CHEMICAL PRINCIPLES (LL) W/ACCESS
CHEMICAL PRINCIPLES (LL) W/ACCESS
7th Edition
ISBN: 9781319421175
Author: ATKINS
Publisher: MAC HIGHER
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Chapter 1, Problem 1E.12E

(a)

Interpretation Introduction

Interpretation:

The ground state configuration of titanium atom has to be predicted.

Concept introduction:

Electronic configuration: The electron configuration is the distribution of electrons of an atom or molecule in atomic or molecular orbitals.

Electrons occupy the lowest energy orbitals. The increasing order of orbital energy is s, p, d and f.    The lowest energy orbital is ls.

The energy order of the orbital for the first three periods is as follows,

    ls, 2s, 2p, 3s, and 3p.

The orbital which is closer to the nucleus has lower energy; therefore the 2s is lower in energy than 3s.   Accordingly 2s is lower in energy than 2p and 4s is lower in energy than 3d.

In general, the orbitals can hold maximum of two electrons, the two electrons must have opposite spin.

The subshell ordering by Aufbau principle is given below,

    ls, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, ...

(a)

Expert Solution
Check Mark

Explanation of Solution

The electronic configuration is depends on the electrons in the atom.  The titanium atom has 22 electrons.

The subshell ordering by Aufbau principle is given below;

    ls, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, ...

Therefore ground state electronic configuration of titanium atom is ls2 2s2 2p6 3s23p6 4s2 3d2 (22electrons).

Titanium belongs to Period 4 and Group 4.  It has a argon core with additional four valence electron.

The electronic configuration also can be written as follows,

    [Ar] 4s2 3d2.

(b)

Interpretation Introduction

Interpretation:

The ground state configuration of chromium atom has to be predicted.

Concept introduction:

Refer to part (a).

(b)

Expert Solution
Check Mark

Explanation of Solution

The electronic configuration is depends on the electrons in the atom.  The chromium atom has 24 electrons.

The subshell ordering by Aufbau principle is given below;

    ls, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, ...

Therefore ground state electronic configuration of chromium atom is ls2 2s2 2p6 3s23p6 4s1 3d5 (24electrons).

Chromium belongs to Period 4 and Group 6.  It has a argon core with four additional valence electrons.

The electronic configuration also can be written as follows,

    [Ar] 4s1 3d5.

(c)

Interpretation Introduction

Interpretation:

The ground state configuration of europium atom has to be predicted.

Concept introduction:

Refer to part (a).

(c)

Expert Solution
Check Mark

Explanation of Solution

The electronic configuration is depends on the electrons in the atom.  The europium atom has 63 electrons.

The subshell ordering by Aufbau principle is given below;

    ls, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, ...

Therefore ground state electronic configuration of europium atom is ls2 2s2 2p6 3s2 3p54s2 3d10 4p6 5s2 4d10 5p6 6s2 4f7 (63electrons).

The electronic configuration also can be written as follows,

    [Xe] 4f7 6s2.

(d)

Interpretation Introduction

Interpretation:

The ground state configuration of krypton atom has to be predicted.

Concept introduction:

Refer to part (a).

(d)

Expert Solution
Check Mark

Explanation of Solution

The electronic configuration is depends on the electrons in the atom.  The krypton atom has 36 electrons.

The subshell ordering by Aufbau principle is given below;

    ls, 2s, 2p, 3s, 3p, 4s, 3d, 4p, 5s, 4d, 5p, 6s, 4f, 5d, 6p, 7s, 5f, 6d, 7p, 8s, ...

Therefore ground state electronic configuration of rubidium atom is ls2 2s2 2p6 3s2 3p6 4s2 3d10 4p6 (36electrons).

Rubidium belongs to Period 4 and Group 18.  It has argon core with eighteen additional valence electron.

The electronic configuration also can be written as follows,

    [Ar] 3d104s24p6.

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Boron, atomic number 5, occurs naturally as two isotopes, 10B and 11B, with natural abundances of 19.9% and 80.1%, respectively.(a) In what ways do the two isotopes differ from each other? Does the electronic configuration of 10B differ from that of 11B? (b) Drawthe orbital diagram for an atom of 11B. Which electrons are the valence electrons? (c) Indicate three ways in which the 1s electrons inboron differ from its 2s electrons. (d) Elemental boron reacts with fluorine to form BF3, a gas. Write a balanced chemical equation forthe reaction of solid boron with fluorine gas. (e) ΔHf° for BF31g2 is -1135.6 kJ>mol. Calculate the standard enthalpy change in thereaction of boron with fluorine. (f) Will the mass percentage of F be the same in 10BF3 and 11BF3? If not, why is that the case?
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Chapter 1 Solutions

CHEMICAL PRINCIPLES (LL) W/ACCESS

Ch. 1 - Prob. 1A.8ECh. 1 - Prob. 1A.9ECh. 1 - Prob. 1A.10ECh. 1 - Prob. 1A.11ECh. 1 - Prob. 1A.12ECh. 1 - Prob. 1A.13ECh. 1 - Prob. 1A.14ECh. 1 - Prob. 1A.15ECh. 1 - Prob. 1A.16ECh. 1 - Prob. 1A.17ECh. 1 - Prob. 1B.1ASTCh. 1 - Prob. 1B.1BSTCh. 1 - Prob. 1B.2ASTCh. 1 - Prob. 1B.2BSTCh. 1 - Prob. 1B.3ASTCh. 1 - Prob. 1B.3BSTCh. 1 - Prob. 1B.4ASTCh. 1 - Prob. 1B.4BSTCh. 1 - Prob. 1B.5ASTCh. 1 - Prob. 1B.5BSTCh. 1 - Prob. 1B.1ECh. 1 - Prob. 1B.2ECh. 1 - Prob. 1B.3ECh. 1 - Prob. 1B.4ECh. 1 - Prob. 1B.5ECh. 1 - Prob. 1B.6ECh. 1 - Prob. 1B.7ECh. 1 - Prob. 1B.8ECh. 1 - Prob. 1B.9ECh. 1 - Prob. 1B.10ECh. 1 - Prob. 1B.11ECh. 1 - Prob. 1B.12ECh. 1 - Prob. 1B.13ECh. 1 - Prob. 1B.14ECh. 1 - Prob. 1B.15ECh. 1 - Prob. 1B.16ECh. 1 - Prob. 1B.17ECh. 1 - Prob. 1B.18ECh. 1 - Prob. 1B.19ECh. 1 - Prob. 1B.21ECh. 1 - Prob. 1B.22ECh. 1 - Prob. 1B.23ECh. 1 - Prob. 1B.24ECh. 1 - Prob. 1B.25ECh. 1 - Prob. 1B.26ECh. 1 - Prob. 1B.27ECh. 1 - Prob. 1B.28ECh. 1 - Prob. 1C.1ASTCh. 1 - Prob. 1C.1BSTCh. 1 - Prob. 1C.1ECh. 1 - Prob. 1C.2ECh. 1 - Prob. 1C.3ECh. 1 - Prob. 1C.7ECh. 1 - Prob. 1D.1ASTCh. 1 - Prob. 1D.1BSTCh. 1 - Prob. 1D.2ASTCh. 1 - Prob. 1D.2BSTCh. 1 - Prob. 1D.1ECh. 1 - Prob. 1D.2ECh. 1 - Prob. 1D.3ECh. 1 - Prob. 1D.4ECh. 1 - Prob. 1D.5ECh. 1 - Prob. 1D.6ECh. 1 - Prob. 1D.7ECh. 1 - Prob. 1D.9ECh. 1 - Prob. 1D.10ECh. 1 - Prob. 1D.11ECh. 1 - Prob. 1D.12ECh. 1 - Prob. 1D.13ECh. 1 - Prob. 1D.14ECh. 1 - Prob. 1D.15ECh. 1 - Prob. 1D.16ECh. 1 - Prob. 1D.17ECh. 1 - Prob. 1D.18ECh. 1 - Prob. 1D.19ECh. 1 - Prob. 1D.20ECh. 1 - Prob. 1D.21ECh. 1 - Prob. 1D.22ECh. 1 - Prob. 1D.23ECh. 1 - Prob. 1D.24ECh. 1 - Prob. 1D.25ECh. 1 - Prob. 1D.26ECh. 1 - Prob. 1E.1ASTCh. 1 - Prob. 1E.1BSTCh. 1 - Prob. 1E.2ASTCh. 1 - Prob. 1E.2BSTCh. 1 - Prob. 1E.1ECh. 1 - Prob. 1E.2ECh. 1 - Prob. 1E.3ECh. 1 - Prob. 1E.4ECh. 1 - Prob. 1E.5ECh. 1 - Prob. 1E.7ECh. 1 - Prob. 1E.8ECh. 1 - Prob. 1E.9ECh. 1 - Prob. 1E.10ECh. 1 - Prob. 1E.11ECh. 1 - Prob. 1E.12ECh. 1 - Prob. 1E.13ECh. 1 - Prob. 1E.14ECh. 1 - Prob. 1E.15ECh. 1 - Prob. 1E.16ECh. 1 - Prob. 1E.17ECh. 1 - Prob. 1E.18ECh. 1 - Prob. 1E.19ECh. 1 - Prob. 1E.20ECh. 1 - Prob. 1E.21ECh. 1 - Prob. 1E.22ECh. 1 - Prob. 1E.23ECh. 1 - Prob. 1E.24ECh. 1 - Prob. 1E.25ECh. 1 - Prob. 1E.26ECh. 1 - Prob. 1F.1ASTCh. 1 - Prob. 1F.1BSTCh. 1 - Prob. 1F.2ASTCh. 1 - Prob. 1F.2BSTCh. 1 - Prob. 1F.3BSTCh. 1 - Prob. 1F.1ECh. 1 - Prob. 1F.2ECh. 1 - Prob. 1F.3ECh. 1 - Prob. 1F.4ECh. 1 - Prob. 1F.5ECh. 1 - Prob. 1F.6ECh. 1 - Prob. 1F.7ECh. 1 - Prob. 1F.8ECh. 1 - Prob. 1F.10ECh. 1 - Prob. 1F.11ECh. 1 - Prob. 1F.12ECh. 1 - Prob. 1F.13ECh. 1 - Prob. 1F.14ECh. 1 - Prob. 1F.15ECh. 1 - Prob. 1F.17ECh. 1 - Prob. 1F.18ECh. 1 - Prob. 1F.19ECh. 1 - Prob. 1F.22ECh. 1 - Prob. 1.1ECh. 1 - Prob. 1.2ECh. 1 - Prob. 1.3ECh. 1 - Prob. 1.9ECh. 1 - Prob. 1.10ECh. 1 - Prob. 1.11ECh. 1 - Prob. 1.12ECh. 1 - Prob. 1.13ECh. 1 - Prob. 1.14ECh. 1 - Prob. 1.15ECh. 1 - Prob. 1.17ECh. 1 - Prob. 1.19ECh. 1 - Prob. 1.21ECh. 1 - Prob. 1.22ECh. 1 - Prob. 1.23ECh. 1 - Prob. 1.24ECh. 1 - Prob. 1.25ECh. 1 - Prob. 1.27ECh. 1 - Prob. 1.28ECh. 1 - Prob. 1.31E
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