CUET · CHEMISTRY · PYQ PAPER 2023
Read the passage carefully and answer the Questions
In the formation of coordination complexes, if the inner \(d\) orbital \((n - 1)d\) is used in hybridisation, the complex, is called an inner orbital or low spin or spin paired complex. And if it uses outer \(d\) orbital \((nd)\) in hybridisation (like \(sp^3d^2\)), it is called outer orbital or high spin or spin free complex. The degeneracy of the \(d\) orbitals has been removed due to ligand electron-metal electron repulsions in the complex. This splitting of the degenerate levels due to the presence of ligands in a definite geometry is termed as crystal field splitting. For coordination complexes, the magnetic moment is determined by the number of unpaired electrons and is calculated by using the 'spin-only' formula, \(\mu = \sqrt{n(n+2)}\) where \(n\) is the number of unpaired electrons and \(\mu\) is the magnetic moment in units of Bohr magneton \((BM)\). The coordination compounds are of great importance. These compounds are widely present in the mineral, plant and animal worlds and are known to play many important functions in the area of analytical chemistry, metallurgy, biological systems, industry and medicine.
Fe(II) octahedral complexes can show different 'spin only' magnetic moments for arrangements in the presence of weak and strong ligands. The 'spin only' magnetic moments of these complexes with weak and strong ligands, respectively, are
- A \(1.0\ BM\) and \(4.90\ BM\)
- B \(4.90\ BM\) and \(0\ BM\)
- C \(5.92\ BM\) and \(1.73\ BM\)
- D \(1.73\ BM\) and \(5.92\ BM\)
Answer & Solution
Correct Answer
(B) \(4.90\ BM\) and \(0\ BM\)
Step-by-step Solution
Detailed explanation
\(Fe^{2+}\) electron configuration: \(d^6\) For weak ligand (high spin): \(n=4\) \(\mu = \sqrt{4(4+2)} = \sqrt{24} = 4.90\ BM\) For strong ligand (low spin): \(n=0\) \(\mu = \sqrt{0(0+2)} = 0\ BM\)
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