Chemistry: An Atoms First Approach
Chemistry: An Atoms First Approach
2nd Edition
ISBN: 9781305079243
Author: Steven S. Zumdahl, Susan A. Zumdahl
Publisher: Cengage Learning
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Chapter 20, Problem 23E

(a)

Interpretation Introduction

Interpretation: The electron configuration of each given ion is to be stated.

Concept introduction: Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are s, p, d and f .

The metals from group 3-12 are known as first row transition metals. The general form of valence electron configuration of these metals is,

[Nobelgas]nfxndynsz

When element lose an electron, a positive charge is accumulated on it whereas when an element gains an electron, negative charge is accumulated on it.

To write: The electron configuration of Ti,Ti2+ and Ti4+ .

(a)

Expert Solution
Check Mark

Answer to Problem 23E

Answer

The electron configuration of Ti is [Ar]3d24s2 . The electron configuration of Ti2+ is [Ar]3d2 . The electron configuration of Ti4+ is [Ar]3d0 .

Explanation of Solution

Explanation

Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are s,p,d and f .

The metals from group 3-12 are known as first row transition metals. The general form of valence electron configuration of these metals is,

[Nobelgas]nfxndynsz

Where,

  • n is the shell of principal orbital.
  • x is the number of electrons in f orbital.
  • y is the number of electrons in d orbital.
  • z is the number of electrons in s orbital.

Titanium is a group 4 element. The atomic number of Titanium (Ti) is 22 and its electronic configuration is,

[Ar]3d24s2

Titanium loses two electrons to acquire two positive charges and becomes Ti2+ . The electronic configuration of Ti2+ is,

Ti2+=Ti2e

Substitute the electron configuration of (Ti) in the above equation.

Ti2+=Ti2e=[Ar]3d24s22e=[Ar]3d2

Titanium loses four electrons to acquire four positive charges and becomes Ti4+ . The electronic configuration of Ti4+ is,

Ti4+=Ti4e

Substitute the electron configuration of (Ti) in the above equation.

Ti4+=Ti4e=[Ar]3d24s24e=[Ar]3d0

(b)

Interpretation Introduction

Interpretation: The electron configuration of each given ion is to be stated.

Concept introduction: Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are s, p, d and f .

The metals from group 3-12 are known as first row transition metals. The general form of valence electron configuration of these metals is,

[Nobelgas]nfxndynsz

When element lose an electron, a positive charge is accumulated on it whereas when an element gains an electron, negative charge is accumulated on it.

To write: The electron configuration of Re,Re2+ and Re3+ .

(b)

Expert Solution
Check Mark

Answer to Problem 23E

Answer

The electron configuration of Re is [Xe]4f145d56s2 . The electron configuration of Re2+ is [Xe]4f145d5 . The electron configuration of Re3+ is [Xe]4f145d4 .

Explanation of Solution

Explanation

Rhenium is a group 7 element. The atomic number of Rhenium (Re) is 75 and its electronic configuration is,

[Xe]4f145d56s2

Rhenium loses two electrons to acquire two positive charges and becomes Re2+ . The electronic configuration of Re2+ is,

Re2+=Re2e

Substitute the electron configuration of (Re) in the above equation.

Re2+=Re2e=[Xe]4f145d56s22e=[Xe]4f145d5

Rhenium loses three electrons to acquire three positive charges and becomes Re3+ . The electronic configuration of Re3+ is,

Re3+=Re3e

Substitute the electron configuration of (Re) in the above equation.

Re3+=Re3e=[Xe]4f145d56s23e=[Xe]4f145d4

(c)

Interpretation Introduction

Interpretation: The electron configuration of each given ion is to be stated.

Concept introduction: Valence electrons are filled in four principal orbitals according to the energy levels. The four orbitals are s, p, d and f .

The metals from group 3-12 are known as first row transition metals. The general form of valence electron configuration of these metals is,

[Nobelgas]nfxndynsz

When element lose an electron, a positive charge is accumulated on it whereas when an element gains an electron, negative charge is accumulated on it.

To write: The electron configuration of Ir,Ir2+ and Ir3+ .

(c)

Expert Solution
Check Mark

Answer to Problem 23E

Answer

The electron configuration of Ir is [Xe]4f145d76s2 . The electron configuration of Ir2+ is [Xe]4f145d7 . The electron configuration of Ir3+ is [Xe]4f145d6 .

Explanation of Solution

Explanation

Iridium is a group 9 element. The atomic number of Iridium (Ir) is 77 and its electronic configuration is,

[Xe]4f145d76s2

Iridium loses two electrons to acquire two positive charges and becomes Ir2+ . The electronic configuration of Ir2+ is,

Ir2+=Ir2e

Substitute the electron configuration of (Ir) in the above equation.

Ir2+=Ir2e=[Xe]4f145d76s22e=[Xe]4f145d7

Iridium loses three electrons to acquire three positive charges and becomes Ir3+ . The electronic configuration of Ir3+ is,

Ir3+=Ir3e

Substitute the electron configuration of (Ir) in the above equation.

Ir3+=Ir3e=[Xe]4f145d76s23e=[Xe]4f145d6

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Chapter 20 Solutions

Chemistry: An Atoms First Approach

Ch. 20 - Prob. 1ALQCh. 20 - Prob. 2ALQCh. 20 - Prob. 3ALQCh. 20 - Prob. 4ALQCh. 20 - Prob. 5QCh. 20 - Four different octahedral chromium coordination...Ch. 20 - Prob. 7QCh. 20 - Prob. 8QCh. 20 - Prob. 9QCh. 20 - Prob. 10QCh. 20 - Prob. 11QCh. 20 - Prob. 12QCh. 20 - Prob. 13QCh. 20 - Prob. 14QCh. 20 - Prob. 15QCh. 20 - Prob. 16QCh. 20 - Prob. 17QCh. 20 - Prob. 18QCh. 20 - Prob. 19QCh. 20 - Prob. 20QCh. 20 - Prob. 21ECh. 20 - Prob. 22ECh. 20 - Prob. 23ECh. 20 - Prob. 24ECh. 20 - Prob. 25ECh. 20 - Prob. 26ECh. 20 - Prob. 27ECh. 20 - Prob. 28ECh. 20 - Prob. 29ECh. 20 - Prob. 30ECh. 20 - Prob. 31ECh. 20 - A coordination compound of cobalt(III) contains...Ch. 20 - Prob. 33ECh. 20 - Prob. 34ECh. 20 - Prob. 35ECh. 20 - Prob. 36ECh. 20 - Prob. 37ECh. 20 - Give formulas for the following complex ions. a....Ch. 20 - Prob. 39ECh. 20 - Prob. 40ECh. 20 - Prob. 41ECh. 20 - Amino acids can act as ligands toward transition...Ch. 20 - Prob. 43ECh. 20 - Prob. 44ECh. 20 - Prob. 45ECh. 20 - Prob. 46ECh. 20 - Prob. 47ECh. 20 - Prob. 48ECh. 20 - Prob. 49ECh. 20 - Prob. 50ECh. 20 - Prob. 51ECh. 20 - Prob. 52ECh. 20 - The CrF64 ion is known to have four unpaired...Ch. 20 - Prob. 54ECh. 20 - Prob. 55ECh. 20 - The complex ion Fe(CN)63 is paramagnetic with one...Ch. 20 - Prob. 57ECh. 20 - Prob. 58ECh. 20 - Prob. 59ECh. 20 - Prob. 60ECh. 20 - The wavelength of absorbed electromagnetic...Ch. 20 - The complex ion NiCL42 has two unpaired electrons,...Ch. 20 - How many unpaired electrons are present in the...Ch. 20 - Prob. 64ECh. 20 - Prob. 65ECh. 20 - Prob. 66ECh. 20 - Prob. 67ECh. 20 - Prob. 68ECh. 20 - Silver is sometimes found in nature as large...Ch. 20 - Prob. 70ECh. 20 - Prob. 71AECh. 20 - The compound cisplatin, Pt(NH3)2Cl2, has been...Ch. 20 - Prob. 73AECh. 20 - Prob. 74AECh. 20 - Prob. 75AECh. 20 - Prob. 76AECh. 20 - Prob. 77AECh. 20 - Name the following coordination compounds. a....Ch. 20 - Prob. 79AECh. 20 - Prob. 80AECh. 20 - Carbon monoxide is toxic because it binds more...Ch. 20 - For the process Co(NH3)5Cl2++ClCO(NH3)4Cl2++NH3...Ch. 20 - Prob. 83CWPCh. 20 - Which of the following molecules exhibit(s)...Ch. 20 - Prob. 85CWPCh. 20 - Prob. 86CWPCh. 20 - Prob. 87CWPCh. 20 - Which of the following statement(s) is( are) true?...Ch. 20 - Prob. 89CPCh. 20 - Consider the pseudo-octahedral complex ion of...Ch. 20 - Prob. 91CPCh. 20 - Prob. 92CPCh. 20 - Prob. 93CPCh. 20 - Prob. 94CPCh. 20 - Prob. 95CPCh. 20 - Prob. 96CPCh. 20 - Prob. 97CPCh. 20 - Prob. 98IPCh. 20 - Prob. 99IPCh. 20 - Prob. 100IPCh. 20 - Prob. 101MP
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