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Question

In the first row high-spin transition metal complexes $[M(H_2O)_6]Cl_2$ with $d^5$ and $d^7$ metal ions, the $d-d$ transitions are

The correct answer is
spin- forbidden for $d^5$ and spin -allowed for $d^7$

Understanding $d-d$ Transition Spin Rules

The nature of $d-d$ transitions (spin-allowed or spin-forbidden) depends on whether the total spin multiplicity changes during the electronic excitation. The spin selection rule states that transitions are allowed only if $\Delta S = 0$. We analyze the spin states for the given $d^5$ and $d^7$ high-spin configurations in an octahedral field.

High-Spin $d^5$ Configuration Analysis

For a high-spin $d^5$ metal ion in an octahedral complex, the electron configuration is $t_{2g}^3 e_g^2$. All five electrons are unpaired.

  • Number of unpaired electrons = 5.
  • Total spin quantum number, $S = \frac{\text{Number of unpaired electrons}}{2} = \frac{5}{2}$.
  • Spin multiplicity = $2S + 1 = 2(\frac{5}{2}) + 1 = 6$. This corresponds to a sextet state.
  • In $d-d$ transitions, an electron moves between $d$ orbitals. For $d^5$ high-spin, the ground state is a sextet ($^6S$). The possible excited states typically do not have the same sextet multiplicity.
  • Since the transition involves a change in spin multiplicity ($\Delta S \neq 0$), the $d-d$ transitions are spin-forbidden.

High-Spin $d^7$ Configuration Analysis

For a high-spin $d^7$ metal ion in an octahedral complex, the electron configuration is $t_{2g}^5 e_g^2$. We determine the number of unpaired electrons:

  • In the $t_{2g}$ orbitals ($t_{2g}^5$): Two orbitals are paired, one is unpaired ($\uparrow$). So, 1 unpaired electron from $t_{2g}$.
  • In the $e_g$ orbitals ($e_g^2$): Both orbitals have one unpaired electron ($\uparrow, \uparrow$). So, 2 unpaired electrons from $e_g$.
  • Total number of unpaired electrons = 1 + 2 = 3.
  • Total spin quantum number, $S = \frac{3}{2}$.
  • Spin multiplicity = $2S + 1 = 2(\frac{3}{2}) + 1 = 4$. This corresponds to a quartet state.
  • The ground state term for $t_{2g}^5 e_g^2$ is a quartet state (e.g., $^4A_{2g}$). The excited states arising from $d-d$ transitions (e.g., from $t_{2g}^4 e_g^3$) also result in quartet states.
  • Since the transition occurs between states of the same spin multiplicity ($\Delta S = 0$), the $d-d$ transitions are spin-allowed.

Conclusion on Transitions

Based on the analysis:

  • $d^5$ high-spin: $d-d$ transitions are spin-forbidden.
  • $d^7$ high-spin: $d-d$ transitions are spin-allowed.

Therefore, the correct description is that $d-d$ transitions are spin-forbidden for $d^5$ and spin-allowed for $d^7$.

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Important Questions from Electronic Spectra

  1. The UV-visible spectrum of $[Ni(en)_3]^{2+}$ (en = ethylenediamine) shows absorbance maxima at $11200 \text{ cm}^{-1}$, $18350 \text{ cm}^{-1}$, and $29000 \text{ cm}^{-1}$.
    Absorbance maximumElectronic transition
    (a) $11200 \text{ cm}^{-1}$(i) $^3A_{2g} \to ^3T_{1g} (F)$
    (b) $18350 \text{ cm}^{-1}$(ii) $^3A_{2g}\to^3T_{2g}$
    (c) $29000 \text{ cm}^{-1}$(iii) $^3A_{2g}\to^3T_{1g} (P)$

    [Given: Atomic number of Ni = 28]
    The correct match(es) between absorbance maximum and electronic transition is/are
  2. In aqueous solution of $K_4[Fe(CN)_6]$, the allowed transition(s) is (are)
  3. The $VO_4^{3-}$, $CrO_4^{2-}$ and $MnO_4^-$ ions exhibit intense ligand to metal charge transfer transition. The wavelengths of this transition follow the order
  4. The lowest energy d $\rightarrow$ d transition of the complexes follow the order
  5. The intense red color of $[Fe(bpy)_3]^{2+}$ (bpy = 2,2'-bipyridine) is due to
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