CUET · CHEMISTRY · PYQ PAPER 2023
Read the passage carefully and answer the Questions:
According to Valence Bond theory, the metal atom or ion under the influence of ligands can use its \((n-1) d, n s, n p\) or \(n s, n p, n d\) orbitals for hybridisation to yield a set of equivalent orbitals of definite geometry such as octahedral, tetrahedral, square planar and so on. These hybridised orbitals are allowed to overlap with ligand orbitals that can donate electron pairs for bonding.
While the VB theory, to a larger extent, explains the formation, structures and magnetic behaviour of coordination compounds, it suffers from the many shortcomings. The Crystal field theory (CFT) is an electrostatic model which considers the metal-ligand bond to be ionic arising purely from electrostatic interactions between the metal ion and the ligand. The five \(d\) orbitals in an isolated gaseous metal atom/ion have same energy, i.e., they are degenerate. This degeneracy is maintained if a spherically symmetrical field of negative charges surrounds the metal atom/ion. However, when this negative field is due to ligands in a complex, it becomes asymmetrical and the degeneracy of the \(d\) orbitals is lifted. It results in splitting of the \(d\) orbitals. The pattern of splitting depends upon the nature of the crystal field.
The spin only magnetic moment of \(\left[ MnBr _4\right]^{2-}\) is 5.9 BM. Predict the geometry of the complex ion?
- A tetrahedral
- B square planar
- C trigonal pyramidal
- D distorted octahedral
Answer & Solution
Correct Answer
(A) tetrahedral
Step-by-step Solution
Detailed explanation
\(Mn + 4(-1) = -2 \Rightarrow Mn = +2\) \(Mn^{2+} \Rightarrow 3d^5\) \(\mu = \sqrt{n(n+2)} \Rightarrow 5.9 = \sqrt{n(n+2)} \Rightarrow n \approx 5\) For \(Mn^{2+}\) (\(d^5\)) with \(n=5\) and coordination number 4, the complex is high-spin. The geometry is tetrahedral.
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