Explanation The periodic table is used to identify the element using the atomic number concept. Its valance electrons are used to find out the nearest noble gas configuration. Cations and anions are formed by lose or gain of electrons to acquire nearest stable noble gas configuration. The formula of compound is formed by crisscross method in such a way that the resultant compound is neutral. For this, charge of one atom will become number of another atom and vice versa. (a) Electron arrangement given are: 2,8,1 and 2,7 Step 1: Finding elements from the given electron arrangement. Here, 2,8,1 gives an atomic number of 11 which corresponds to the element sodium (Na). Here, 2,7 gives an atomic number of 9 which corresponds to the element fluorine (F). Step 2: Generation of cations and anions of the elements: Na (11); 2,8,1 can easily donate 1 electron to form nearest noble gas configuration of Ne (10) as 2,8. Hence, the cation will be formed by sodium atom as follows: Na → Na + + e − F (9); 2,7 can better accept 1 electron than donating 7 electrons to form nearest noble gas configuration of Ne (10) as 2,8. Hence, the anion will be formed by fluorine as follows: F + e − → F − Step 3: Compound formed by Na + and F − ions: The charge on Na + is +1 and that on F − is -1. They will combine in 1:1 ratio and by crisscross method, the formula of compound will be NaF . The name of Na + is sodium ion and that of F − is fluoride ion thus, name of compound will be sodium fluoride. (b) Given electron arrangements are: 2,8,8,2 and 2,8,6 Step 1: Finding elements from the given electron arrangement. Here, 2,8,8,2 gives an atomic number 20 which corresponds to the element calcium (Ca). Here, 2,8,6 gives an atomic number 16 which corresponds to the element sulfur (S). Step 2: Generation of cations and anions of the elements: Ca (20);2,8,8,2 can better donate 2 electrons to form nearest noble gas configuration of Ar (18) as 2,8,8. Hence, the cation will be formed by calcium as follows: Ca → Ca + 2 + 2 e − S (16);2,8,6 can better accept 2 electrons to form nearest noble gas configuration of Ar (18) as 2,8,8

13th Edition

ISBN: 9781292228860

Author: Timberlake

Publisher: PEARSON

See similar textbooks

Videos

Question

Chapter 6, Problem 6.100UTC

Interpretation Introduction

To determine: To identify the given anion, cation, compound and its name

Expert Solution & Answer

Want to see the full answer?

Check out a sample textbook solution

See solution

Chapter 6, Problem 6.99UTC

Chapter 6, Problem 6.101UTC

Students have asked these similar questions

Choose the Lewis structure for the compound below: H2CCHOCH2CH(CH3)2 HH H :d H H H C. Η H H HH H H H H. H H H HH H H H H H- H H H C-H H H HHHH

Each of the highlighted carbon atoms is connected to hydrogen atoms.

く Complete the reaction in the drawing area below by adding the major products to the right-hand side. If there won't be any products, because nothing will happen under these reaction conditions, check the box under the drawing area instead. Note: if the products contain one or more pairs of enantiomers, don't worry about drawing each enantiomer with dash and wedge bonds. Just draw one molecule to represent each pair of enantiomers, using line bonds at the chiral center. More... No reaction. Explanation Check O + G 1. Na O Me Click and drag to start drawing a structure. 2. H + 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Center | Accessibility 000 Ar P

Chapter 6 Solutions

CHEMISTRY

Ch. 6.1 - State the number of electrons that be must be lost...Ch. 6.1 - State the number of electrons that must be gained...Ch. 6.1 - State the number of electrons lost or gained when...Ch. 6.1 - Prob. 6.4PP Ch. 6.1 - Write the symbols for the ions with the following...Ch. 6.1 - Write the symbols for the ions with the following...Ch. 6.1 - Prob. 6.7PP Ch. 6.1 - State the number of protons and electrons in each...Ch. 6.1 - Write the symbol for the ion of each of the...Ch. 6.1 - Write the symbol for the ion of each of the...

Knowledge Booster

Learn more about

Need a deep-dive on the concept behind this application? Look no further. Learn more about this topic, chemistry and related others by exploring similar questions and additional content below.