Chemistry: Principles and Practice
Chemistry: Principles and Practice
3rd Edition
ISBN: 9780534420123
Author: Daniel L. Reger, Scott R. Goode, David W. Ball, Edward Mercer
Publisher: Cengage Learning
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Chapter 8, Problem 8.33QE

(a)

Interpretation Introduction

Interpretation:

The cation with the electron configuration 1s22s22p63s23p1 and 1+ charge has to be determined.

Concept Introduction:

The electronic configuration is defined as the distribution of electrons in various atomic orbitals of the atom. The electrons that are present in an outermost orbital are known as valence electrons whereas those present in the orbitals with lower quantum numbers are called core electrons. The general outer electronic configuration of sblock elements is ns12, that of pblock elements is ns2np16, that of dblock elements is (n1)d110ns02 and that of fblock elements is (n2)f114(n1)d010ns2.

Electrons are filled in orbitals in accordance with three rules: Aufbau principle, Hund’s rule, and Pauli’s exclusion principle. Aufbau principle states that electrons are filled in the orbitals from lower to higher energy level as follows:

  1s<2s<2p<3s<3p<4s<3d<4p<5s<4d<5p<6s<4f<5d<6p<7s

Hund’s rule states that initially each orbital is singly occupied and then pairing occurs and Pauli’s exclusion principle states that the spin of two electrons in one orbital is always different.

(a)

Expert Solution
Check Mark

Explanation of Solution

The electron configuration of the given ion is 1s22s22p63s23p1. Since this ion has a charge of 1+, it is formed by the removal of one electron from the valence shell of the neutral atom. The electron configuration of the neutral atom is 1s22s22p63s23p2. It has a total of 14 electrons so its atomic number is 14. According to the periodic table, the element with atomic number 14 is silicon (Si) so the ion with 1+ charge is Si+.

(b)

Interpretation Introduction

Interpretation:

The cation with the electron configuration 1s22s22p63s1 and 1+ charge has to be determined.

Concept Introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

The electron configuration of the given ion is 1s22s22p63s1. Since this ion has a charge of 1+, it is formed by the removal of one electron from the valence shell of the neutral atom. The electron configuration of the neutral atom is 1s22s22p63s2. It has a total of 12 electrons so its atomic number is 12. According to the periodic table, the element with atomic number 12 is magnesium (Mg) so the ion with 1+ charge is Mg+.

(c)

Interpretation Introduction

Interpretation:

The cation with the electron configuration 1s22s22p63s23p64s23d10 and 1+ charge has to be determined.

Concept Introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

The electron configuration of the given ion is 1s22s22p63s23p64s23d10. Since this ion has a charge of 1+, it is formed by the removal of one electron from the valence shell of the neutral atom. The electron configuration of the neutral atom is 1s22s22p63s23p64s23d104p1. It has a total of 31 electrons so its atomic number is 31. According to the periodic table, the element with atomic number 31 is gallium (Ga) so the ion with 1+ charge is Ga+.

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Chapter 8 Solutions

Chemistry: Principles and Practice

Ch. 8 - Prob. 8.11QECh. 8 - Prob. 8.12QECh. 8 - Prob. 8.13QECh. 8 - Prob. 8.14QECh. 8 - Prob. 8.15QECh. 8 - Prob. 8.16QECh. 8 - Prob. 8.17QECh. 8 - Prob. 8.18QECh. 8 - Explain why the electron affinity of lithium is...Ch. 8 - Prob. 8.20QECh. 8 - Prob. 8.21QECh. 8 - Prob. 8.22QECh. 8 - Prob. 8.23QECh. 8 - Prob. 8.24QECh. 8 - Prob. 8.25QECh. 8 - Prob. 8.26QECh. 8 - Prob. 8.27QECh. 8 - Prob. 8.28QECh. 8 - Prob. 8.29QECh. 8 - Prob. 8.30QECh. 8 - Prob. 8.31QECh. 8 - Prob. 8.32QECh. 8 - Prob. 8.33QECh. 8 - Prob. 8.34QECh. 8 - Prob. 8.35QECh. 8 - Prob. 8.36QECh. 8 - Prob. 8.37QECh. 8 - Prob. 8.38QECh. 8 - Write the symbols for a cation and an anion that...Ch. 8 - Prob. 8.40QECh. 8 - Prob. 8.41QECh. 8 - What neutral atoms are isoelectronic with the...Ch. 8 - Prob. 8.43QECh. 8 - Prob. 8.44QECh. 8 - Prob. 8.45QECh. 8 - Prob. 8.46QECh. 8 - Prob. 8.47QECh. 8 - Prob. 8.48QECh. 8 - Prob. 8.49QECh. 8 - Prob. 8.50QECh. 8 - Prob. 8.51QECh. 8 - Prob. 8.52QECh. 8 - Prob. 8.53QECh. 8 - Prob. 8.54QECh. 8 - Prob. 8.55QECh. 8 - Of the atoms with the electron configurations...Ch. 8 - Prob. 8.57QECh. 8 - Prob. 8.58QECh. 8 - Prob. 8.59QECh. 8 - Prob. 8.60QECh. 8 - Prob. 8.61QECh. 8 - Prob. 8.62QECh. 8 - Prob. 8.63QECh. 8 - Prob. 8.64QECh. 8 - Prob. 8.65QECh. 8 - Prob. 8.66QECh. 8 - Prob. 8.67QECh. 8 - Prob. 8.68QECh. 8 - Prob. 8.69QECh. 8 - Prob. 8.70QECh. 8 - What is the electron configuration of the Ba3+...Ch. 8 - Prob. 8.72QECh. 8 - Prob. 8.73QECh. 8 - Prob. 8.74QECh. 8 - Prob. 8.75QECh. 8 - Prob. 8.76QECh. 8 - Prob. 8.77QECh. 8 - Prob. 8.78QECh. 8 - Prob. 8.79QECh. 8 - Prob. 8.80QECh. 8 - Prob. 8.81QECh. 8 - Prob. 8.82QECh. 8 - Prob. 8.83QECh. 8 - Prob. 8.84QECh. 8 - Prob. 8.85QECh. 8 - Prob. 8.86QECh. 8 - Prob. 8.87QECh. 8 - Prob. 8.88QECh. 8 - Prob. 8.89QECh. 8 - Prob. 8.90QECh. 8 - Palladium, with an electron configuration of [Kr]...Ch. 8 - Prob. 8.92QECh. 8 - Prob. 8.93QECh. 8 - Prob. 8.94QECh. 8 - Prob. 8.95QECh. 8 - Prob. 8.96QECh. 8 - Prob. 8.97QECh. 8 - Prob. 8.98QECh. 8 - Arrange the elements lithium, carbon, and oxygen...Ch. 8 - Prob. 8.100QECh. 8 - Prob. 8.101QECh. 8 - Prob. 8.102QECh. 8 - Prob. 8.103QECh. 8 - Prob. 8.104QECh. 8 - Prob. 8.105QECh. 8 - Prob. 8.106QECh. 8 - Prob. 8.107QE
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