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Chapter 21: Problem 51

How many unpaired electrons are in the following complex ions? a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case) b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\)

Short Answer

Expert verified
The number of unpaired electrons in the given complex ions are: a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) (low-spin case): 0 unpaired electrons b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\): 2 unpaired electrons c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\): 1 unpaired electron

Step by step solution

01

Identify the oxidation state for each metal

In each complex, we first need to determine the oxidation state of the central metal atom: a. \(\mathrm{Ru}\left(\mathrm{NH}_{3}\right)_{6}^{2+}\) : The total charge of the complex is 2+, and ammonia (NH3) is neutral, so the oxidation state of Ru is +2. b. \(\mathrm{Ni}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) : The total charge of the complex is 2+, and water (H2O) is neutral, so the oxidation state of Ni is +2. c. \(\mathrm{V}(\mathrm{en})_{3}{ }^{3+}\) : The total charge of the complex is 3+, and ethylenediamine (en) is neutral, so the oxidation state of V is +3.
02

Write the electron configurations

Next, we will write the electron configurations of the central metal atoms in their corresponding oxidation states: a. Ru+2: Ru has an atomic number of 44, and its electron configuration is [Kr] 4d7 5s1. After losing 2 electrons, we have [Kr] 4d6. b. Ni+2: Ni has an atomic number of 28, and its electron configuration is [Ar] 3d8 4s2. After losing 2 electrons, we have [Ar] 3d8. c. V+3: V has an atomic number of 23, and its electron configuration is [Ar] 3d3 4s2. After losing 3 electrons, we have [Ar] 3d2.
03

Determine the number of unpaired electrons

For each electron configuration, count the number of unpaired electrons in the d orbitals: a. [Kr] 4d6 (low-spin case): In the low-spin case, electrons preferentially occupy the lower energy orbitals first. Here, the d orbitals are completely filled before any electron jumps into a higher energy orbital. Thus, there are no unpaired electrons, and the answer is 0. b. [Ar] 3d8: The d orbitals fill in the following order: 鈫戔啌鈫戔啌鈫戔啌鈫戔啈, so there are 2 unpaired electrons. c. [Ar] 3d2: The d orbitals fill in the following order: 鈫戔啌鈫�, so there is 1 unpaired electron.
04

Final Answers

The number of unpaired electrons in the complex ions are: a. Ru(NH3)6^2+ (low-spin case): 0 unpaired electrons b. Ni(H2O)6^2+: 2 unpaired electrons c. V(en)3^3+: 1 unpaired electron

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Most popular questions from this chapter

. Which is more likely to be paramagnetic, \(\mathrm{Fe}(\mathrm{CN})_{6}^{4-}\) or \(\mathrm{Fe}\left(\mathrm{H}_{2} \mathrm{O}\right)_{6}^{2+}\) ? Explain.

BAL is a chelating agent used in treating heavy metal poisoning. It acts as a bidentate ligand. What type of linkage isomers are possible when BAL coordinates to a metal ion?

Hemoglobin (abbreviated \(\mathrm{Hb}\) ) is a protein that is responsible for the transport of oxygen in the blood of mammals. Each hemoglobin molecule contains four iron atoms that serve as the binding sites for \(\mathrm{O}_{2}\) molecules. The oxygen binding is \(\mathrm{pH}\) dependent. The relevant equilibrium reaction is $$\mathrm{HbH}_{4}{ }^{4+}(a q)+4 \mathrm{O}_{2}(g) \rightleftharpoons \mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}(a q)+4 \mathrm{H}^{+}(a q)$$ Use Le Ch芒telier's principle to answer the following. a. What form of hemoglobin, \(\mathrm{HbH}_{4}^{4+}\) or \(\mathrm{Hb}\left(\mathrm{O}_{2}\right)_{4}\), is favored in the lungs? What form is favored in the cells? b. When a person hyperventilates, the concentration of \(\mathrm{CO}_{2}\) in the blood decreases. How does this affect the oxygenbinding equilibrium? How does breathing into a paper bag help to counteract this effect? (Hint: \(\mathrm{CO}_{2}\) reacts with water to produce carbonic acid.) c. When a person has suffered a cardiac arrest, an injection of a sodium bicarbonate solution is given. Why is this step necessary?

Draw all the geometrical isomers of \(\left[\mathrm{Cr}(\mathrm{en})\left(\mathrm{NH}_{3}\right)_{2} \mathrm{BrCl}\right]^{+} .\) Which of these isomers also have an optical isomer? Draw the various isomers.

Amino acids can act as ligands toward transition metal ions. The simplest amino acid is glycine \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2} \mathrm{H}\right)\). Draw a structure of the glycinate anion \(\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2}^{-}\right)\) acting as a bidentate ligand. Draw the structural isomers of the square planar complex \(\mathrm{Cu}\left(\mathrm{NH}_{2} \mathrm{CH}_{2} \mathrm{CO}_{2}\right)_{2}\)
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