English, asked by purabi0707, 3 months ago

in crystal field of Pbr5 the d orbital with highest energy is?

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Answered by Anonymous
5

When the ligands approach the central metal ion, d- or f-subshell degeneracy is broken due to the static electric field.

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Answered by divu45
1

your answer is option 2nd

In Crystal Field Theory, it is assumed that the ions are simple point charges (a simplification). When applied to alkali metal ions containing a symmetric sphere of charge, calculations of bond energies are generally quite successful. The approach taken uses classical potential energy equations that take into account the attractive and repulsive interactions between charged particles (that is, Coulomb's Law interactions).

E∝q1q2r(1)

with

E the bond energy between the charges and

q1 and q2 are the charges of the interacting ions and

r is the distance separating them.

This approach leads to the correct prediction that large cations of low charge, such as K+ and Na+ , should form few coordination compounds. For transition metal cations that contain varying numbers of d electrons in orbitals that are NOT spherically symmetric, however, the situation is quite different. The shape and occupation of these d-orbitals then becomes important in an accurate description of the bond energy and properties of the transition metal compound.

When examining a single transition metal ion, the five d-orbitals have the same energy (Figure 1 ). When ligands approach the metal ion, some experience more opposition from the d-orbital electrons than others based on the geometric structure of the molecule. Since ligands approach from different directions, not all d-orbitals interact directly. These interactions, however, create a splitting due to the electrostatic environment.

For example, consider a molecule with octahedral geometry. Ligands approach the metal ion along the x , y , and z axes. Therefore, the electrons in the dz2 and dx2−y2 orbitals (which lie along these axes) experience greater repulsion. It requires more energy to have an electron in these orbitals than it would to put an electron in one of the other orbitals. This causes a splitting in the energy levels of the d-orbitals. This is known as crystal field splitting. For octahedral complexes, crystal field splitting is denoted by Δo (or Δoct ). The energies of the dz2 and dx2−y2 orbitals increase due to greater interactions with the ligands. The dxy , dxz , and dyz orbitals decrease with respect to this normal energy level and become more stable.

Figure 2a.jpg

Figure 1 : (a) Distributing a charge of −6 uniformly over a spherical surface surrounding a metal ion causes the energy of all five d orbitals to increase due to electrostatic repulsions, but the five d orbitals remain degenerate. Placing a charge of −1 at each vertex of an octahedron causes the d orbitals to split into two groups with different energies: the dx2−y2 and dz2 orbitals increase in energy, while the, dxy, dxz, and dyz orbitals decrease in energy. The average energy of the five d orbitals is the same as for a spherical distribution of a −6 charge, however. Attractive electrostatic interactions between the negatively charged ligands and the positively charged metal ion (far right) cause all five d orbitals to decrease in energy but does not affect the splittings of the orbitals. (b) The two eg orbitals (left) point directly at the six negatively charged ligands, which increases their energy compared with a spherical distribution of negative charge. In contrast, the three t2g orbitals (right) point between the negatively charged ligands, which decreases their energy compared with a spherical distribution of charge.

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