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Identify the complex in which only one d...

Identify the complex in which only one d orbital is used in the hybridisation

A

`[Ni(CN)_4]^(2-)`

B

`[Fe(CN)_6]^(3-)`

C

`[Co(en)_3]^(3+)`

D

`[FeF_6]^(3-)`

Text Solution

AI Generated Solution

The correct Answer is:
To solve the question of identifying the complex in which only one d orbital is used in the hybridization, we will analyze the given options step by step. ### Step 1: Identify the oxidation state of the central metal atom. We need to determine the oxidation state of the central metal atom in the complex. For the complex given in the question, nickel cyanide (Ni(CN)₄)²⁻, we can calculate the oxidation state of nickel (Ni). - The cyanide ion (CN) has a charge of -1. Since there are four cyanide ligands, the total negative charge contributed by the ligands is -4. - The overall charge of the complex is -2. Therefore, we can set up the equation: \[ x + (-1) \times 4 = -2 \] Solving for x (the oxidation state of Ni): \[ x - 4 = -2 \implies x = +2 \] ### Step 2: Determine the electronic configuration of the metal in that oxidation state. Nickel has an atomic number of 28. The electronic configuration of neutral nickel is: \[ \text{Ni: } [Ar] 4s^2 3d^8 \] In the +2 oxidation state, two electrons are removed from the 4s orbital: \[ \text{Ni}^{2+}: [Ar] 3d^8 \] ### Step 3: Analyze the ligand field and pairing of electrons. The ligand in this case is cyanide (CN⁻), which is a strong field ligand. Strong field ligands cause pairing of electrons in the d orbitals. - In the 3d subshell, we have 8 electrons. The strong field ligand will pair up the electrons in the 3d orbitals, resulting in the following configuration: \[ \text{3d: } ↑↓ ↑↓ ↑↓ ↑↓ ↑ (3d_{xy}, 3d_{xz}, 3d_{yz}, 3d_{x^2-y^2}, 3d_{z^2}) \] ### Step 4: Determine the hybridization. To form a complex with four cyanide ligands, we need to hybridize the available orbitals. The configuration shows that all the 3d electrons are paired, and we will also use the 4s and 4p orbitals for hybridization. - The hybridization for this complex can be determined as follows: - One 3d orbital (from the paired 3d electrons) - One 4s orbital - Two 4p orbitals This results in a dsp² hybridization, which is characteristic of square planar complexes. ### Conclusion: In this case, only one d orbital is involved in the hybridization (the one that is paired). Therefore, the answer to the question is: **The complex in which only one d orbital is used in the hybridization is Ni(CN)₄²⁻.**
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