
Which of the following is/are characteristics of a tetrahedral complex?
A.${d_{{x^2} - {y^2}}}$ and ${d_{{z^2}}}$ orbital are low energy orbitals
B.Most tetrahedral complexes are high spin
C.Crystal field splitting is found double in octahedral complexes
D.Splitting pattern in the tetrahedral complex is just opposite of that in octahedral complexes.
Answer
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Hint: In tetrahedral complexes, ligand approaches the central metal in between the axes of a tetrahedron. The $d$ orbital that is in the direction of the ligand experiences higher repulsion and it loses its degeneracy.
Complete step by step solution:
In a tetrahedral complex, the metal ion lies at the centre of the tetrahedron and ligands are present on its corners. None of the $d - $ orbitals in tetrahedral complexes point directly towards the ligands and hence have a splitting of energy less than that in the octahedral complex. Thus, option C is incorrect.
The ${d_{{x^2} - {y^2}}}$ and ${d_{{z^2}}}$ orbitals lie between the ligands and comparatively farther from ligands than the other three orbitals. Thus they become more stable and their energies lower. Meanwhile in octahedral complexes, the ${d_{{x^2} - {y^2}}}$ and ${d_{{z^2}}}$ form higher energy orbitals.
Thus, option A and D are correct.
Tetrahedral complexes have small energy barriers between $d - $ orbitals . Thus they have crystal field stabilization energy less than pairing energy and most tetrahedral complexes are high spin.
Thus, the correct options are A, B and D.
Note:
In free metal ions, all $d - $ orbitals are at the same energy level. These orbitals that have the same energy are known as degenerate orbitals. However, when a ligand approaches the metal, the electrons are repelled by the lone pairs of ligand. As a result, the energy of the $d - $ orbital increases. The orbitals that lie in the same direction of the ligands experience a greater repulsion. Whereas, the orbitals lying away from the ligands experience a relatively lower repulsion. So, the five degenerate $d - $ orbitals converted into different sets of orbitals having different energies is called crystal splitting.
Complete step by step solution:
In a tetrahedral complex, the metal ion lies at the centre of the tetrahedron and ligands are present on its corners. None of the $d - $ orbitals in tetrahedral complexes point directly towards the ligands and hence have a splitting of energy less than that in the octahedral complex. Thus, option C is incorrect.
The ${d_{{x^2} - {y^2}}}$ and ${d_{{z^2}}}$ orbitals lie between the ligands and comparatively farther from ligands than the other three orbitals. Thus they become more stable and their energies lower. Meanwhile in octahedral complexes, the ${d_{{x^2} - {y^2}}}$ and ${d_{{z^2}}}$ form higher energy orbitals.
Thus, option A and D are correct.
Tetrahedral complexes have small energy barriers between $d - $ orbitals . Thus they have crystal field stabilization energy less than pairing energy and most tetrahedral complexes are high spin.
Thus, the correct options are A, B and D.
Note:
In free metal ions, all $d - $ orbitals are at the same energy level. These orbitals that have the same energy are known as degenerate orbitals. However, when a ligand approaches the metal, the electrons are repelled by the lone pairs of ligand. As a result, the energy of the $d - $ orbital increases. The orbitals that lie in the same direction of the ligands experience a greater repulsion. Whereas, the orbitals lying away from the ligands experience a relatively lower repulsion. So, the five degenerate $d - $ orbitals converted into different sets of orbitals having different energies is called crystal splitting.
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