Chapter 23, Problem 23.88P

(a)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

  ${\left[Cr{(H_{2}O)}_6\right]}^{3+}$

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

  $\left[noblegas\right]n{s}^2\left(n-1\right)d^x.$

The spectrochemical series is

  $I^{-}<C{l}^{-}<F^{-}<O{H}^{-}<H_{2}O<SC{N}^{-}<en<N{O}_2^{-}<C{N}^{-}<CO$

Explanation of Solution

Electron configuration of $Cr:\left[Ar\right]4s^{1}3d^5$

Charge on $Cr$: The aqua ligands are neutral, so the charge on $Cris+3.$

Electron configuration of $C{r}^{3+}:\left[Ar\right]3d^3$

Six ligands indicate an octahedral arrangement.  Using Hund’s rule, fill the lower energy

$t_{2}g$ orbitals first, filling empty orbitals before pairing electrons within an orbital.

(b)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

  ${\left[Cu{(H_{2}O)}_4\right]}^{2+}$

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

  $\left[noblegas\right]n{s}^2\left(n-1\right)d^x.$

The spectrochemical series is

  $I^{-}<C{l}^{-}<F^{-}<O{H}^{-}<H_{2}O<SC{N}^{-}<en<N{O}_2^{-}<C{N}^{-}<CO$

Explanation of Solution

Electron configuration of $Cu:\left[Ar\right]4s^{1}3d^{10}$

Charge on $Cu$: The aqua ligands are neutral, so $Cu$ has a +2 charge.

Electron configuration of $C{u}^{2+}:\left[Ar\right]3d^9$

Four ligands and a $d^9$ configuration indicate a square planar geometry (only filled d sublevel ions exhibit tetrahedral geometry).  Use Hund’s rule to fill in the nine d electrons.  Thus, the correct orbital-energy splitting diagram shows one unpaired electron.   

(c)

Interpretation Introduction

Interpretation:

For the given complex the orbital splitting diagram has to be drawn using spectrochemial series.

  ${\left[Fe{F}_6\right]}^{3-}$

Concept introduction:

The element in the periodic table and count its position in the respective transition series. These elements are in Periods 5 and 6, so the general configuration is

  $\left[noblegas\right]n{s}^2\left(n-1\right)d^x.$

The spectrochemical series is

  $I^{-}<C{l}^{-}<F^{-}<O{H}^{-}<H_{2}O<SC{N}^{-}<en<N{O}_2^{-}<C{N}^{-}<CO$

Explanation of Solution

Electron configuration of $Fe:\left[Ar\right]4s^{2}3d^6$

Charge on $Fe$: Each fluoride ligand has a –1 charge for a total charge of –6, so $Fe$ has a +3 charge to make the overall complex charge equal to –3.

Electron configuration of $F{e}^{3+}:\left[Ar\right]3d^5$

Six ligands indicate an octahedral arrangement.  Use Hund’s rule to fill the orbitals.

F– is a weak-field ligand, so the splitting energy, Δ, is not large enough to overcome the resistance to electron pairing.  The electrons remain unpaired, and the complex is called high-spin.