sulfate is a negatively charged ligand, contributing a charge of -2.
- Since the overall charge of the complex is 0, the oxidation state of the central transition metal (Fe) must be +2 to balance the charge.
- Therefore, the transition metal in this complex is in the +2 oxidation state.

c) H[Co(CO)4]:
- In this complex, we have four carbonyl (CO) ligands.
- Carbonyl is a neutral ligand, so it does not contribute any charge.
- Since there is no other ligand or counterion present in the complex, the overall charge of the complex is +1.
- Therefore, the oxidation state of the central transition metal (Co) must be -1 to balance the charge.
- Therefore, the transition metal in this complex is in the -1 oxidation state.

d) K4[Ni(CN)4]:
- In this complex, we have four cyanide (CN) ligands.
- Cyanide is a negatively charged ligand, contributing a charge of -1 each. Since there are four cyanide ligands, the total charge contribution is -4.
- Since the overall charge of the complex is -4 (as indicated by the K4 counterions), the oxidation state of the central transition metal (Ni) must be 0 to balance the charge.
- Therefore, the transition metal in this complex is in the 0 oxidation state.

Conclusion:

The transition metal in the complex K4[Ni(CN)4] is in the zero oxidation state.

The silver-producing reaction in this demonstration is one that is commonly used in basic organic laboratory classes to identify aldehydes. The reaction called "The Silver Mirror Test" or Tollens' Test is accomplished by mixing aqueous silver nitrate with aqueous ammonia to produce a solution known as Tollens' reagent. Although this solution contains only a very weak oxidizing agent, it is strong enough to oxidize the aldehyde functional group. As this oxidation occurs, silver is reduced from the +1 oxidation state to metallic silver. This metallic silver is deposited on the walls of the test tube producing a reflective "mirror".

Understanding Coordination Number
The coordination number of a metal ion in a complex refers to the number of ligand donor atoms that are bonded to it. In the case of the complex [Cr(NH3)3(H2O)3], we need to analyze the ligands attached to chromium (Cr).
Identifying Ligands
- The complex contains two types of ligands:
- Ammonia (NH3), which is a neutral ligand.
- Water (H2O), which is also a neutral ligand.
Counting the Ligands
- There are three NH3 ligands.
- There are three H2O ligands.
Calculating the Coordination Number
- To find the coordination number, we simply add the number of donor atoms from each ligand type:
- From NH3: 3 ligands x 1 donor atom each = 3
- From H2O: 3 ligands x 1 donor atom each = 3
- Total coordination number = 3 (from NH3) + 3 (from H2O) = 6
Conclusion
Thus, the coordination number of Cr in the complex [Cr(NH3)3(H2O)3] is 6. This means that option 'B' is correct. The coordination number reflects how many atoms are directly bonded to the metal ion, which in this case is chromium.