Coordination Chemistry

Step 1: Definition
A ligand is a chemical species that donates a pair of electrons to a central metal atom or ion to form a coordinate bond.
Step 2: Requirement
To act as a Lewis base (electron donor), the species must possess at least one lone pair (unshared pair) of electrons.
Step 3: Conclusion
While many ligands are anions or small molecules, the absolute requirement is the presence of an unshared electron pair.
Final Answer: (B)

Step 1: Concept
An outer orbital complex uses higher energy 'nd' orbitals for hybridization (usually ).
Step 2: Analysis of (A), (B), (C)
- , , and use orbitals ( ) and undergo hybridization, making them inner orbital complexes.
Step 3: Analysis of (D)
- In , has a configuration. The two vacant orbitals required for octahedral geometry cannot be provided by the subshell even with pairing. Thus, it uses , , and orbitals ( hybridization).
Step 4: Conclusion
is an outer orbital complex.
Final Answer: (D)

Step 1: Understanding the Concept:
This is the cyanide process for extraction of silver.

Step 2: Detailed Explanation:
The reaction is used in the extraction of silver from argentite (Ag S).
.

Step 3: Final Answer:
Silver (Ag)

Step 1: Understanding the Concept:
Colour in transition metal ions arises from d-d transitions, which require unpaired electrons or vacant d-orbitals.
Step 2: Detailed Explanation:
\begin{itemize} \item Cu : d configuration, one unpaired electron, coloured (blue). \item Cu : d configuration, fully filled, no d-d transitions possible, colourless. \item Ni : d configuration, two unpaired electrons, coloured (green). \item Fe : d configuration, five unpaired electrons, coloured (yellow/brown). \end{itemize}
Step 3: Final Answer:
Cu is colourless, option (B).

Step 1: Understanding the Concept:
Outer orbital complexes (high spin) use outer d-orbitals ( ) for hybridization, while inner orbital complexes (low spin) use inner ( )d orbitals. This depends on the ligand field strength.
Step 2: Detailed Explanation:}}
\begin{itemize} \item (A) : Fe (d ), CN is strong field ligand, pairing occurs, uses inner d orbitals (d ), inner orbital complex. \item (B) : Mn (d ), CN strong field, pairing occurs, uses inner d orbitals, inner orbital complex. \item (C) : Co (d ), NH is strong field for Co , pairing occurs, uses inner d orbitals (d ), inner orbital complex. \item (D) : Ni (d ), NH is moderate field but for Ni , it forms outer orbital complex using sp d hybridization (outer d orbitals). \end{itemize}
Step 3: Final Answer:
is an outer orbital complex, option (D).

Step 1: Understanding the Concept:
Dichromate ions in alkaline medium convert to chromate ions.
Step 2: Detailed Explanation:}}
Potassium dichromate (K Cr O ) in aqueous solution exists as Cr O ions. When heated with an alkali (caustic soda, NaOH), the dichromate ion reacts to form chromate ion (CrO ). The color changes from orange (dichromate) to yellow (chromate).
Step 3: Final Answer:
K Cr O on heating with aqueous caustic soda gives chromate ions (CrO ), option (A).

Step 1: Understanding the Concept:
Effective Atomic Number (EAN) = Atomic number of metal - Oxidation state + 2 × (Coordination number).
Step 2: Detailed Explanation:
In [Cu(NH ) ]SO , the complex ion is [Cu(NH ) ] . NH is a neutral ligand. Oxidation state of Cu: Let x be the oxidation state. x + 0 = +2 ⇒ x = +2. Cu (Z = 29) in +2 state has 27 electrons. Each NH donates 2 electrons. Total electrons donated = 4 × 2 = 8. EAN = 27 + 8 = 35.
Step 3: Final Answer:
The effective atomic number is 35, option (C).

Step 1: Understanding the Concept:
Geometrical isomerism (cis-trans) occurs in compounds with restricted rotation, typically due to a double bond, where each carbon of the double bond has two different substituents.
Step 2: Detailed Explanation:
\begin{itemize} \item (A) 1-butene: CH =CH-CH CH . One carbon of the double bond has two H atoms, so no geometrical isomerism. \item (B) 1,2-dibromobutene: Br-CH=CBr-CH CH . Both carbons of the double bond have different substituents (H and Br on one, Br and CH CH on the other), so it shows cis-trans isomerism. \item (C) Propene: CH -CH=CH . One carbon has two H atoms, so no geometrical isomerism. \item (D) iso-butylene: (CH ) C=CH . One carbon has two CH groups, the other has two H atoms, so no geometrical isomerism. \end{itemize}
Step 3: Final Answer:
1,2-dibromobutene shows geometrical isomerism, option (B).

Step 1: Understanding the Concept:
Coordination number is the number of ligand donor atoms directly bonded to the metal.
Step 2: Detailed Explanation:
Coordination number counts only sigma bonds between metal and ligand donor atoms.
Step 3: Final Answer:
Coordination number is the number of ligands bonded by sigma bonds.

Step 1: Understanding the Concept:
Bidentate ligands have two donor atoms.
Step 2: Detailed Explanation:
Ethylene diamine has two donor nitrogen atoms, so it is bidentate.
Step 3: Final Answer:
Ethylene diamine is a bidentate ligand.

Step 1: Understanding the Concept:
Linkage isomerism occurs when a ligand can coordinate through different donor atoms.
Step 2: Detailed Explanation:
The nitro group can coordinate through N or O, showing linkage isomerism.
Step 3: Final Answer:
The isomerism is linkage isomerism.

Step 1: Understanding the Concept:
Magnetic moment BM, where is number of unpaired electrons.
Step 2: Detailed Explanation:
has configuration with 5 unpaired electrons, giving highest magnetic moment.
Step 3: Final Answer:
has the highest magnetic moment.

Step 1: Understanding the Concept:
Low spin complexes are favoured by strong field ligands and heavier transition metals.
Step 2: Detailed Explanation:
4d and 5d series elements have larger crystal field splitting, favouring low spin complexes.
Step 3: Final Answer:
Both 4d and 5d series form low spin complexes.

Step 1: Understanding the Concept:
MA B type complexes show geometrical isomerism.
Step 2: Detailed Explanation:
Cis form (violet), trans form (green).
Step 3: Final Answer:
Geometrical isomerism.