Absorption Spectrum Analysis of [Cr(NH$_3$)$_6$]$^{3+}$

The question asks about the ground term and the spin-allowed electronic transitions observed in the absorption spectrum of the coordination complex [Cr(NH$_3$)$_6$]$^{3+}$. This complex is octahedral, and the central metal ion is Chromium in the +3 oxidation state.

Chromium(III) d Configuration

Chromium (Cr) has the electron configuration [Ar] 3d$^5$ 4s$^1$. In the [Cr(NH$_3$)$_6$]$^{3+}$ complex, Chromium is in the +3 oxidation state (Cr$^{3+}$). Cr$^{3+}$ loses three electrons (one from 4s and two from 3d), resulting in a 3d$^3$ electron configuration.

Free Ion Ground Term for d$^3$

To find the free ion ground term for the d$^3$ configuration, we apply Hund's rules:

• Place the three electrons in the 5 d orbitals individually with parallel spins to maximize spin multiplicity. The m$_l$ values are +2, +1, 0, -1, -2.
• Electrons occupy orbitals with m$_l$ = +2, +1, 0, each with spin +1/2.
• The total spin angular momentum, S = 1/2 + 1/2 + 1/2 = 3/2. The spin multiplicity is 2S+1 = 2(3/2) + 1 = 4.
• The total orbital angular momentum, L = $\sum$m$_l$ = (+2) + (+1) + (0) = +3.
• An L value of 3 corresponds to an F term.

Therefore, the free ion ground term for a d$^3$ configuration is $^4$F.

Electronic Transitions in Octahedral Field (d$^3$)

In an octahedral ligand field, the free ion terms split into different terms according to their symmetry. For a d$^3$ configuration in an octahedral field (O$_h$ symmetry), the $^4$F ground term splits into $^4$A$_{2g}$, $^4$T$_{2g}$, and $^4$T$_{1g}$(F). The lowest energy term among these is $^4$A$_{2g}$, which is the ground state term in the octahedral complex.

The absorption spectrum shows bands corresponding to electronic transitions from the ground state to higher energy states. Spin-allowed transitions are those where the spin multiplicity does not change (ΔS = 0). Since the ground state is a quartet (spin multiplicity 4, $^4$A$_{2g}$), the spin-allowed transitions will be to other quartet terms.

According to the Tanabe-Sugano diagram for d$^3$ in an octahedral field, the lowest energy spin-allowed transitions from the ground state $^4$A$_{2g}$ are:

1. $^4$A$_{2g}$ → $^4$T$_{2g}$ (lowest energy, corresponds to 10Dq)
2. $^4$A$_{2g}$ → $^4$T$_{1g}$(F) (next higher energy)
3. $^4$A$_{2g}$ → $^4$T$_{1g}$(P) (highest energy among common spin-allowed transitions)

The absorption spectrum of [Cr(NH$_3$)$_6$]$^{3+}$ shows two bands in the visible region, around 475 nm and 365 nm. These correspond to the first two spin-allowed transitions from the ground state:

• The band at 475 nm (longer wavelength, lower energy) corresponds to the transition $^4$A$_{2g}$ → $^4$T$_{2g}$.
• The band at 365 nm (shorter wavelength, higher energy) corresponds to the transition $^4$A$_{2g}$ → $^4$T$_{1g}$(F).

Therefore, the ground term (free ion) is $^4$F, and the spin-allowed transitions observed in the spectrum are $^4$A$_{2g}$ → $^4$T$_{2g}$ and $^4$A$_{2g}$ → $^4$T$_{1g}$(F).

Comparing this with the given options, the correct option states the ground term is $^4$F and the transitions are $^4$A$_{2g}$ → $^4$T$_{2g}$ and $^4$A$_{2g}$ → $^4$T$_{1g}$(F).