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Which of the following complexes is (are) square planar? (a) \(\left[\mathrm{Ti}(\mathrm{CN})_{4}\right]^{2-}\) (b) \(\left[\mathrm{Ni}(\mathrm{CN})_{4}\right]^{2-}\) (c) \(\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}\) (d) \(\left[\mathrm{Pt}(\mathrm{CN})_{4}\right]^{2-}\)

Short Answer

Expert verified
(b) and (d) are square planar.

Step by step solution

01

Analyze Electron Configuration of the Metal

Identify the oxidation state and electron configuration of each metal ion. For square planar complexes, the metal often has a \(d^8\) electron configuration.
02

Determine Oxidation State and Configuration for (a) Ti Complex

In \([\mathrm{Ti}(\mathrm{CN})_4]^{2-}\), Ti starts as \(4+\) in \(\mathrm{Ti}^{4+}\) with configuration \(1s^2 2s^2 2p^6 3s^2 3p^6\). Ti has no d electrons, so this complex cannot be square planar.
03

Determine Oxidation State and Configuration for (b) Ni Complex

In \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\), Ni starts as \(+2\). \(\mathrm{Ni}^{2+}\) has a \(d^8\) configuration, which is favorable for a square planar geometry.
04

Determine Oxidation State and Configuration for (c) Zn Complex

In \([\mathrm{Zn}(\mathrm{CN})_4]^{2-}\), Zn starts as \(+2\). \(\mathrm{Zn}^{2+}\) is \(d^{10}\), and does not favor a square planar configuration.
05

Determine Oxidation State and Configuration for (d) Pt Complex

In \([\mathrm{Pt}(\mathrm{CN})_4]^{2-}\), Pt is typically \(+2\). \(\mathrm{Pt}^{2+}\) has a \(d^8\) configuration, which is known to favor square planar geometry for platinum complexes.
06

Conclusion

Nickel and platinum complexes with \(d^8\) electron configurations can be square planar. Therefore, \([\mathrm{Ni}(\mathrm{CN})_4]^{2-}\) and \([\mathrm{Pt}(\mathrm{CN})_4]^{2-}\) are square planar.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Electron Configuration
Understanding electron configuration is key to predicting the geometry of metal complexes. Electron configuration describes the distribution of electrons within an atom or ion's orbitals. For transition metals, electrons fill orbitals following specific rules designed to optimize electron stability. These include the Aufbau principle, Hund's rule, and the Pauli exclusion principle.
  • The Aufbau principle states electrons fill from the lowest available energy levels to higher ones.
  • Hund's rule suggests electrons will fill degenerate orbitals singly first to the same spin before doubling up.
  • The Pauli exclusion principle dictates that no two electrons in the same atom can have identical quantum numbers.
For complexes, understanding the metal's electron configuration helps to predict how it will interact with ligands, which in turn affects the resulting geometry.
Oxidation State
The oxidation state of a metal in a complex provides insight into its electron holding capacity following ligand interactions. It is a hypothetical charge that indicates the degree of oxidation (loss of electrons) of an atom. When determining a metal's oxidation state in a metal complex:
  • Consider the neutral state of the metal and the formal charges of surrounding ligands.
  • Use the summation of oxidation states within a complex to achieve the net charge of the entire compound.
For example, in the complex \( [\text{Ni}(\text{CN})_4]^{2-} \), Ni's oxidation state is \(+2\) because cyanides are neutral ligands. Knowing the oxidation state aids in determining electron configurations, such as identifying if \( d^8 \) configuration occurs.
d8 Electron Configuration
The \( d^8 \) electron configuration is critical in predicting a square planar geometry for metal complexes. This specific arrangement of electrons involves having eight electrons distributed across the five d orbitals. Metal complexes with \( d^8 \) typically adopt a square planar structure due to electronic and steric factors which allow for the minimization of repulsion between ligand and metal electron pairs. Metals like nickel (\( ext{Ni}^{2+} \)) and platinum (\( ext{Pt}^{2+} \)) are classic examples where \( d^8 \) configuration leads to this geometry.
- Nickel with its \( d^8 \) configuration aligns in two paired and four unpaired electrons favoring a complete planar dispersion.- Platinum complexes are more stable in square planar configurations due to strong ligand field stabilization energy achieved.Thus, examining the electron configuration of a metal ion is vital to understanding its geometrical preferences.
Metal Complexes
Metal complexes, also known as coordination compounds, consist of a central metal atom or ion surrounded by ligands that are molecules or ions bonded through coordinate covalent bonds. These complexes exhibit various geometric structures, such as octahedral, tetrahedral, and square planar, influenced by factors like electron configuration and ligand strength.
In transition metals, the interplay of d-orbitals and ligands dictates the geometry of the complex:
  • Ligands can donate pairs of electrons, stabilizing certain d-orbital configurations.
  • The choice of ligands and their electrical field affects the shape and form of these complexes.
The coordination number, type of ligand, and electron distribution within metal complexes are critical to understanding their complex behaviors and functionalities, delivering insights into their reactivity and stability. This makes metal complexes an essential concept in both inorganic chemistry and materials science.

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Most popular questions from this chapter

Determine the number of unpaired electrons in the following tetrahedral complexes. All tetrahedral complexes are high spin. (a) \(\left[\mathrm{FeCl}_{4}\right]^{2-}\) (c) \(\left[\mathrm{MnCl}_{4}\right]^{2-}\) (b) \(\mathrm{Na}_{2}\left[\mathrm{CoCl}_{4}\right]\) (d) \(\left(\mathrm{NH}_{4}\right)_{2}\left[\mathrm{ZnCl}_{4}\right]\)

The following equations represent various ways of obtaining transition metals from their compounds. Balance each equation. (a) \(\mathrm{Cr}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})+\mathrm{Cr}(\mathrm{s})\) (b) \(\operatorname{TiCl}_{4}(\ell)+\operatorname{Mg}(\mathrm{s}) \longrightarrow \operatorname{Ti}(\mathrm{s})+\operatorname{Mg} \mathrm{Cl}_{2}(\mathrm{s})\) (c) \(\left[\mathrm{Ag}(\mathrm{CN})_{2}\right]^{-}(\mathrm{aq})+\mathrm{Zn}(\mathrm{s}) \longrightarrow\) \(\mathrm{Ag}(\mathrm{s})+\left[\mathrm{Zn}(\mathrm{CN})_{4}\right]^{2-}(\mathrm{aq})\) (d) \(\mathrm{Mn}_{3} \mathrm{O}_{4}(\mathrm{s})+\mathrm{Al}(\mathrm{s}) \longrightarrow \mathrm{Mn}(\mathrm{s})+\mathrm{Al}_{2} \mathrm{O}_{3}(\mathrm{s})\)

A How many geometric isomers of the complex \(\left[\mathrm{Cr}(\mathrm{dmen})_{3}\right]^{3+}\) can exist? (dmen is the bidentate ligand 1,1-dimethylethylenediamine.) $$ \left(\mathrm{CH}_{3}\right)_{2} \ddot{\mathrm{N}} \mathrm{CH}_{2} \mathrm{CH}_{2} \ddot{\mathrm{N}} \mathrm{H}_{2} $$ 1,1-Dimethylethylenediamine, dmen

Give the name or formula for each ion or compound, as appropriate. (a) dichlorotetraaquachromium(III) chloride (b) \(\left[\mathrm{Cr}\left(\mathrm{NH}_{3}\right)_{5} \mathrm{SO}_{4}\right] \mathrm{Cl}\) (c) sodium tetrachlorocobaltate( (11) (d) \(\left[\mathrm{Fe}\left(\mathrm{C}_{2} \mathrm{O}_{4}\right)_{3}\right]^{3-}\)

How many geometric isomers are possible for the square planar complex \(\left[\mathrm{Pt}\left(\mathrm{NH}_{3}\right)(\mathrm{CN}) \mathrm{Cl}_{2}\right]^{-} ?^{-}\)

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