Solutions to Chemistry Homework Assignment on Coordination Chemistry

This document provides detailed solutions to a chemistry homework assignment. The solutions cover various topics including the absence of isomerism in a compound, the lack of color in zinc compounds explained by the absence of electron excitation due to fully occupied d-orbitals. The assignment includes calculations involving the solubility product (Ksp) to determine the minimum concentration of ions in solutions of barium fluoride and calcium sulfate. It explores the relationship between ligand strength, crystal field splitting, and the color of coordination complexes, explaining why blue light has higher energy than purple light. Furthermore, the document provides solutions to buffer calculations, determining the pH and molar ratios required for a dimethylamine buffer, and calculating the crystal field splitting energy from the absorbance spectrum of a complex. The solutions are comprehensive and provide step-by-step explanations.

Question 2
a.
b. No possible isomerism
c.
Question 3
Tt does not have incomplete d-orbitals/ does not form an ion with incomplete d-orbitals.Transition
metals, on the other hand have either incomplete d-orbitals or can give rise to an ion with incomplete
d-orbitals. What this means, is that once in solution, the H2O ligands (their lone electron pairs) cause
repulsion while forming a dative bond with the central metal cation. This repulsion causes d-orbital
splitting, that results in certain d-orbitals being higher in energy than others. Now the interesting bit
comes in: the d-orbital electron is able to be excited to a higher energy d-orbital (which has the
available space) which it achieves via photon absorption, hence giving the compound its color. Zinc
does not have this "privilege," as it has fully occupied d-orbitals (even when its Zn2+), therefore
there is no electron excitation and no color.
Question 4
a.
Given, [Ba2+] = 0.035 M
Ba(NO3)2 (aq) + 2 NaF (aq) -----------------> BaF2 (s) + 2 NaNO3
Partial dissociation of BaF2 is :
BaF2 (s) <--------------------> Ba2+ (aq) + 2 F- (aq)
Expression of Ksp is :
Ksp = [Ba2+].[F-]2
[F-]2 = Ksp / [Ba2+]
[F-] = (2.45 x 10-5 / 0.035)1/2
[F-] = 7 x 10-4 M
Hence, minimum concentration = 7 x 10-4 M
b.
. Given, [Ca2+] = 0.085 M
CaI2 (aq) + K2SO4 (aq) -----------------> CaSO4(s) + 2 KI (aq)
Partial dissociation of CaSO4(s) is :
CaSO4(s) <----------------> Ca2+(aq) + SO42-(aq)

Expression of Ksp is :
Ksp = [Ca2+].[SO42-]
[SO42-] = Ksp / [Ca2+]
[SO42-] = (7.10 x 10-5 / 8.5 x 10-2)
[SO42-] = 8.35 x 10-4 M
Hence, minimum concentration = 8.35 x 10-4 M
c. Because of equal valences,
minimum concentration= 1. 77 ×10−10
0 . 0018 =9. 833 ×10−8 M
Question 8
Question 9

Blue color (A) has strong ligand because different ligands are associated with either high or
low spin—a "strong field" ligand results in a large ∆o and a low spin configuration, while
a "weak field" ligand results in a small ∆o and a high spin configuration since blue light
has a much higher frequency and more energy than purple light. The diagram for the 2
cases are shown below
Purple
blue
Question 10
HNO3 + NH3 → NH4NO
Kw =¿
¿
pH=-log5.5556−10=9.26
From the problem statement,
Question 11
Expected to have high spin.
The Fe2+ ion in [Fe(H2O)6]2+ has outer electronic configuration of 3d6 . The complex is a
high spin complex. It contains 4 unpaired electrons

Question 12
a. The buffer is basic (Acid/base=0.5)
b. Greater than 9.23
c. pH = p K A + log Cbase
Cacid =10+ log ( 0.5 )=9.699
Question 15
For dimethylamine:
Acidity (pKa) 10.64
Basicity (pKb) 3.36
At this point have a short think: If the molar ratio of dimethylamine : dimethylammonium chloride was 1 :1 you
would be at the half equivalence point , and pH would equal pKa . The pH would be 10.64.
But you want a buffer with pH = 10.43. Therefore there must be a molar excess of the dimethylammonium chloride.
Use the H-H equation:
pH = pKa + log ( [[base]/[salt]]
10.43 = 10.64 + log ([base[/[salt])
log ([base[/]salt]) = 10-43 - 10.64
log ([base]/[salt] = -0.21
[base]/[salt] = 10^-0.21
[base]/[salt] = 0.62
The buffer must be prepared by mixing 0.62mol dimethylamine and 1.0mol dimethylammonium chloride.
Mass ratio as required:
Molar mass dimethylammonium chloride 81.55g/mol
Molar mass dimethylamine = 45.08g/mol - 0.62 mol = 27.95g
Mass ratio of dimethylammonium chloride : dimethylamine = 81.55 : 27.95
OR Ratio = 81.55/27.95 = 2.92:1.0
Question 16
a. Red colour (intensely coloured)
b.
The given absorbance spectrum shows the complex absorbs 480 nm( not accurate)
480 nm = 10000000 / 480 cm-1
480 nm = 20,833.33 cm-1
Since 1 kJ mol-1 = 83.7 cm-1

Thus, crystal field splitting energy = 20833.33 / 83.7
= 248.9047 kJ mol-1
= 248 kJ mol-1
c.