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Chapter 23: Problem 56

Obtain the distribution of \(d\) electrons in the complex ions listed below, using crystal field theory. Each ion is either tetrahedral or square planar. On the basis of the number of unpaired electrons (given in parentheses), decide the correct geometry. a. \(\mathrm{Pt}\left(\mathrm{NH}_{3}\right)_{2}\left(\mathrm{NO}_{2}\right)_{2}^{2+}(2)\) b. \(\mathrm{MnCl}_{4}{ }^{2-}(5)\) c. \(\mathrm{NiCl}_{4}{ }^{2-}\) d. \(\mathrm{AuF}_{4}^{-}(0)\)

Short Answer

Expert verified
a) Square planar, b) Tetrahedral, c) Tetrahedral, d) Square planar.

Step by step solution

01

Determine Electronic Configuration

First, determine the electronic configuration of the metal ions in each complex. For Pt, Mn, Ni, and Au, start with the electronic configuration of the neutral atom. Then adjust the number of electrons for the oxidation state indicated by the charge on the complex.
02

Calculate Number of d Electrons

Based on the oxidation state, calculate the number of remaining d electrons for each metal ion:- Pt in \( ext{Pt(NH}_{3})_{2}( ext{NO}_{2})_{2}^{2+}\) is Pt(II), with 8 d electrons.- Mn in \( ext{MnCl}_{4}^{2-}\) is Mn(II), with 5 d electrons.- Ni in \( ext{NiCl}_{4}^{2-}\) is Ni(II), with 8 d electrons.- Au in \( ext{AuF}_{4}^{-}\) is Au(III), with 8 d electrons.
03

Apply Crystal Field Theory (CFT)

Use CFT to determine the splitting of d orbitals. For tetrahedral complexes, the d orbitals split into t2 and e levels without inversion center effects, and for square planar, they split into a more complex pattern. Identify possible configurations for electrons under each geometry.
04

Match Geometry with Unpaired Electrons

Match the number of unpaired electrons provided: - **Pt(NH鈧�)鈧�(NO鈧�)鈧偮测伜**, 2 unpaired: Square planar (d鈦�, two unpaired). - **MnCl鈧劼测伝**, 5 unpaired: Tetrahedral (all occupied singly). - **NiCl鈧劼测伝**: Will be determined by known data. - **AuF鈧勨伝**, 0 unpaired: Square planar (d鈦�, completely paired).
05

Assign Geometry

Finally, confirm the correct geometry for each: - Pt(NH鈧�)鈧�(NO鈧�)鈧偮测伜 is square planar. - MnCl鈧劼测伝 is tetrahedral. - NiCl鈧劼测伝 is likely tetrahedral based on typical structure (4 unpaired electrons). - AuF鈧勨伝 is square planar.

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Key Concepts

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

d orbitals
Let's delve into the world of d orbitals, crucial players in crystal field theory. In coordination complexes, metal ions typically have their d orbitals split by the electric field created by interacting ligands. These d orbitals have specific shapes and are designated as:
  • dxy
  • dyz
  • dzx
  • d虫虏-测虏
  • d锄虏

In a free ion, these five d orbitals are degenerate, meaning they have the same energy. However, upon forming complexes, the symmetry of the ligand's arrangement can cause the d orbitals to split into different energy levels.
Understanding this splitting is key for predicting the geometric and electronic properties of various complexes, including predicting the number of unpaired electrons.
tetrahedral complexes
Tetrahedral complexes are one type of coordination complex where the central metal atom is surrounded by four ligands at the corners of a tetrahedron. These complexes create a weak field due to poor direct overlap with the central metal atom鈥檚 d orbitals.
The splitting pattern of d orbitals in a tetrahedral field results in two sets:
  • t2 level: consisting of three degenerate orbitals (dxy, dyz, dzx)
  • e level: consisting of two degenerate orbitals (d虫虏-测虏, d锄虏)

In tetrahedral complexes, the energy difference between these levels ( &Delta_t) ) is often smaller compared to other geometries, leading to high-spin configurations with more unpaired electrons. Consequently, for MnCl鈧劼测伝, which has 5 unpaired electrons, a tetrahedral structure is appropriate.
square planar complexes
Square planar complexes are another class of coordination complexes where one might see significant d orbital splitting due to the arrangement of four ligands in a square plane around the central metal ion.
Here, the d orbital splitting results in:
  • Lowest energy: d锄虏
  • Next: dxy
  • Then: dyz, dzx
  • Highest energy: d虫虏-测虏

This large splitting can result in low-spin configurations, often leading to fewer unpaired electrons. For example, Pt(NH鈧�)鈧�(NO鈧�)鈧偮测伜 has two unpaired electrons fitting a square planar setting, while AuF鈧勨伝, with no unpaired electrons, also suits this configuration.
unpaired electrons
Unpaired electrons significantly impact the magnetic properties of metal complexes. These electrons are those not paired with another electron with an opposite spin in an orbital.
In crystal field theory, the number of unpaired electrons in a metal complex can help determine its geometry since different arrangements cause different splitting of d orbitals:
  • More unpaired electrons often indicate a tetrahedral complex (high-spin)
  • Fewer or no unpaired electrons often suggest square planar or octahedral complexes (low-spin)

For instance, MnCl鈧劼测伝, with 5 unpaired electrons, favors the high-spin tetrahedral geometry, whereas AuF鈧勨伝, with 0 unpaired electrons, aligns better with the low-spin square planar geometry.
electronic configuration
Electronic configuration is the arrangement of electrons around the nucleus of an atom or ion in specific orbitals. When determining the properties of coordination complexes, understanding the electronic configuration becomes vital.
To determine a metal ion鈥檚 configuration in a complex, begin with its neutral atomic configuration and adjust for its charge:
  • Subtract electrons for positive charges (oxidation state)
  • Add electrons for negative charges

For example, in the complexes discussed:
  • Pt is in a +2 oxidation state, leading to 8 d electrons
  • Mn is in a +2 oxidation state, resulting in 5 d electrons
  • Ni, also in a +2 state, has 8 d electrons
  • Au, in a +3 state, has 8 d electrons
Understanding these configurations helps predict the properties and the geometry of complexes through electron arrangement in d orbitals.

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Most popular questions from this chapter

Consider the complex ion \(\mathrm{Mn}\left(\mathrm{NH}_{3}\right)_{2}\left(\mathrm{H}_{2} \mathrm{O}\right)_{3}(\mathrm{OH})^{2+}\). a. What is the oxidation state of the metal atom? b. Give the formula and name of each ligand in the ion. c. What is the coordination number of the metal atom? d. What would be the charge on the complex if all ligands were chloride ions?

Draw cis-trans structures of any of the following square planar or octahedral complexes that exhibit geometric isomerism. Label the drawings cis or trans. a. \(\left[\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{2} \mathrm{Cl}_{2}\right]\) b. \(\mathrm{Pd}\left(\mathrm{NH}_{3}\right){ }_{3} \mathrm{Cl}^{+}\) c. \(\mathrm{Pd}\left(\mathrm{NH}_{3}\right)_{4}^{2}+\) d. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{4} \mathrm{Br}_{2}^{+}\)

For each of the following complexes, determine the oxidation state of the transition-metal atom. a. \(\left[\mathrm{CoCl}(\mathrm{en})_{2}\left(\mathrm{NO}_{2}\right)\right] \mathrm{NO}_{2}\) b. \(\mathrm{PtCl}_{4}^{2-}\) c. \(\mathrm{K}_{3}\left[\mathrm{Cr}(\mathrm{CN})_{6}\right]\) d. \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{5}(\mathrm{OH})^{2+}\)

The \(\mathrm{Co}(\mathrm{en})_{3}{ }^{3+}\) ion has a maximum absorption at \(470 \mathrm{~nm}\). What color do you expect for this ion?

Rust spots on clothes can be removed by dissolving the rust in oxalic acid. The oxalate ion forms a stable complex with \(\mathrm{Fe}^{3+}\). Using an electron-dot formula, indicate how an oxalate ion bonds to the metal ion.
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