Textbook Question
Complete the exercises below. Draw the crystal-field energy-level diagrams and show the placement of electrons for each of the following complexes:
a. [VCl6]3–,
b. [FeF6]3– (a high-spin complex),
24. Transition Metals and Coordination Compounds
Magnetic Properties of Complex Ions: Octahedral Complexes
Multiple Choice
Which of the following complex ions is/are diamagnetic in nature?
I. [Mn(Br4)]2– II. [V(NO2)4]4– III. [Zn(NH3)4]2+ IV. [Sc(H2O)6]3+
A
I, II, and IV
B
II and III
C
I and II
D
I, III, and IV
E
iii and Iv
Verified step by step guidance1
Step 1: Understand the concept of diamagnetism. Diamagnetic substances have all their electrons paired and do not have any unpaired electrons. This results in no net magnetic moment.
Step 2: Determine the electron configuration of the central metal ion in each complex. For [Mn(Br4)]2–, Mn is in the +2 oxidation state, for [V(NO2)4]4–, V is in the +4 oxidation state, for [Zn(NH3)4]2+, Zn is in the +2 oxidation state, and for [Sc(H2O)6]3+, Sc is in the +3 oxidation state.
Step 3: Calculate the number of unpaired electrons for each metal ion. Mn2+ has 5 unpaired electrons, V4+ has 1 unpaired electron, Zn2+ has 0 unpaired electrons, and Sc3+ has 0 unpaired electrons.
Step 4: Identify which complexes are diamagnetic. Since diamagnetic complexes have no unpaired electrons, [Zn(NH3)4]2+ and [Sc(H2O)6]3+ are diamagnetic.
Step 5: Conclude that the complexes [Zn(NH3)4]2+ and [Sc(H2O)6]3+ are diamagnetic, corresponding to options III and IV.
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