| List-I (Complex) | List-II (Hybridisation of central metal ion) |
|---|---|
| (A) [CoF_6]^3- | (I) d^2sp^3 |
| (B) [NiCl_4]^2- | (II) sp^3 |
| (C) [Co(NH_3)_6]^3+ | (III) sp^3d^2 |
| (D) [Ni(CN)_4]^2- | (IV) dsp^2 |
Syllabus Analysis & Trend Mapping
| List-I (Complex) | List-II (Hybridisation of central metal ion) |
|---|---|
| (A) [CoF_6]^3- | (I) d^2sp^3 |
| (B) [NiCl_4]^2- | (II) sp^3 |
| (C) [Co(NH_3)_6]^3+ | (III) sp^3d^2 |
| (D) [Ni(CN)_4]^2- | (IV) dsp^2 |
| List-I (Complex) | List-II (Hybridisation & Magnetic character) | (A) [mathrmMnBr_4]^2- | (I) d^2sp^3 & diamagnetic | (B) [mathrmFeF_6]^3- | (II) sp^3d^2 & paramagnetic | (C) [mathrmCo(C_2O_4)_3]^3- | (III) sp^3 & diamagnetic | (D) [mathrmNi(CO)_4] | (IV) sp^3 & paramagnetic
Choose the correct answer from the options given below:
Solution### Related Formula
Valence Bond Theory utilizes orbital hybridization configurations (sp^3, d^2sp^3, sp^3d^2) along with ligand strengths to predict net magnetic parameters.
### Core Logic
Let us analyze each coordination system:
* (A) [mathrmMnBr_4]^2-: mathrmMn^2+ (3d^5), weak field mathrmBr^-
ightarrow no pairing. Hybridization = sp^3 (4 unpaired electrons, paramagnetic)
ightarrow (IV).
* (B) [mathrmFeF_6]^3-: mathrmFe^3+ (3d^5), weak field mathrmF^-
ightarrow outer orbital complex. Hybridization = sp^3d^2 (paramagnetic)
ightarrow (II).
* (C) [mathrmCo(C_2O_4)_3]^3-: mathrmCo^3+ (3d^6), chelating oxalate induces strong field pairing
ightarrow t_2g^6 e_g^0. Hybridization = d^2sp^3 (diamagnetic)
ightarrow (I).
* (D) [mathrmNi(CO)_4]: mathrmNi^0 (3d^8 4s^2), strong field mathrmCO forces 4s electrons into 3d
ightarrow 3d^10. Hybridization = sp^3 (diamagnetic)
ightarrow (III) .
Correct match matches option (4): (A)-(IV), (B)-(II), (C)-(I), (D)-(III).
### Pattern Recognition
Strong field neutral carbonyl ligands like mathrmCO trigger absolute shift of s-valence pairs into the inner d shell completely matching diamagnetic criteria.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds
Q67
jee_main_2024_01_february_morning
Colour in Coordination Compounds
Given below are two statements:
Statement (I): A solution of [Ni(H_2O)_6]^2+ is green in colour.
Statement (II): A solution of [Ni(CN)_4]^2- is colourless.
In the light of the above statements, choose the most appropriate answer from the options given below:
Solution### Core Logic
[Ni(H_2O)_6]^2+: Water is a weak field ligand. Ni^2+ is a 3d^8 system. In an octahedral weak field, it has 2 unpaired electrons (t_2g^6 e_g^2). Due to the presence of unpaired electrons, d-d transition is possible, making the solution green in colour.
[Ni(CN)_4]^2-: CN^- is a strong field ligand. The complex is square planar (dsp^2 hybridization). All 8 electrons are paired in the lower energy d-orbitals. Because there are no unpaired electrons (diamagnetic), d-d transition does not fall in the visible region, and it is colourless.
### Step 1: Evaluate Statements
Statement I is correct (Green due to unpaired electrons).
Statement II is correct (Colourless as it is diamagnetic).
### Pattern Recognition
Strong field ligands (like CN^-) with d^8 metal ions (Ni^2+, Pd^2+, Pt^2+) generally force pairing, creating square planar, diamagnetic, and often colourless complexes unless charge transfer occurs.
### Evaluation Rubric / Model Answer
null
### Chapter Mix
Class 12 Chemistry: Coordination Compounds More Coordination Compounds Questions — jee_main_2024_29_january_eveningPractice all Coordination Compounds previous-year questions →
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