Log out Go to App
Go to App

Learning Materials

Features

Discover

Chapter 3: Problem 108

Write the formula for each of the following compounds: a. zinc chloride b. \(\operatorname{tin}(1 \mathrm{V})\) fluoride c. calcium nitride d. aluminum sulfide e. mercury(I) selenide f. silver iodide

Short Answer

Expert verified
a. \( ZnCl_2 \) b. \( SnF_4 \) c. \( Ca_3N_2 \) d. \( Al_2S_3 \) e. \( Hg_2Se \) f. \( AgI \)

Step by step solution

01

a. Zinc Chloride

1. Identify ions: Zinc (Zn) has a charge of +2 and chloride (Cl) has a charge of -1. 2. Write formula: Since the charges are +2 and -1, we need two chloride ions to balance the charge of one zinc ion. The formula for zinc chloride is ZnCl2.
02

b. Tin(IV) Fluoride

1. Identify ions: Tin(IV) (Sn) has a charge of +4 and fluoride (F) has a charge of -1. 2. Write formula: Since the charges are +4 and -1, we need four fluoride ions to balance the charge of one tin ion. The formula for tin(IV) fluoride is SnF4.
03

c. Calcium Nitride

1. Identify ions: Calcium (Ca) has a charge of +2 and nitride (N) has a charge of -3. 2. Write formula: Since the charges are +2 and -3, we need three calcium ions to balance with two nitride ions. The formula for calcium nitride is Ca3N2.
04

d. Aluminum Sulfide

1. Identify ions: Aluminum (Al) has a charge of +3 and sulfide (S) has a charge of -2. 2. Write formula: Since the charges are +3 and -2, we need two aluminum ions to balance with three sulfide ions. The formula for aluminum sulfide is Al2S3.
05

e. Mercury(I) Selenide

1. Identify ions: Mercury(I) (Hg) has a charge of +1 and selenide (Se) has a charge of -2. 2. Write formula: Since the charges are +1 and -2, we need two mercury ions to balance the charge of one selenide ion. Note that mercury(I) is diatomic: Hg2^2+. The formula for mercury(I) selenide is Hg2Se.
06

f. Silver Iodide

1. Identify ions: Silver (Ag) has a charge of +1 and iodide (I) has a charge of -1. 2. Write formula: Since the charges are +1 and -1, we need one silver ion to balance the charge of one iodide ion. The formula for silver iodide is AgI.

Unlock Step-by-Step Solutions & Ace Your Exams!

  • Full Textbook Solutions

    Get detailed explanations and key concepts

  • Unlimited Al creation

    Al flashcards, explanations, exams and more...

  • Ads-free access

    To over 500 millions flashcards

  • Money-back guarantee

    We refund you if you fail your exam.

Start your free trial

Over 30 million students worldwide already upgrade their learning with Vaia!

One App. One Place for Learning.

All the tools & learning materials you need for study success - in one app.

Get started for free

Most popular questions from this chapter

Benzene \(\left(\mathrm{C}_{6} \mathrm{H}_{6}\right)\) consists of a six- membered ring of carbon atoms with one hydrogen bonded to each carbon. Write Lewis structures for benzene, including resonance structures.

\(\mathrm{SF}_{6}, \mathrm{ClF}_{5,}\) and \(\mathrm{XeF}_{4}\) are three compounds whose central atoms do not follow the octet rule. Draw Lewis structures for these compounds.

What noble gas has the same electron configuration as each of the ions in the following compounds? a. cesium sulfide b. strontium fluoride c. calcium nitride d. aluminum bromide

The most common exceptions to the octet rule are compounds or ions with central atoms having more than eight electrons around them. \(\mathrm{PF}_{5}, \mathrm{SF}_{4}, \mathrm{ClF}_{3},\) and \(\mathrm{Br}_{3}^{-}\) are examples of this type of exception. Draw the Lewis structure for these compounds or ions. Which elements, when they have to, can have more than eight electrons around them? How is this rationalized?

Give three ions that are isoelectronic with neon. Place these ions in order of increasing size.
See all solutions

Recommended explanations on Chemistry Textbooks

Ionic and Molecular Compounds

Read Explanation

Chemical Analysis

Read Explanation

Making Measurements

Read Explanation

Inorganic Chemistry

Read Explanation

Physical Chemistry

Read Explanation

Organic Chemistry

Read Explanation
View all explanations

What do you think about this solution?

We value your feedback to improve our textbook solutions.

Study anywhere. Anytime. Across all devices.

Sign-up for free