Coordination Compounds

Step 1: Understanding Optical Activity in Coordination Compounds
- Optical activity arises in coordination compounds when they lack a plane of symmetry and can exist in non-superimposable mirror image forms (enantiomers).
- This typically happens when the ligands are arranged in a chiral manner around the metal center.
Step 2: Structure of
- Ethylenediamine ( ) is a bidentate ligand, meaning it forms two coordinate bonds with the central metal ion.
- The complex has a chiral octahedral structure due to the three ethylenediamine ligands binding in a way that restricts symmetry.
Step 3: Optical Isomerism
- The presence of three bidentate ethylenediamine ligands in a cis-arrangement leads to two non-superimposable mirror image forms: Delta (Δ) and Lambda (Λ) isomers.
- This results in optical activity because the compound can rotate plane-polarized light.
Step 4: Conclusion
- is optically active as it exhibits chirality.
- Unlike simple metal complexes with monodentate ligands, this complex does not have a plane of symmetry.

Step 1: Understanding the Coordination Compound
- The given complex contains tris(ethane-1,2-diamine), which means three molecules of ethylenediamine (en) ligand are coordinated to the central metal cobalt(III).
- The oxidation state of cobalt is +3 in this complex.
Step 2: Determining the Formula
- Cobalt (III) sulfate implies that sulfate ( ) is the counter ion.
- Since cobalt is in the +3 oxidation state and the ethylenediamine ligand (en) is neutral, the charge on the complex ion must be +3.
- To balance the charge, two complex cations are needed with three sulfate anions ( ).
Step 3: Final Formula
The correct chemical formula of the given coordination compound is:
Step 4: Conclusion
- Ethylenediamine (en) is a bidentate ligand, meaning it forms two bonds with cobalt.
- The correct formula is , ensuring charge neutrality.

Step 1: Understanding Hybridization
Hybridization in coordination complexes is determined based on valence bond theory (VBT) and ligand field theory (LFT). The electronic configuration of Nickel (Ni, Z = 28) is:
Step 2: Hybridization of
- CO is a strong field ligand and causes low-spin or square planar geometry.
- Ni is in zero oxidation state (0).
- The valence shell configuration of Ni⁰ is: - CO ligand donates electron pairs, leading to sp³ hybridization and tetrahedral geometry.
Step 3: Hybridization of
- CN^- is a strong field ligand, which causes low-spin pairing.
- Ni is in ⁺² oxidation state, so the valence shell is: - The CN^- ligands cause electron pairing in the d-orbitals, leading to dsp² hybridization and square planar geometry.
Step 4: Conclusion
- has sp³ hybridization (tetrahedral).
- has dsp² hybridization (square planar).

Step 1: Identifying the Reaction Type
- The given reaction involves a simple alkane (butane, ).
- Alkanes undergo free radical substitution reactions in the presence of halogens and UV light.
Step 2: Reaction Mechanism
The reaction follows a free radical chain mechanism consisting of three steps: 1. Initiation Step: Formation of chlorine radicals under UV light 2. Propagation Step: Substitution of a hydrogen atom in butane 3. Termination Step: Combination of free radicals to stop the chain reaction
Step 3: Conclusion
- The reagent required for this reaction is chlorine ( ) and ultraviolet (UV) light.
- The UV light helps break the Cl-Cl bond, generating chlorine radicals that drive the reaction.

Step 1: Understanding Magnetic Moment
The spin-only magnetic moment ( ) of an ion is calculated using the formula: where is the number of unpaired electrons.
Step 2: Electronic Configuration of
- The atomic number of manganese (Mn) is 25.
- Its electronic configuration: .
- For (after losing two electrons from ): .
- Since has all unpaired electrons, .
Step 3: Calculation of Magnetic Moment
Substituting into the formula:
Step 4: Conclusion
Among the given ions, has the highest spin-only magnetic moment of 5.92 BM due to the presence of five unpaired electrons.