Coordination Compounds

(i)
(ii)
(iii)
(iv)

Cu(OH) dissolves in aqueous ammonia due to complex formation.
In excess NH , Cu forms the deep-blue ammine complex:

This complex is soluble, so the precipitate dissolves.
In NaOH, no such soluble cuprate complex is formed (Cu(OH) is not appreciably amphoteric),
hence Cu(OH) remains as an insoluble blue precipitate.

Step 1: Assign charges.
- Let oxidation state of Ni = .
- Each CN ligand carries a charge of –1.
- Total number of ligands = 4.

Step 2: Write the charge balance equation.

Step 3: Consider Ni oxidation state.
Nickel generally shows +2 oxidation state in complexes. So, .

Step 4: Substitute values.
Step 5: Final Answer.
Hence, the value of in [Ni(CN) ] is –2.

Step 1: Analyze the complex.
The given compound is .
- Inside the coordination sphere: .
- Outside the coordination sphere: 2 Cl ions.

Step 2: Dissociation in aqueous solution.

Step 3: Count total ions.
- 1 cation:
- 2 anions: Cl ions
Total = ions
Step 4: Final Answer.
Thus, the total number of ions in solution is 3.

(i) Hexa aqua chromium(III) chloride
(Complex cation charge balanced by chloride ions.)
(ii) Dichlorido diammin e platinum(II)
(Neutral complex: Pt with two neutral ammine and two chlorido ligands.)
(iii) Hexa ammine platinum(IV) chloride
(Pt with six neutral ammine ligands gives a complex cation, balanced by Cl .)
(iv) Sodium penta cyano nitrosyl ferrate(II)
(Anion charge since Fe , five CN and NO ; hence Na counterions.)

In this question, we are dealing with the complex , which is an octahedral coordination complex. The central metal ion is iron (Fe), surrounded by two ammonia (NH ) molecules and four cyanide (CN ) ions. The coordination number of the metal ion is 6, forming an octahedral geometry. Geometrical Isomerism in Octahedral Complexes:
Geometrical isomerism arises in octahedral complexes when there are different possible spatial arrangements of ligands. In this case, the complex can have cis and trans isomers based on the arrangement of the ligands. 1. Cis-Isomer: In the cis configuration, the two NH molecules are placed next to each other, i.e., they occupy adjacent positions (90° apart) in the octahedral arrangement. The four cyanide ions are placed at the remaining positions, opposite each other. This creates a symmetrical arrangement of the ligands in the octahedral complex. The formula for this isomer can be represented as: 2. Trans-Isomer: In the trans configuration, the two NH molecules are placed opposite each other, i.e., they occupy opposite positions (180° apart) in the octahedral arrangement. The four cyanide ions occupy the remaining positions. This arrangement is different from the cis-isomer and results in a distinct geometric shape. The formula for this isomer is: Conclusion: - The cis-isomer has the two NH molecules adjacent to each other, while in the trans-isomer, the NH molecules are opposite each other. These two different spatial arrangements of the ligands lead to geometrical isomerism in the complex.

The complex consists of a coordination sphere and 2 chloride ions outside the coordination sphere. When this complex dissolves in water, it dissociates as follows:
Thus, the total number of ions generated is: - 1 ion of ,
- 2 ions of .
The total number of ions = 1 + 2 = 3 ions. Therefore, the correct answer is option (C) 3.

(x) Tetraammine diaqua cobalt (III) chloride
Ligands: 4 ammine (NH ; neutral) and 2 aqua (H O; neutral). Central metal: Co(III) complex cation charge . Chloride is the counter anion ( each), hence three Cl are required.
[2pt] (y) Potassium trioxalato chromate (III)
Oxalato (oxalate) ligand = C O (bidentate). Three such ligands: total charge . Central metal: Cr(III) complex-anion charge . Balance with three K .
[2pt]

To calculate the oxidation number of cobalt in the given complex :

Step 1: Assign charges of known species.
- Potassium (K) has an oxidation state of .
- Carbonyl (CO) is a neutral ligand, so each CO contributes charge.

Step 2: Write the overall charge balance equation.
Let the oxidation number of cobalt be . The complex ion is , since K is outside with .

Step 3: Solve for .
Thus, the oxidation number of cobalt in is .

(i)
(ii)
(iii)
(iv)

Idea of ionisation isomerism:
Two coordination compounds having the same overall composition but differing in the counter-ion outside the coordination sphere (which can be interchanged with a ligand inside) are called ionisation isomers.
Given pair:

In the first salt, is the counter-ion and is bound to Co(III) inside the sphere.
In the second, is the counter-ion and is coordinated to Co(III).
Experimental proof by precipitation tests:
\underline{Test for free sulphate (BaCl test):}
White precipitate of confirms free .
\underline{Test for free chloride (AgNO test):}
Curdy white precipitate of confirms free .

(x) :
- Oxidation state of Co: +3 (since each oxalate contributes -2 charge, and there are three oxalates, the net charge is -6. The overall charge of the complex is -3, so Co must be +3 to balance this).
- Distribution of d-orbitals: Co (d⁶ configuration) will have 6 electrons in its d-orbitals.
- Coordination number of Co: 6 (since there are three oxalate ions, each of which is a bidentate ligand, contributing 2 donor atoms per ligand, for a total coordination of 6).
(y) :
- Oxidation state of Mn: +2 (the complex ion is neutralized by the sulfate ion , indicating Mn is in the +2 oxidation state).
- Distribution of d-orbitals: Mn (d⁵ configuration) will have 5 electrons in its d-orbitals.
- Coordination number of Mn: 6 (since there are six water molecules, each contributing one donor atom, making the coordination number 6).

(x) :
The IUPAC name of this complex is pentaamminechloridocobalt(III) chloride.
Explanation:
- The complex ion consists of cobalt as the central metal ion, five ammonia molecules (NH₃) and one chloride ion (Cl) as ligands.
- The oxidation state of cobalt is +3, and the complex is neutralized by two chloride ions (Cl⁻) outside the coordination sphere.
(y) :
The IUPAC name of this complex is potassium tetracyanidonickelate(II).
Explanation:
- The complex ion consists of nickel as the central metal ion, four cyanide (CN⁻) ligands.
- The oxidation state of nickel is +2, and the complex is neutralized by two potassium ions (K⁺).
(c)(ii)

Step 1: Analyze

• Oxidation State of Ni: . So, .
• Configuration: Ni ( ) is . is .
• Ligand Nature: is a weak field ligand (Spectrochemical series). It does not cause pairing of electrons.
• Orbital Filling: The electrons remain arranged as 3 pairs and 2 unpaired electrons according to Hund's rule.
• Hybridization: (Tetrahedral).
• Result: Due to the presence of 2 unpaired electrons, it is Paramagnetic.

Step 2: Analyze

• Oxidation State of Ni: . So, neutral Ni.
• Configuration: Ni ( ) is .
• Ligand Nature: is a strong field ligand. It causes strong repulsion and forces electrons to pair up.
• Rearrangement: Under the influence of , the two electrons are pushed into the orbitals to pair up with the unpaired electrons. The configuration becomes .
• Hybridization: (Tetrahedral).
• Result: All electrons are paired ( ). Due to the absence of unpaired electrons, it is Diamagnetic.