D Block Elements

To determine the ground state and the maximum number of spin-allowed electronic transitions possible for a tetrahedral complex, we need to consider the electronic configuration and crystal field theory.

1. **Electronic Configuration and Ground State:**
   • ion is derived from which has the electronic configuration . As , it loses two electrons, resulting in .
   • In a tetrahedral complex, the crystal field splitting pattern is such that the orbitals (higher energy) are above the orbitals (lower energy).
   • gives a high spin tetrahedral configuration because tetrahedral field does not provide enough splitting to pair electrons (weak field ligands).
   • The ground state term for a high-spin configuration in a tetrahedral field is .
2. **Determining Spin-Allowed Transitions:**
   • Spin-allowed transitions for any dⁿ configuration are governed by selection rules which allow transitions between terms of the same spin multiplicity.
   • For a configuration in a tetrahedral field, the possible terms based on Hund's rules and considering weak field are , , and .
   • Transitions possible from the ground state are to and , thus counting as three transitions: , , and a transition back to itself, .

Therefore, the correct answer is and 3 spin-allowed transitions. This matches the option and 3.

To determine the number of unpaired electrons in the complex , we must first consider the properties of the cobalt ion involved. In this complex, the cobalt is in the +2 oxidation state.

Step 1: Determine the electronic configuration of Co

Cobalt (Co), with an atomic number of 27, has the electron configuration: .

Step 2: Determine the electronic configuration of

When Co loses two electrons to form , these electrons are removed from the 4s orbital and one from the 3d orbital. Thus, the electron configuration for is: .

Step 3: Analyze the electronic configuration of

The -orbitals can hold a maximum of 10 electrons. In the case of , the electrons will fill according to Hund’s rule: first singly in each orbital before any pairing occurs.

Considering the five -orbitals, the arrangement is as follows (where ↑ represents an unpaired electron and ↑↓ represents a paired electron):

- Orbital 1: ↑↑
- Orbital 2: ↑↑
- Orbital 3: ↑↑
- Orbital 4: ↑
- Orbital 5: vacant

Conclusion

This configuration results in three unpaired electrons.

Hence, the number of unpaired electrons in is 3.

To determine the total degeneracy (number of microstates) for a Ti³⁺ ion in spherical symmetry, we need to understand the electron configuration and apply the formula for calculating microstates. A Ti³⁺ ion is derived from a titanium atom (atomic number 22) by removing three electrons, giving an electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 3d¹. The electron configuration shows that there is one electron in the 3d subshell. The number of microstates for a given number of electrons (n) in a d-orbital is calculated using the formula: total microstates = (2l + 1)! / [n!(2l + 1 - n)!], where l = 2 for d-orbitals. Plugging the values into the formula, we have: total microstates = 5! / [1!(5 - 1)!] = 5! / [1!×4!] = 5 × 4! / 4! = 5. The calculated number of microstates for a single electron in a 3d orbital is 5. A singly filled d-orbital has 10 microstates due to spin multiplicity, as each orbital has two possible spin orientations. Thus, the total degeneracy for Ti³⁺ is 10.

To determine the correct order of energy levels for the d orbitals in a square planar complex, it is important to understand the geometry and electron configuration involved.

Square planar complexes are typically formed by d⁸ metal ions. In such complexes, there is a distinct splitting of the d orbitals due to the ligand field effect.

The splitting pattern of d orbitals in a square planar complex is primarily influenced by both the nature of ligands and the metal center. However, the typical energy order of the d orbitals when influenced by the ligand field is:

1. The orbital experiences the highest energy because it points directly towards the surrounding ligands on the x and y axes.
2. The orbital has lesser energy than because it is in the plane but not directly facing the ligands.
3. The orbital is next, as it is oriented along the z-axis, further from the ligand plane, leading to less direct interaction with ligands.
4. Lastly, the and orbitals have the lowest energy because they are oriented between the ligand axes, leading to minimal direct overlap.

Therefore, the order of increasing energy is:

This matches the correct answer: .

All other options do not correctly represent the energy ordering of d orbitals in a square planar complex, validating our correct option choice. For example:

• The option incorrectly places below .
• The option incorrectly separates and .
• The option gives an incorrect overall sequence.

• (A) The rate is not necessarily faster in the case of SF₆ than in [AlF₆]³⁻, because the size and charge of the metal ion as well as the nature of the ligands play a significant role in the rate of substitution reactions.
• (B) The rate of substitution reactions generally increases as the size of the metal ion increases, meaning [Mg(H₂O)₆]²⁺ will have a slower rate than [Sr(H₂O)₆]²⁺, not the other way around.
• (C) The rate of water exchange is faster in [Ni(H₂O)₆]²⁺ compared to [Co(NH₃)₅(H₂O)]²⁺ due to the difference in the coordination environment and the charge distribution around the central metal ion.
• (D) The rate of substitution is generally faster in [Cr(H₂O)₆]²⁺ than in [Cr(H₂O)₆]³⁺, because the higher charge on the chromium ion in [Cr(H₂O)₆]³⁺ results in stronger ligand binding and slower exchange.

Thus, the correct answers are (C) and (D).

For Sc²⁺, the electronic configuration is [Ar]3d¹, and the d-electrons are in the t_2g and e_g orbitals in an octahedral field.

1. The number of d-d transitions is determined by the number of unpaired electrons. In Sc²⁺, there is only one unpaired electron in the 3d¹ configuration, so only 1 transition is possible.
2. The nature of the transition is ”spin allowed” because a single electron can undergo a spin-flip transition without violating any selection rules.

Thus, the number and nature of d-d transitions for Sc²⁺ are 1 and spin allowed.

The correct answer is (A)

Among the given complexes, the one with the largest molar absorptivity in the visible region of the electronic absorption spectrum is usually the one that has the most intense d-d transitions or charge transfer absorptions.

• [Ti(H₂O)₆]³⁺ has weak transitions in the visible region.
• [NiCl₄]²⁻ has d-d transitions that absorb light, but they are not as strong in the visible region.
• [CrO₄]²⁻ has a strong absorption due to charge transfer between the metal and ligand, which leads to the largest
• molar absorptivity in the visible region.
• [Mn(H₂O)₆]²⁺ also has weak d-d transitions in the visible region.

Therefore, the complex [CrO₄]²⁻ exhibits the largest molar absorptivity in the visible region.

Thus, the correct answer is (C).

To determine the oxidation states of manganese (Mn) in the compounds P and Q:

1. In compound P, Na[Mn(CO)₅] :
   • In the Na[Mn(CO)₅] complex, carbon monoxide (CO) is a neutral ligand, and sodium (Na) is Na⁺.
   • Since the overall charge of the complex is neutral, the oxidation state of Mn must be +1.
2. In compound Q, [CH₃Mn(CO)₅] :
   • The methyl group (CH₃) is also neutral, and the CO ligands remain neutral.
   • The oxidation state of Mn in this complex must be +1, as the overall charge of the complex is zero.

Thus, the oxidation states of Mn in P and Q are +1 and +1, respectively

In [Mn(H₂O)₆]²⁺, the Mn²⁺ ion has the electronic configuration [Ar]3d⁵. Due to this configuration, the d-d transitions are symmetry forbidden because the 3d⁵ configuration is stable and does not easily undergo transitions.

In [Ti(H₂O)₆]³⁺, the Ti³⁺ ion has the electronic configuration [Ar]3d¹. The single electron in the 3d¹ configuration can undergo d-d transitions, making them symmetry allowed.

Thus, the d-d transitions in [Mn(H₂O)₆]²⁺ are symmetry forbidden, while those in [Ti(H₂O)₆]³⁺ are symmetry allowed.

The correct answer is (B)

Ionization isomers are compounds that contain the same components, but the ions are arranged differently in the crystal structure, resulting in the exchange of ions between the cation and anion.

• [Co(NH₃)₆(SO₄)]Br and [Co(NH₃)₆]BrSO₄ are examples of ionization isomers because the sulfate and bromide ions have swapped positions between the cation and the anion in the two complexes.
• Linkage isomers involve ligands that can coordinate to the metal through different atoms, which is not the case here.
• Optical isomers are mirror images of each other, which is not applicable here because the two complexes are not non-superimposable mirror images.
• Coordination isomers refer to different coordination spheres around the metal center, which also does not apply in this case.

Thus, the correct answer is (A) ionization isomers.