Which of the following statement is/are correct for complex [NiCl4]2-? (A) Ni has oxidation state +2 (B) Cl is a weak field ligand (C) Compound is paramagnetic (D) dsp2 hybridisation (E) Low spin complex Choose the correct answer from the options given below:
(A), (B) and (C) only
Let's carefully examine each statement regarding the complex ion [NiCl<sub>4</sub>]<sup>2-</sup> to determine which ones are correct.
To find the oxidation state of Nickel (Ni) in the complex [NiCl<sub>4</sub>]<sup>2-</sup>, we know the overall charge of the complex is -2 and the charge of each chloride ion (Cl<sup>-</sup>) is -1.
Let the oxidation state of Ni be \(x\). The equation is:
\(x + 4 \times (-1) = -2\)
\(x - 4 = -2\)
\(x = -2 + 4\)
\(x = +2\)
So, the oxidation state of Ni in [NiCl<sub>4</sub>]<sup>2-</sup> is +2. Statement (A) is correct.
Ligands are classified as strong field or weak field based on their position in the spectrochemical series. The spectrochemical series lists common ligands in order of increasing crystal field splitting energy. Chloride ion (Cl<sup>-</sup>) is located towards the lower end of the spectrochemical series, making it a weak field ligand. Weak field ligands cause smaller splitting of d orbitals.
Statement (B) is correct.
Paramagnetism is determined by the presence of unpaired electrons. First, find the number of d electrons for the central metal ion.
The complex [NiCl<sub>4</sub>]<sup>2-</sup> is typically tetrahedral due to the bulky nature of chloride ligands and Ni(II) being a d<sup>8</sup> ion with weak field ligands. In a tetrahedral crystal field, the five d orbitals split into a lower energy set (e) and a higher energy set (t<sub>2</sub>).
Filling 8 electrons in tetrahedral splitting (e<sup>4</sup> t<sub>2</sub><sup>4</sup>):
Total number of unpaired electrons = 0 (from e) + 2 (from t<sub>2</sub>) = 2.
Since there are unpaired electrons, the compound [NiCl<sub>4</sub>]<sup>2-</sup> is paramagnetic.
Statement (C) is correct.
dsp2 hybridisation leads to a square planar geometry. [NiCl<sub>4</sub>]<sup>2-</sup> with the d<sup>8</sup> Ni<sup>2+</sup> ion and weak field Cl<sup>-</sup> ligands forms a tetrahedral complex. Tetrahedral complexes typically involve sp3 hybridisation.
dsp2 hybridisation occurs in d<sup>8</sup> complexes with strong field ligands that cause pairing of electrons, vacating an inner d orbital (specifically \(3d_{x^2-y^2}\)) for hybridisation with 4s and two 4p orbitals. Since Cl<sup>-</sup> is a weak field ligand, pairing is not forced, and a tetrahedral geometry is favoured with sp3 hybridisation.
Statement (D) is incorrect.
The terms 'low spin' and 'high spin' typically apply clearly to octahedral and sometimes square planar complexes, particularly for d<sup>4</sup> to d<sup>7</sup> configurations where the choice between pairing and filling higher energy orbitals depends on the relative magnitudes of crystal field splitting energy ($\Delta$) and pairing energy (P). In tetrahedral complexes, the crystal field splitting energy ($\Delta_t$) is much smaller than the pairing energy (P) ($\Delta_t \approx \frac{4}{9}\Delta_o$, and $\Delta_o$ itself might be less than P for weak field ligands). This means electrons tend to occupy orbitals singly as much as possible before pairing, characteristic of high spin complexes.
For a d<sup>8</sup> configuration in a tetrahedral field, the electron distribution (e<sup>4</sup> t<sub>2</sub><sup>4</sup>) naturally results in 2 unpaired electrons regardless of whether the splitting is large or small relative to pairing energy. The concept of 'low spin' (forced pairing) doesn't apply in the same way. However, if classified, tetrahedral complexes with weak field ligands are considered high spin due to the small splitting.
Statement (E) is incorrect.
| Statement | Description | Correct? |
|---|---|---|
| (A) | Ni oxidation state is +2 | Yes |
| (B) | Cl is a weak field ligand | Yes |
| (C) | Compound is paramagnetic | Yes (2 unpaired electrons) |
| (D) | dsp2 hybridisation | No (typically sp3 for tetrahedral) |
| (E) | Low spin complex | No (considered high spin due to weak field/small splitting) |
Based on our analysis, statements (A), (B), and (C) are correct for the complex [NiCl<sub>4</sub>]<sup>2-</sup>.
| Concept | Explanation | Relevance to [NiCl<sub>4</sub>]<sup>2-</sup> |
|---|---|---|
| Oxidation State | The charge on the central metal atom in a complex. | Calculated as +2 for Ni. |
| Ligand Field Strength | How much a ligand splits the d orbitals (weak field vs. strong field). | Cl<sup>-</sup> is a weak field ligand, causing small splitting. |
| Crystal Field Theory (CFT) | Explains bonding, magnetic properties, and color based on d orbital splitting. | Used to determine electron configuration in split d orbitals and predict paramagnetism. |
| Paramagnetism/Diamagnetism | Paramagnetic substances have unpaired electrons and are attracted by a magnetic field. Diamagnetic substances have all paired electrons and are repelled. | Presence of unpaired electrons in the split d orbitals of Ni<sup>2+</sup> determines paramagnetism. |
| Hybridisation & Geometry | Mixing of atomic orbitals to form hybrid orbitals for bonding, determining the complex's shape. | dsp2 for square planar, sp3 for tetrahedral. [NiCl<sub>4</sub>]<sup>2-</sup> is typically sp3/tetrahedral. |
| High Spin/Low Spin | Relates to electron distribution in d orbitals based on the balance between crystal field splitting energy ($\Delta$) and pairing energy (P). High spin means electrons occupy higher energy orbitals before pairing; low spin means pairing occurs first. | Tetrahedral complexes are generally high spin due to small $\Delta_t$. For d<sup>8</sup>, the unpaired electrons are fixed at 2 regardless of spin state definition, but it aligns with high spin behaviour. |
Understanding coordination complexes involves several key concepts:
Match List-I with List-II:
| List-I | List-II |
|---|---|
| (A) Diamagnetic solid | (I) CrO₂ |
| (B) Ferromagnetic solid | (II) Fe₃O₄ |
| (C) Antiferromagnetic solid | (III) NaCl |
| (D) Ferrimagnetic solid | (IV) MnO |
Choose the correct answer from the options given below:
[NiCl₂(PPh₃)₂] is named as:
Inner orbital complex among the following is:
(A) [Co(NH₃)₆]³⁺
(B) [CoF₆]³⁻
(C) [Ni(CN)4]²⁻
(D) [MnCl₆]³⁻
(E) [FeF₆]³⁻
Choose the correct answer from the options given below:
Which will form the most stable complex?
How many Cr-O bonds in dichromate ions are of the same bond length and are in resonance?