Co Ordination Compounds Question 100
Question: Nickel (Z=28) combines with a uninegative monodentate ligand to form a diamagnetic complex $ {{[NiL_4]}^{2-}} $ . The hybridisation involved and the number of unpaired electrons present in the complex are respectively:
Options:
A) $ sp^{3} $ , two
B) $ dsp^{2} $ , zero
C) $ dsp^{2} $ , one
D) $ sp^{3} $ , zero
Show Answer
Answer:
Correct Answer: B
Solution:
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First, let’s consider the oxidation state of Ni in the complex:
- The complex has a 2- charge overall
- There are four uninegative ligands (L-)
- So, the oxidation state of Ni must be +2 to balance the charge
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Ni(II) has an electron configuration of [Ar] $3d^8$
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The complex is diamagnetic, which means all electrons are paired. This is a crucial piece of information.
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For all electrons to be paired in a $d^8$ system, we need a strong-field ligand situation, which results in a low-spin complex.
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In a low-spin $d^8$ complex, all 8 d electrons will pair up in the lower energy d orbitals.
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The complex has a coordination number of 4 (four ligands).
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For a coordination number of 4, there are two possible geometries:
- Tetrahedral ($sp^3$ hybridization)
- Square planar ($dsp^2$ hybridization)
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Tetrahedral complexes are typically high-spin, while square planar complexes are typically low-spin.
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Given that this is a low-spin complex (as it’s diamagnetic), it must be square planar.
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Square planar geometry involves $dsp^2$ hybridization.
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In $dsp^2$ hybridization, all electrons are paired, resulting in zero unpaired electrons.
Therefore, the correct answer is:
B) $dsp^2$, zero
The complex has $dsp^2$ hybridization and zero unpaired electrons.
BETA
