JEE Challenger
More from Coordination Compounds
ChemMedAdv 2018 • Paper 2

Match Hybrid Orbitals with Transition Metal Complexes

Match each set of hybrid orbitals given in LIST-I with the corresponding complex(es) provided in LIST-II.

LIST–ILIST–IIP. dsp21. [FeF6]4−Q. sp32. [Ti(H2O)3Cl3]R. sp3d23. [Cr(NH3)6]3+S. d2sp34. [FeCl4]2−5. Ni(CO)46. [Ni(CN)4]2−\begin{array}{ll} \textbf{LIST--I} & \textbf{LIST--II} \\ \textbf{P.}\ \text{dsp}^2 & \textbf{1.}\ [\text{FeF}_6]^{4-} \\ \textbf{Q.}\ \text{sp}^3 & \textbf{2.}\ [\text{Ti}(\text{H}_2\text{O})_3\text{Cl}_3] \\ \textbf{R.}\ \text{sp}^3\text{d}^2 & \textbf{3.}\ [\text{Cr}(\text{NH}_3)_6]^{3+} \\ \textbf{S.}\ \text{d}^2\text{sp}^3 & \textbf{4.}\ [\text{FeCl}_4]^{2-} \\ & \textbf{5.}\ \text{Ni}(\text{CO})_4 \\ & \textbf{6.}\ [\text{Ni}(\text{CN})_4]^{2-} \end{array}

Choose the correct matching option:

Options

A

P→5;Q→4,6;R→2,3;S→1\mathbf{P} \rightarrow 5;\quad \mathbf{Q} \rightarrow 4, 6;\quad \mathbf{R} \rightarrow 2, 3;\quad \mathbf{S} \rightarrow 1

B

P→5,6;Q→4;R→3;S→1,2\mathbf{P} \rightarrow 5, 6;\quad \mathbf{Q} \rightarrow 4;\quad \mathbf{R} \rightarrow 3;\quad \mathbf{S} \rightarrow 1, 2

C

P→6;Q→4,5;R→1;S→2,3\mathbf{P} \rightarrow 6;\quad \mathbf{Q} \rightarrow 4, 5;\quad \mathbf{R} \rightarrow 1;\quad \mathbf{S} \rightarrow 2, 3

Correct
D

P→4,6;Q→5,6;R→1,2;S→3\mathbf{P} \rightarrow 4, 6;\quad \mathbf{Q} \rightarrow 5, 6;\quad \mathbf{R} \rightarrow 1, 2;\quad \mathbf{S} \rightarrow 3

Topics & Concepts

Coordination CompoundsCoordination CompoundsHybridization

Step-by-Step Solution

To determine the correct matching between the hybrid orbitals in LIST-I and the coordination complexes in LIST-II, we analyze the electronic configuration, oxidation state, coordination number, and hybridization of the central metal atom/ion for each complex:

  1. [FeF6]4−[\text{FeF}_6]^{4-}:

    • Oxidation state of Fe=+2\text{Fe} = +2.
    • Electronic configuration of Fe2+=[Ar] 3d6\text{Fe}^{2+} = [\text{Ar}]\,3\text{d}^6.
    • Since F−\text{F}^- is a weak field ligand, electron pairing does not occur in the 3d3\text{d} subshell.
    • Coordination number is 66, so the complex uses outer orbitals for bonding, resulting in an outer orbital octahedral complex with sp3d2\text{sp}^3\text{d}^2 hybridization.
    • Therefore, 1→R1 \rightarrow \mathbf{R}.

  2. [Ti(H2O)3Cl3][\text{Ti}(\text{H}_2\text{O})_3\text{Cl}_3]:

    • Oxidation state of Ti=+3\text{Ti} = +3.
    • Electronic configuration of Ti3+=[Ar] 3d1\text{Ti}^{3+} = [\text{Ar}]\,3\text{d}^1.
    • Coordination number is 66. Two empty inner 3d3\text{d} orbitals are readily available for bonding regardless of the ligand field strength.
    • The hybridization is inner orbital octahedral, which is d2sp3\text{d}^2\text{sp}^3.
    • Therefore, 2→S2 \rightarrow \mathbf{S}.

  3. [Cr(NH3)6]3+[\text{Cr}(\text{NH}_3)_6]^{3+}:

    • Oxidation state of Cr=+3\text{Cr} = +3.
    • Electronic configuration of Cr3+=[Ar] 3d3\text{Cr}^{3+} = [\text{Ar}]\,3\text{d}^3.
    • Coordination number is 66. With three electrons residing in the t2gt_{2g} orbitals, two vacant 3d3\text{d} (ege_g) orbitals are available for hybridization.
    • The hybridization is inner orbital octahedral, d2sp3\text{d}^2\text{sp}^3.
    • Therefore, 3→S3 \rightarrow \mathbf{S}.

  4. [FeCl4]2−[\text{FeCl}_4]^{2-}:

    • Oxidation state of Fe=+2\text{Fe} = +2.
    • Electronic configuration of Fe2+=[Ar] 3d6\text{Fe}^{2+} = [\text{Ar}]\,3\text{d}^6.
    • Coordination number is 44. Since Cl−\text{Cl}^- is a weak field ligand, it forms a high-spin tetrahedral complex.
    • The hybridization is sp3\text{sp}^3.
    • Therefore, 4→Q4 \rightarrow \mathbf{Q}.

  5. Ni(CO)4\text{Ni}(\text{CO})_4:

    • Oxidation state of Ni=0\text{Ni} = 0.
    • Electronic configuration of Ni0=[Ar] 3d84s2\text{Ni}^0 = [\text{Ar}]\,3\text{d}^8 4\text{s}^2.
    • CO\text{CO} is a strong field ligand which forces the 4s4\text{s} electrons into the 3d3\text{d} orbitals, fully pairing all ten 3d3\text{d} electrons (3d103\text{d}^{10}).
    • Coordination number is 44. Hybridization involves the empty 4s4\text{s} and 4p4\text{p} orbitals, giving a tetrahedral geometry with sp3\text{sp}^3 hybridization.
    • Therefore, 5→Q5 \rightarrow \mathbf{Q}.

  6. [Ni(CN)4]2−[\text{Ni}(\text{CN})_4]^{2-}:

    • Oxidation state of Ni=+2\text{Ni} = +2.
    • Electronic configuration of Ni2+=[Ar] 3d8\text{Ni}^{2+} = [\text{Ar}]\,3\text{d}^8.
    • CN−\text{CN}^- is a strong field ligand which causes pairing of the two unpaired 3d3\text{d} electrons, leaving one 3d3\text{d} orbital vacant.
    • Coordination number is 44. This gives a square planar geometry with dsp2\text{dsp}^2 hybridization.
    • Therefore, 6→P6 \rightarrow \mathbf{P}.

Summary of Matching:

  • P (dsp2)→6\mathbf{P}\ (\text{dsp}^2) \rightarrow 6
  • Q (sp3)→4,5\mathbf{Q}\ (\text{sp}^3) \rightarrow 4, 5
  • R (sp3d2)→1\mathbf{R}\ (\text{sp}^3\text{d}^2) \rightarrow 1
  • S (d2sp3)→2,3\mathbf{S}\ (\text{d}^2\text{sp}^3) \rightarrow 2, 3

Thus, the correct option is (C).

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