Coordination Compounds

is more stable than , which means that the complex with is more stable and less likely to precipitate compared to the complex with .

So, the correct option is (C): ,

contains different ligands ( , , and ) and phosphine ligands ( ). The different arrangements of these ligands around the central atom can result in various isomers. In this case, you can have geometric (cis-trans) isomers due to the presence of both cis and trans configurations of and ligands with respect to each other, leading to two isomers. Additionally, you can have optical isomers (enantiomers) because of the chiral nature of the ligands, which can give you another two isomers.

So, the correct option is (C):

The correct option is(A): Ni(CO)₄ – Tetrahedral, paramagnetic.
Geometry: Ni(CO)₄ is not tetrahedral. It has a square planar geometry. The nickel atom (Ni) is bonded to four carbon monoxide (CO) ligands in a flat, square arrangement.
Magnetic Properties: Ni(CO)₄ is indeed paramagnetic. It has unpaired electrons in its molecular orbital diagram, making it attracted to an external magnetic field.
So, the statement that Ni(CO)₄ is tetrahedral is incorrect; it has a square planar geometry.

The correct option is(C): [Co(CO)₅NH₃]⁺².

Bonds: In this complex, the bond between the cobalt atom (Co) and the ammonia molecule (NH₃) is a (sigma) bond. Sigma bonds are formed by the head-on overlap of atomic orbitals.

Bonds: The five carbon monoxide (CO) ligands in the complex are bonded to the cobalt atom via (pi) bonds. Pi bonds result from the side-to-side overlap of p orbitals. Each CO ligand contributes one π bond to the complex, and since there are five CO ligands, there are a total of five bonds.

So, [Co(CO)₅NH₃]⁺² contains both and bonds,

The correct option is (B):6.
The coordination number of Ni in [Ni[(C₂O₄)₃]^-4 is 6. This is due to the bidentate nature of C₂O^3-₄ ligands, and there are three such ligands present. Therefore, the coordination number is calculated as 2 (bidentate) x 3 (ligands) = 6.

Cobalt has an atomic number of 27. The electron configuration for a neutral cobalt atom (Co) is: 1s² 2^s2 2^p6 3s² 3p⁶ 4s² 3d⁷.
When cobalt forms the complex , it loses three electrons to become : 1s² 2s² 2^p6 3s² 3p⁶ 4s⁰ 3d⁶.
Now, you can count the number of unpaired electrons in the 3d orbitals. In this case, there are 4 unpaired electrons.

So, the correct answer is (C) : 4

The catalytic activity of transition metals is often associated with their ability to exhibit variable oxidation states. This variability in oxidation states allows them to readily undergo redox reactions, which are a common feature of many catalytic processes. Transition metals can easily donate and accept electrons in their various oxidation states, making them effective catalysts in many chemical reactions.

So, the correct option is (D): Variable oxidation states.

Geometric isomerism occurs in octahedral complexes when there are at least two different types of ligands or when there is a bidentate ligand present. Geometric isomerism is not possible when all ligands are the same type and are monodentate. In such cases, the ligands can only occupy adjacent positions in the octahedral coordination sphere.
Among the options given:
(1) [MA₂B₄] has two types of ligands (A and B), so it can exhibit geometric isomerism.
(2) [MA₃B₃] has two types of ligands (A and B), so it can exhibit geometric isomerism.
(3) [MA₄B₂] has two types of ligands (A and B), so it can exhibit geometric isomerism.
(4) [MA₅B] has only one type of ligand (A), and they are all monodentate. Therefore, it will not show geometric isomerism.

So, the correct answer is (D): [MA₅B].

[Cr(H₂O)₆]Cl₃, with a chromium atom having an atomic number of 24, exhibits a magnetic moment measuring 3.83 B.M. The correct arrangement of chromium's 3d electrons within the complex is as follows: 3d¹_xy, 3d¹_zy, 3d¹_xz.
The magnetic moment value of 3.83 B.M. is indicative of the presence of 3 unpaired electrons in the system. This determination was made using the formula 3.83 = n(n+2), which yields a value of n=3. All these unpaired electrons are situated in the t²g orbitals, corresponding to 3d¹_xy, 3d¹_yz, and 3d¹_xz.