The question asks us to assign the electronic transitions corresponding to the observed bands in the spectrum of an aqueous solution of [Ni(H2O)6]²⁺. Let's analyze the complex.

[Ni(H2O)6]²⁺ is an octahedral complex. The central metal ion is Ni²⁺. Nickel (Ni) has an atomic number of 28. The electronic configuration of Ni is [Ar] 3d⁸ 4s². For Ni²⁺, the two 4s electrons are lost, resulting in a 3d⁸ configuration.

In an octahedral ligand field, a d⁸ configuration has a ³A_2g ground state term symbol according to the Tanabe-Sugano diagram. The spin-allowed electronic transitions from the ground state occur to other triplet states. For a d⁸ octahedral complex, the three spin-allowed transitions in order of increasing energy (and thus decreasing wavelength) are:

• $\nu_1: {}^3\text{T}_{2g} \leftarrow {}^3\text{A}_{2g}$ (lowest energy)
• $\nu_2: {}^3\text{T}_{1g}\text{(F)} \leftarrow {}^3\text{A}_{2g}$ (middle energy)
• $\nu_3: {}^3\text{T}_{1g}\text{(P)} \leftarrow {}^3\text{A}_{2g}$ (highest energy)

The electronic spectrum of [Ni(H2O)6]²⁺ shows three bands at different wavelengths:

• Band A at ~400 nm
• Band B at ~690 nm
• Band C at ~1070 nm

Remember that energy is inversely proportional to wavelength. The band at the longest wavelength (1070 nm) corresponds to the lowest energy transition ($\nu_1$). The band at the shortest wavelength (400 nm) corresponds to the highest energy transition ($\nu_3$). The band in between (690 nm) corresponds to the middle energy transition ($\nu_2$).

Let's match the bands to the transitions based on energy order:

• Band C (~1070 nm) is the lowest energy band. This corresponds to the $\nu_1$ transition: ${}^3\text{T}_{2g} \leftarrow {}^3\text{A}_{2g}$.
• Band B (~690 nm) is the middle energy band. This corresponds to the $\nu_2$ transition: ${}^3\text{T}_{1g}\text{(F)} \leftarrow {}^3\text{A}_{2g}$.
• Band A (~400 nm) is the highest energy band. This corresponds to the $\nu_3$ transition: ${}^3\text{T}_{1g}\text{(P)} \leftarrow {}^3\text{A}_{2g}$.

The question asks for the transitions assigned to A, B, and C, respectively. Therefore, the assignments are:

• A: ${}^3\text{T}_{1g}\text{(P)} \leftarrow {}^3\text{A}_{2g}$
• B: ${}^3\text{T}_{1g}\text{(F)} \leftarrow {}^3\text{A}_{2g}$
• C: ${}^3\text{T}_{2g} \leftarrow {}^3\text{A}_{2g}$

Comparing this with the given options, we look for the sequence: T_1g(P) $\leftarrow$ A_2g, T_1g $\leftarrow$ A_2g, and T_2g $\leftarrow$ A_2g. The options often omit the spin multiplicity (3) and sometimes the (F) label for clarity or simplicity in presenting the choices.

Let's examine the options provided:

Option 1: T_1g(P) $\leftarrow$ A_2g, T_2g $\leftarrow$ A_2g, and T_1g $\leftarrow$ A_2g (Incorrect order for B and C)

Option 2: T_1g(P) $\leftarrow$ A_2g, T_1g $\leftarrow$ A_2g, and T_2g $\leftarrow$ A_2g (Correct order and transitions, where T_1g without (P) refers to T_1g(F))

Option 3: T_2g $\leftarrow$ A_2g, T_1g $\leftarrow$ A_2g, and T_1g(P) $\leftarrow$ A_2g (Incorrect order for A and C)

Option 4: T_1g $\leftarrow$ A_2g, T_2g $\leftarrow$ A_2g, and T_1g(P) $\leftarrow$ A_2g (Incorrect order for A, B, and C)

Based on our analysis of the d⁸ octahedral complex and the energy ordering of the transitions, Option 2 correctly assigns the transitions to bands A, B, and C respectively.

Electronic Spectrum Analysis for [Ni(H2O)6]²⁺

The electronic spectrum arises from the absorption of light promoting electrons from the ground state to excited states. For [Ni(H2O)6]²⁺, a d⁸ system in an octahedral field, the ground state is ³A_2g. The transitions observed are spin-allowed transitions to the excited triplet states ³T_2g, ³T_1g(F), and ³T_1g(P).

The energy of these transitions increases in the order $\nu_1 < \nu_2 < \nu_3$.

• $\nu_1$: ³T_2g <-- ³A_2g
• $\nu_2$: ³T_1g(F) <-- ³A_2g
• $\nu_3$: ³T_1g(P) <-- ³A_2g

The observed bands A (~400 nm), B (~690 nm), and C (~1070 nm) correspond to these transitions based on their wavelengths. Shorter wavelength means higher energy.

• C (~1070 nm) is the lowest energy band, so it is $\nu_1$.
• B (~690 nm) is the middle energy band, so it is $\nu_2$.
• A (~400 nm) is the highest energy band, so it is $\nu_3$.

Therefore, the transitions for A, B, and C respectively are $\nu_3$, $\nu_2$, and $\nu_1$.

A: ³T_1g(P) <-- ³A_2g

B: ³T_1g(F) <-- ³A_2g

C: ³T_2g <-- ³A_2g

Comparing this to the structure of the options, the correct assignment sequence for A, B, and C is T_1g(P) <-- A_2g, T_1g <-- A_2g, and T_2g <-- A_2g, where the second T_1g transition is understood to be the one arising from the ³F term (i.e., ³T_1g(F)).

Assignments for Bands A, B, and C

Based on the energy ordering and the Tanabe-Sugano diagram for a d⁸ octahedral complex, the transitions for the bands are:

• Band A (~400 nm): Corresponds to the highest energy transition, $\nu_3$, which is T_1g(P) $\leftarrow$ A_2g.
• Band B (~690 nm): Corresponds to the middle energy transition, $\nu_2$, which is T_1g(F) $\leftarrow$ A_2g (often simply written as T_1g $\leftarrow$ A_2g when T_1g(P) is explicitly mentioned).
• Band C (~1070 nm): Corresponds to the lowest energy transition, $\nu_1$, which is T_2g $\leftarrow$ A_2g.

Thus, the correct assignment sequence for A, B, and C is T_1g(P) $\leftarrow$ A_2g, T_1g $\leftarrow$ A_2g, T_2g $\leftarrow$ A_2g.