Werner's theory and Valence Bond - Free MCQ Practice Test with solutions,

Cyanide ion is strong field ligand because it is a pseudohalide ion, pseudohalide ions are stronger coordinating ligand & they have the ability to form bond from pseudohalide to the metal and bond (from the metal to pseudohalide).

• An outer orbital complex uses the 4d or 5d orbitals for bonding rather than the 3d orbitals.

• In [Ni(NH₃​)₆​]²⁺, NH_3​ is a weak field ligand, which does not cause pairing of 3d electrons. As a result, the 4d orbitals are utilized, leading to sp³d² hybridization (outer orbital complex).

• Other options ([Co(NH₃​)₆​]³⁺, [Mn(CN)₆​]³⁻, [Fe(CN)₆​]⁴⁻) use 3d orbitals due to strong field ligands or electron pairing, making them inner orbital complexes.

[SbF₅]^2-  has sp³d  hybridization and shows square pyramidal geometry

AgNO₃ + Cl⁻ → AgCl + NO₃⁻
AgCl + 2NH₃ → [Ag(NH₃)₂]Cl

The spectrochemical series arranges ligands based on their ability to split the d-orbital energy levels in a coordination complex. Stronger field ligands cause a greater splitting (Δ) compared to weaker field ligands.

1. Option A: NH₃ > H₂O
   Correct: Ammonia (NH₃) is a stronger field ligand than water (H₂O) in the spectrochemical series.

2. Option B: F⁻ > C₂O₄²⁻
   Incorrect: Oxalate (C₂O₄²⁻) is a stronger field ligand than fluoride (F⁻). This statement is incorrect.

3. Option C: NCS⁻ > SCN⁻
   Correct: When bonded through nitrogen (NCS⁻), the ligand is stronger than when bonded through sulfur (SCN⁻).

4. Option D: en > EDTA
   Correct: Ethylenediamine (en) is a stronger field ligand compared to EDTA in a single coordination site due to its chelating nature.

μ_s = √n(n + 2) = 5.9 B. M.
⇒ μ² = n(n + 2) = 34.81
⇒ n = 5
Thus, Mn will have five unpaired electrons so Mn will be in +2 state with Br^- being weak- field ligands.
Thus, x = 2 for [MnBr₄]^2-

[Mn(CN)₆]³⁻ and [Fe(CN)₆]³⁻ are inner orbital complexes and paramagnetic, while [Co(C₂O₄)₃]³⁻ is diamagnetic in nature.

Ti(C₂H₄)₄ is an organometallic compound due to Ti directly attached to C- atom

[Ni(NH₃)₆]²⁺ = octahedral
[Pt(NH₃)₄]²⁺ = square planar
[Zn(NH₃)₄]²⁺ = tetrahedral

In both states (paramagnetic and diamagnetic) of the given complex, Ni exists as Ni²⁺ whose electronic configuration is [Ar]3d⁸4s⁰.

In the above paramagnetic state the geometry of the complex is sp³ giving tetrahedral geometry. The diamagnetic state is achieved by pairing of electrons in 3d orbital.

Thus the geometry of the complex will be dsp² giving square planar geometry.

Fe³⁺ in [Fe(CN)₆]^3-

low spin complex μ = 1.732BM Ni in [Ni(CO)₄]

low and high spin complex is applicable for d⁴ to d⁷ configuration
in

low spin complex μ = 1.732BM

[Ti(H₂O)₆]Cl₄
Coordination number 6 ⇒ octahedral complex Ti is in +4 oxidations state ⇒ no unpaired electrons
⇒ magnetic moment = 0 B.M.

[Cu(NH₃)₄]²⁺ has square planar structure and is paramagnetic. 

Oxidation state of

Since, is a strong field ligand, it will cause pairing of electrons,
So, there is one unpaired electron in 3d-orbital of which makes the compound paramagnetic.

Shape of  [Cu(NH₃)₄]²⁺ is square planar.

[Co(C₂O₄)₃]³⁻ is diamagnetic as oxalate is a strong ligand causing pairing of 3d electrons in Co³⁺ thereby leading to d²sp³ hybridisation.

Octahedral and Diamagnetic

Both are paramagnetic, the only difference is that CN^- is a strong field ligand whereas F^- is a weak field ligand.

Monodentate ligand is uninegative and hence, the oxidation number of Ni in [NiX₄]^2- is +2.
The electronic configuration of Ni²⁺ is [Ar]3d⁸.
If the compound is diamagnetic it means all the electrons are paired. Hence, there will be one 3d orbital empty and one 4s orbital empty.
The hybridisation involved in the formation of complex [NiX₄]^2- is dsp².

Number of unpaired electrons = 2
Geometry = tetrahedral

Prussian blue is a deep blue pigment is  Fe₄[Fe(CN)₆]₃