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Question

Identify transition metal complexes which are not octahedral in shape.

(A) [Co(NH₃)₆]³⁺

(B) [Ni(CO)₄]

(C) [CoCl(NH₃)₅]²⁺

(D) [CoCl₂(NH₃)₄]⁺

(E) [PtCl₄]²⁻

Choose the correct answer from the options given below:

The correct answer is

(B) and (E) only

Understanding Shapes of Transition Metal Complexes

Transition metal complexes exhibit various shapes, primarily determined by the coordination number of the central metal ion and the electronic configuration, which is influenced by the nature of the ligands.

Common shapes for transition metal complexes include:

  • Coordination number 4: Tetrahedral or Square planar.
  • Coordination number 5: Trigonal bipyramidal or Square pyramidal.
  • Coordination number 6: Octahedral.

The question asks us to identify complexes that are not octahedral in shape. Let's analyze each given complex:

Analysis of Each Transition Metal Complex

(A) $[Co(NH₃)₆]$$^{3+}$

  • Central metal: Cobalt (Co)
  • Oxidation state: +3. Since NH₃ is neutral, Co + 6(0) = +3, so Co is in the +3 oxidation state. Cobalt(III) has a d⁶ electronic configuration.
  • Ligands: Six ammonia (NH₃) ligands.
  • Coordination number: 6.
  • Shape: Complexes with coordination number 6 are typically octahedral. Co(III) d⁶ complexes with strong field ligands like NH₃ are low spin and octahedral.

This complex is octahedral.

(B) $[Ni(CO)₄]$

  • Central metal: Nickel (Ni)
  • Oxidation state: 0. Since CO is neutral, Ni + 4(0) = 0, so Ni is in the 0 oxidation state. Nickel(0) has a d¹⁰ electronic configuration.
  • Ligands: Four carbonyl (CO) ligands.
  • Coordination number: 4.
  • Shape: For coordination number 4, shapes can be tetrahedral or square planar. d¹⁰ complexes typically form tetrahedral structures because the d-orbitals are completely filled, and there is no crystal field stabilization energy preference for square planar geometry. Carbonyl is also a strong field ligand, but for d¹⁰, tetrahedral is common.

This complex is tetrahedral, which is not octahedral.

(C) $[CoCl(NH₃)₅]$$^{2+}$

  • Central metal: Cobalt (Co)
  • Oxidation state: +3. Since NH₃ is neutral (0) and Cl has a charge of -1, Co + (-1) + 5(0) = +2, so Co is in the +3 oxidation state. Cobalt(III) has a d⁶ electronic configuration.
  • Ligands: Five ammonia (NH₃) ligands and one chloride (Cl⁻) ligand.
  • Coordination number: 5 + 1 = 6.
  • Shape: Complexes with coordination number 6 are typically octahedral. This is an octahedral complex with two different types of ligands.

This complex is octahedral.

(D) $[CoCl₂(NH₃)₄]$$^{+}$

  • Central metal: Cobalt (Co)
  • Oxidation state: +3. Since NH₃ is neutral (0) and Cl has a charge of -1, Co + 2(-1) + 4(0) = +1, so Co is in the +3 oxidation state. Cobalt(III) has a d⁶ electronic configuration.
  • Ligands: Four ammonia (NH₃) ligands and two chloride (Cl⁻) ligands.
  • Coordination number: 4 + 2 = 6.
  • Shape: Complexes with coordination number 6 are typically octahedral. This is another octahedral complex with two different types of ligands.

This complex is octahedral.

(E) $[PtCl₄]$$^{2-}$

  • Central metal: Platinum (Pt)
  • Oxidation state: +2. Since Cl has a charge of -1, Pt + 4(-1) = -2, so Pt is in the +2 oxidation state. Platinum(II) is a d⁸ ion.
  • Ligands: Four chloride (Cl⁻) ligands.
  • Coordination number: 4.
  • Shape: For coordination number 4, shapes can be tetrahedral or square planar. For d⁸ ions, especially with second and third-row transition metals like Pt(II), square planar geometry is highly favored due to significant crystal field stabilization energy.

This complex is square planar, which is not octahedral.

Summary of Complex Shapes

Let's summarize the shapes we determined:

Complex Metal Oxidation State d Configuration Coordination Number Shape Octahedral?
$[Co(NH₃)₆]$$^{3+}$ Co +3 d⁶ 6 Octahedral Yes
$[Ni(CO)₄]$ Ni 0 d¹⁰ 4 Tetrahedral No
$[CoCl(NH₃)₅]$$^{2+}$ Co +3 d⁶ 6 Octahedral Yes
$[CoCl₂(NH₃)₄]$$^{+}$ Co +3 d⁶ 6 Octahedral Yes
$[PtCl₄]$$^{2-}$ Pt +2 d⁸ 4 Square planar No

The complexes that are not octahedral are $[Ni(CO)₄]$ (B) and $[PtCl₄]$$^{2-}$ (E).

Conclusion

Based on our analysis, the transition metal complexes which are not octahedral in shape are $[Ni(CO)₄]$ and $[PtCl₄]$$^{2-}$. This corresponds to options (B) and (E).

Revision Table: Coordination Complex Shapes

Coordination Number Common Shapes Examples (Non-Octahedral)
2 Linear $[Ag(NH₃)₂]$$^{+}$
3 Trigonal planar $[HgI₃]$$^{-}$
4 Tetrahedral $[Ni(CO)₄]$, $[ZnCl₄]$$^{2-}$
4 Square planar $[PtCl₄]$$^{2-}$, $[Ni(CN)₄]$$^{2-}$
5 Trigonal bipyramidal, Square pyramidal $[Fe(CO)₅]$
6 Octahedral $[Co(NH₃)₆]$$^{3+}$, $[Cr(H₂O)₆]$$^{3+}$

Additional Information on Transition Metal Complex Geometry

The shape of a transition metal complex is a fundamental property influencing its reactivity, physical properties, and spectroscopic behavior. While coordination number is the primary determinant of geometry, factors like the metal's electronic configuration (especially d-electron count), the size of the metal ion, and the nature of the ligands (steric and electronic effects) play crucial roles.

  • Crystal Field Theory (CFT): CFT helps explain why certain geometries are favored for specific d-electron configurations. For instance, the strong stabilization of the d$_{x²-y²}$ orbital in square planar geometry makes it highly favorable for d⁸ ions, especially with strong field ligands, compared to tetrahedral geometry where d-orbitals are degenerate.
  • Ligand Field Theory (LFT) / Molecular Orbital Theory (MOT): These more advanced theories provide a more complete picture by considering covalent bonding effects in addition to electrostatic interactions.
  • Isomerism: The geometry of a complex is essential for understanding isomerism, such as cis-trans isomerism in square planar and octahedral complexes, and facial-meridional isomerism in octahedral complexes.

Identifying the correct shape is a key skill in studying coordination chemistry.

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Important Questions from Electrochemistry

  1. The product of complete hydrolysis of XeF₆ in the following reaction is:

    XeF₆ + H₂O → ? HF

  2. In a reaction A and B react to form product. The initial rate of reaction (ro) was determined using different initial concentrations of A and B as shown below:

    A/mol L-1B/mol L-1ro/mol L-1 s-1
    0.100.306.81 × 10-4
    0.100.102.27 × 10-4
    0.200.3013.62 × 10-4

    What is the initial rate of reaction (ro) when the critical concentration of A and B is 0.50 mol/L and 0.50 mol/L, respectively?

  3. In which of the following actinoid elements 6d subshell is vacant?

  4. Which of the following shows both, Frenkel and Schottky defect?

  5. The role of a catalyst is to change:

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