Living By Chemistry: First Edition Textbook
Living By Chemistry: First Edition Textbook
1st Edition
ISBN: 9781559539418
Author: Angelica Stacy
Publisher: MAC HIGHER
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Chapter U2.4, Problem 3E

(a)

Interpretation Introduction

Interpretation:

Lewis dot symbol of Te, I, K, Bi, In and Pb must be drawn. These must also be arranged according to the group number in the periodic table.

Concept Introduction :

Lewis dot symbol is the representation of an element’s atom with the valence electrons shown as dots.

(a)

Expert Solution
Check Mark

Answer to Problem 3E

Lewis dot symbols for all the given elements are drawn below.

  Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 3E , additional homework tip  1

K (1), I (13), Pb (14), Bi (15) Te (16) I (17).

All the elements are arranged according to increasing group number.

Explanation of Solution

K, I, Pb, Bi, Te and I have 1, 3, 4, 5, 6 and 7 valence electrons respectively. The Lewis dot structures are shown as follows:

  Living By Chemistry: First Edition Textbook, Chapter U2.4, Problem 3E , additional homework tip  2

These belong to group 1, 13, 14, 15, 16 and 17 respectively. , all are arranged in increasing group number as follows:

K (1), I (13), Pb (14), Bi (15) Te (16) I (17).

Here, group number are in brackets.

(b)

Interpretation Introduction

Interpretation:

The number of covalent bonds each element would make must be determined.

Concept Introduction :

Number of covalent bonds depends on the valency of the element.

(b)

Expert Solution
Check Mark

Answer to Problem 3E

K, In, Pb, Bi, Te and I will form 1, 3, 4, 3, 2, 1 covalent bonds respectively.

Explanation of Solution

Valency is equal to number of valence electrons (upto 4). When valence electrons are greater than 4 then valency is obtained by subtracting valence electrons from 8.

Thus valency for K, In and Pb are 1, 3 and 4 respectively as there are 1, 3, 4 valence electrons.

These will form 1, 3 and 4 covalent bonds respectively.

Bi, Te and I have 5, 6 and 7 valence electrons respectively. Thus valencies of these elements are 3, 2 and 1 respectively. So these will form 3, 2 and 1 covalent bond respectively.

Chapter U2 Solutions

Living By Chemistry: First Edition Textbook

Ch. U2.2 - Prob. 3ECh. U2.2 - Prob. 4ECh. U2.2 - Prob. 5ECh. U2.2 - Prob. 6ECh. U2.2 - Prob. 7ECh. U2.3 - Prob. 1TAICh. U2.3 - Prob. 1ECh. U2.3 - Prob. 2ECh. U2.3 - Prob. 3ECh. U2.3 - Prob. 4ECh. U2.3 - Prob. 5ECh. U2.4 - Prob. 1TAICh. U2.4 - Prob. 1ECh. U2.4 - Prob. 2ECh. U2.4 - Prob. 3ECh. U2.4 - Prob. 4ECh. U2.4 - Prob. 5ECh. U2.4 - Prob. 6ECh. U2.4 - Prob. 7ECh. U2.4 - Prob. 8ECh. U2.5 - Prob. 1TAICh. U2.5 - Prob. 1ECh. U2.5 - Prob. 2ECh. U2.5 - Prob. 3ECh. U2.5 - Prob. 4ECh. U2.5 - Prob. 5ECh. U2.5 - Prob. 6ECh. U2.5 - Prob. 7ECh. U2.5 - Prob. 8ECh. U2.5 - Prob. 9ECh. U2.6 - Prob. 1TAICh. U2.6 - Prob. 1ECh. U2.6 - Prob. 2ECh. U2.6 - Prob. 3ECh. U2.6 - Prob. 4ECh. U2.6 - Prob. 5ECh. U2.6 - Prob. 6ECh. U2.6 - Prob. 7ECh. U2.6 - Prob. 8ECh. U2.6 - Prob. 9ECh. U2.6 - Prob. 10ECh. U2.6 - Prob. 11ECh. U2.7 - Prob. 1TAICh. U2.7 - Prob. 1ECh. U2.7 - Prob. 2ECh. U2.7 - Prob. 3ECh. U2.7 - Prob. 4ECh. U2.7 - Prob. 5ECh. U2.8 - Prob. 1TAICh. U2.8 - Prob. 1ECh. U2.8 - Prob. 2ECh. U2.8 - Prob. 3ECh. U2.8 - Prob. 4ECh. U2.8 - Prob. 5ECh. U2.8 - Prob. 6ECh. U2.9 - Prob. 1TAICh. U2.9 - Prob. 1ECh. U2.9 - Prob. 2ECh. U2.9 - Prob. 3ECh. U2.9 - Prob. 4ECh. U2.9 - Prob. 5ECh. U2.9 - Prob. 6ECh. U2.9 - Prob. 7ECh. U2.10 - Prob. 1TAICh. U2.10 - Prob. 1ECh. U2.10 - Prob. 2ECh. U2.10 - Prob. 3ECh. U2.10 - Prob. 4ECh. U2.10 - Prob. 5ECh. U2.10 - Prob. 6ECh. U2.10 - Prob. 7ECh. U2.10 - Prob. 8ECh. U2.10 - Prob. 9ECh. U2.11 - Prob. 1TAICh. U2.11 - Prob. 1ECh. U2.11 - Prob. 2ECh. U2.11 - Prob. 3ECh. U2.11 - Prob. 4ECh. U2.11 - Prob. 5ECh. U2.11 - Prob. 6ECh. U2.11 - Prob. 7ECh. U2.12 - Prob. 1TAICh. U2.12 - Prob. 1ECh. U2.12 - Prob. 2ECh. U2.12 - Prob. 3ECh. U2.12 - Prob. 4ECh. U2.12 - Prob. 5ECh. U2.12 - Prob. 6ECh. U2.12 - Prob. 7ECh. U2.13 - Prob. 1TAICh. U2.13 - Prob. 1ECh. U2.13 - Prob. 2ECh. U2.13 - Prob. 3ECh. U2.13 - Prob. 4ECh. U2.13 - Prob. 5ECh. U2.14 - Prob. 1TAICh. U2.14 - Prob. 1ECh. U2.14 - Prob. 2ECh. U2.14 - Prob. 3ECh. U2.14 - Prob. 4ECh. U2.14 - Prob. 5ECh. U2.14 - Prob. 6ECh. U2.14 - Prob. 7ECh. U2.14 - Prob. 8ECh. U2.15 - Prob. 1TAICh. U2.15 - Prob. 1ECh. U2.15 - Prob. 2ECh. U2.15 - Prob. 4ECh. U2.15 - Prob. 5ECh. U2.15 - Prob. 6ECh. U2.16 - Prob. 1TAICh. U2.16 - Prob. 1ECh. U2.16 - Prob. 2ECh. U2.16 - Prob. 3ECh. U2.16 - Prob. 4ECh. U2.16 - Prob. 6ECh. U2.16 - Prob. 7ECh. U2.17 - Prob. 1TAICh. U2.17 - Prob. 1ECh. U2.17 - Prob. 2ECh. U2.17 - Prob. 3ECh. U2.17 - Prob. 4ECh. U2.17 - Prob. 5ECh. U2.17 - Prob. 6ECh. U2.17 - Prob. 7ECh. U2.17 - Prob. 8ECh. U2.17 - Prob. 9ECh. U2.18 - Prob. 1TAICh. U2.18 - Prob. 1ECh. U2.18 - Prob. 2ECh. U2.18 - Prob. 3ECh. U2.18 - Prob. 4ECh. U2.18 - Prob. 5ECh. U2.18 - Prob. 6ECh. U2.18 - Prob. 7ECh. U2.19 - Prob. 1TAICh. U2.19 - Prob. 1ECh. U2.19 - Prob. 2ECh. U2.19 - Prob. 3ECh. U2.19 - Prob. 4ECh. U2.19 - Prob. 5ECh. U2.19 - Prob. 6ECh. U2.19 - Prob. 7ECh. U2.19 - Prob. 8ECh. U2.19 - Prob. 9ECh. U2.20 - Prob. 1TAICh. U2.20 - Prob. 1ECh. U2.20 - Prob. 2ECh. U2.20 - Prob. 3ECh. U2.20 - Prob. 4ECh. U2.20 - Prob. 5ECh. U2.20 - Prob. 6ECh. U2.20 - Prob. 7ECh. U2.20 - Prob. 8ECh. U2.20 - Prob. 9ECh. U2.21 - Prob. 1TAICh. U2.21 - Prob. 1ECh. U2.21 - Prob. 2ECh. U2.21 - Prob. 3ECh. U2.21 - Prob. 4ECh. U2.21 - Prob. 6ECh. U2 - Prob. SI1RECh. U2 - Prob. SI2RECh. U2 - Prob. SI3RECh. U2 - Prob. SI4RECh. U2 - Prob. SI5RECh. U2 - Prob. SII1RECh. U2 - Prob. SII2RECh. U2 - Prob. SII3RECh. U2 - Prob. SII4RECh. U2 - Prob. SII5RECh. U2 - Prob. SIII1RECh. U2 - Prob. SIII2RECh. U2 - Prob. SIII3RECh. U2 - Prob. SIII4RECh. U2 - Prob. SIII5RECh. U2 - Prob. SIII6RECh. U2 - Prob. SIV1ECh. U2 - Prob. SIV2ECh. U2 - Prob. SIV3ECh. U2 - Prob. SIV4ECh. U2 - Prob. SIV5ECh. U2 - Prob. 1RECh. U2 - Prob. 2RECh. U2 - Prob. 3RECh. U2 - Prob. 4RECh. U2 - Prob. 5RECh. U2 - Prob. 6RECh. U2 - Prob. 7RECh. U2 - Prob. 8RE
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