Consider the given pairs of complex ions. The correct pair(s) in which each complex ion exhibits three spin-allowed $d\text{-}d$ transitions (excluding those arising due to distortions) is(are)

Question

Which is the weakest ligand?

Answer 1

The quickest route here is to reduce every complex to a single number, its $d$-electron count, and remember a short rule: only $d^2$, $d^3$, $d^7$ and $d^8$ configurations (with an $F$ ground term) give three spin-allowed $d$-$d$ bands, in EITHER octahedral or tetrahedral geometry, since the same $F$-term correlation diagram is used for all four. Configurations $d^1$, high-spin $d^4$, high-spin $d^6$ and $d^9$ (a $D$ term) give just one band, and high-spin $d^5$ gives none at all, because its $^6S$ ground state has no other sextet state to go to.

Work out the $d$-count for each metal ion in turn:

$\mathrm{V^{3+}} = d^2$, $\mathrm{Ni^{2+}} = d^8$, $\mathrm{Cr^{3+}} = d^3$, $\mathrm{Ti^{3+}} = d^1$, $\mathrm{Mn^{2+}} = d^5$ (HS), $\mathrm{Fe^{3+}} = d^5$ (HS with $\mathrm{F^-}$), $\mathrm{Fe^{2+}} = d^6$ (HS, tetrahedral).

Now sort the four option pairs by this rule:

(A) $\mathrm{[V(H_2O)_6]^{3+}}$ ($d^2$) and $\mathrm{[Ni(H_2O)_6]^{2+}}$ ($d^8$): both in the three-band group. Every complex in this pair qualifies.
(B) $\mathrm{[Cr(H_2O)_6]^{3+}}$ ($d^3$, octahedral) and $\mathrm{[NiCl_4]^{2-}}$ ($d^8$, tetrahedral): both again fall in the three-band group; the tetrahedral geometry does not change which group $d^8$ belongs to, it only changes the exact term labels and band positions, not the count of three.
(C) $\mathrm{[Ti(H_2O)_6]^{3+}}$ ($d^1$) and $\mathrm{[Mn(H_2O)_6]^{2+}}$ ($d^5$, HS): $d^1$ gives one band and high-spin $d^5$ gives zero bands (its ground term $^6A_{1g}$ has no other sextet term to jump to). Neither is a three-band ion.
(D) $\mathrm{[FeF_6]^{3-}}$ ($d^5$, HS) and $\mathrm{[FeCl_4]^{2-}}$ ($d^6$, HS, tetrahedral): the first gives zero bands for the same reason as $\mathrm{Mn^{2+}}$ above, and the second is a $D$-term ion, giving only one band.

Let's summarize:

The "three spin-allowed bands" club is exactly $d^2, d^3, d^7, d^8$, regardless of octahedral or tetrahedral geometry.
High-spin $d^5$ is the special case with zero spin-allowed bands, which is why $\mathrm{Mn^{2+}}$ and high-spin $\mathrm{Fe^{3+}}$ salts look almost colorless.

So the pairs where BOTH complexes show three spin-allowed transitions are (A) and (B).

Answer 2

The quickest route here is to reduce every complex to a single number, its $d$-electron count, and remember a short rule: only $d^2$, $d^3$, $d^7$ and $d^8$ configurations (with an $F$ ground term) give three spin-allowed $d$-$d$ bands, in EITHER octahedral or tetrahedral geometry, since the same $F$-term correlation diagram is used for all four. Configurations $d^1$, high-spin $d^4$, high-spin $d^6$ and $d^9$ (a $D$ term) give just one band, and high-spin $d^5$ gives none at all, because its $^6S$ ground state has no other sextet state to go to.

Work out the $d$-count for each metal ion in turn:

$\mathrm{V^{3+}} = d^2$, $\mathrm{Ni^{2+}} = d^8$, $\mathrm{Cr^{3+}} = d^3$, $\mathrm{Ti^{3+}} = d^1$, $\mathrm{Mn^{2+}} = d^5$ (HS), $\mathrm{Fe^{3+}} = d^5$ (HS with $\mathrm{F^-}$), $\mathrm{Fe^{2+}} = d^6$ (HS, tetrahedral).

Now sort the four option pairs by this rule:

(A) $\mathrm{[V(H_2O)_6]^{3+}}$ ($d^2$) and $\mathrm{[Ni(H_2O)_6]^{2+}}$ ($d^8$): both in the three-band group. Every complex in this pair qualifies.
(B) $\mathrm{[Cr(H_2O)_6]^{3+}}$ ($d^3$, octahedral) and $\mathrm{[NiCl_4]^{2-}}$ ($d^8$, tetrahedral): both again fall in the three-band group; the tetrahedral geometry does not change which group $d^8$ belongs to, it only changes the exact term labels and band positions, not the count of three.
(C) $\mathrm{[Ti(H_2O)_6]^{3+}}$ ($d^1$) and $\mathrm{[Mn(H_2O)_6]^{2+}}$ ($d^5$, HS): $d^1$ gives one band and high-spin $d^5$ gives zero bands (its ground term $^6A_{1g}$ has no other sextet term to jump to). Neither is a three-band ion.
(D) $\mathrm{[FeF_6]^{3-}}$ ($d^5$, HS) and $\mathrm{[FeCl_4]^{2-}}$ ($d^6$, HS, tetrahedral): the first gives zero bands for the same reason as $\mathrm{Mn^{2+}}$ above, and the second is a $D$-term ion, giving only one band.

Let's summarize:

The "three spin-allowed bands" club is exactly $d^2, d^3, d^7, d^8$, regardless of octahedral or tetrahedral geometry.
High-spin $d^5$ is the special case with zero spin-allowed bands, which is why $\mathrm{Mn^{2+}}$ and high-spin $\mathrm{Fe^{3+}}$ salts look almost colorless.

So the pairs where BOTH complexes show three spin-allowed transitions are (A) and (B).

Consider the given pairs of complex ions. The correct pair(s) in which each complex ion exhibits three spin-allowed \(d\text{-}d\) transitions (excluding those arising due to distortions) is(are)

Show Hint

The Correct Option is A, B

Solution and Explanation

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