![Bundle: Chemistry: An Atoms First Approach, Loose-leaf Version, 2nd + OWLv2 with Student Solutions Manual, 4 terms (24 months) Printed Access Card](https://www.bartleby.com/isbn_cover_images/9781337086431/9781337086431_largeCoverImage.gif)
(a)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of of sulfur difluoride
(a)
![Check Mark](/static/check-mark.png)
Explanation of Solution
Sulfur belongs to the Group
Fluorine is a non-metal of the
Oxidation state
In
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(b)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of sulfur hexafluoride
(b)
![Check Mark](/static/check-mark.png)
Explanation of Solution
Sulfur belongs to the Group
Fluorine is a non-metal of the
Oxidation state
In
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(c)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of
(c)
![Check Mark](/static/check-mark.png)
Explanation of Solution
Sodium belongs to the Group
Dihydrogen phosphate is an anion having oxidation state
So, the formula of the given compound is
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(d)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of lithium nitride
(d)
![Check Mark](/static/check-mark.png)
Explanation of Solution
Lithium belongs to the Group
Nitrogen is a non-metal of the
Oxidation state
So, the formula of the given compound is
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(e)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of chromium (III) carbonate.
(e)
![Check Mark](/static/check-mark.png)
Explanation of Solution
Chromium belongs to the Group
Carbonate is an anion having oxidation state
So, the formula of the given compound is
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(f)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of tin (II) fluoride.
(f)
![Check Mark](/static/check-mark.png)
Explanation of Solution
Tin belongs to the Group
Fluorine is a non-metal of the
Oxidation state
So, the formula of the given compound is
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(g)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of ammonium acetate.
(g)
![Check Mark](/static/check-mark.png)
Explanation of Solution
The oxidation state of ammonium
Acetate
So, the formula of the given compound is
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(h)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of ammonium hydrogen sulfate.
(h)
![Check Mark](/static/check-mark.png)
Explanation of Solution
The oxidation state of ammonium
Hydrogen sulfate
So, the formula of the given compound is
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(i)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of cobalt (III) nitrate.
(i)
![Check Mark](/static/check-mark.png)
Explanation of Solution
The oxidation state of cobalt
Nitrate
So, the formula of the given compound is
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(j)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of mercury (I) chloride.
(j)
![Check Mark](/static/check-mark.png)
Explanation of Solution
The oxidation state of mercury
Chloride
Two mercury and chlorine ions is joined to form mercury (I) chloride.
So, the formula of the given compound is
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(k)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of potassium chlorate.
(k)
![Check Mark](/static/check-mark.png)
Explanation of Solution
The oxidation state of potassium
Chlorate
So, the formula of the given compound is
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
(l)
Interpretation: The formula of the binary compound formed from the given pairs of elements in each case is to be stated.
Concept introduction: The oxidation state of an element corresponds to the group number of that element. In case of non-metals, the oxidation state can be calculated as,
To determine: The formula of sodium hydride.
(l)
![Check Mark](/static/check-mark.png)
Explanation of Solution
The oxidation state of sodium
Hydride
So, the formula of the given compound is
The formula of ionic compound is given such that positive ion (cation) always written first and negative ion written second.
Want to see more full solutions like this?
Chapter 3 Solutions
Bundle: Chemistry: An Atoms First Approach, Loose-leaf Version, 2nd + OWLv2 with Student Solutions Manual, 4 terms (24 months) Printed Access Card
- H2SO4 (cat.), H₂O 100 °C NH₂arrow_forwardX Draw the major products of the elimination reaction below. If elimination would not occur at a significant rate, check the box under the drawing area instead. ది www. Cl + OH Elimination will not occur at a significant rate. Click and drag to start drawing a structure.arrow_forwardNonearrow_forward
- 1A H 2A Li Be Use the References to access important values if needed for this question. 8A 3A 4A 5A 6A 7A He B C N O F Ne Na Mg 3B 4B 5B 6B 7B 8B-1B 2B Al Si P 1B 2B Al Si P S Cl Ar K Ca Sc Ti V Cr Mn Fe Co Ni Cu Zn Ga Ge As Se Br Kr Rb Sr Y Zr Nb Mo Tc Ru Rh Pd Ag Cd In Sn Sb Te I Xe * Cs Ba La Hf Ta W Re Os Ir Pt Au Hg Tl Pb Bi Po At Rn Fr Ra Ac Rf Ha ****** Ce Pr Nd Pm Sm Eu Gd Tb Dy Ho Er Tm Yb Lu Th Pa U Np Pu Am Cm Bk Cf Es Fm Md No Lr Analyze the following reaction by looking at the electron configurations given below each box. Put a number and a symbol in each box to show the number and kind of the corresponding atom or ion. Use the smallest integers possible. cation anion + + Shell 1: 2 Shell 2: 8 Shell 3: 1 Shell 1 : 2 Shell 2 : 6 Shell 1 : 2 Shell 2: 8 Shell 1: 2 Shell 2: 8arrow_forwardNonearrow_forwardIV. Show the detailed synthesis strategy for the following compounds. a. CH3CH2CH2CH2Br CH3CH2CCH2CH2CH3arrow_forward
- Do the electrons on the OH participate in resonance with the ring through a p orbital? How many pi electrons are in the ring, 4 (from the two double bonds) or 6 (including the electrons on the O)?arrow_forwardPredict and draw the product of the following organic reaction:arrow_forwardNonearrow_forward
- Redraw the molecule below as a skeletal ("line") structure. Be sure to use wedge and dash bonds if necessary to accurately represent the direction of the bonds to ring substituents. Cl. Br Click and drag to start drawing a structure. : ☐ ☑ Parrow_forwardK m Choose the best reagents to complete the following reaction. L ZI 0 Problem 4 of 11 A 1. NaOH 2. CH3CH2CH2NH2 1. HCI B OH 2. CH3CH2CH2NH2 DII F1 F2 F3 F4 F5 A F6 C CH3CH2CH2NH2 1. SOCl2 D 2. CH3CH2CH2NH2 1. CH3CH2CH2NH2 E 2. SOCl2 Done PrtScn Home End FA FQ 510 * PgUp M Submit PgDn F11arrow_forwardNonearrow_forward
- Living By Chemistry: First Edition TextbookChemistryISBN:9781559539418Author:Angelica StacyPublisher:MAC HIGHERIntroductory Chemistry: A FoundationChemistryISBN:9781337399425Author:Steven S. Zumdahl, Donald J. DeCostePublisher:Cengage Learning
- World of Chemistry, 3rd editionChemistryISBN:9781133109655Author:Steven S. Zumdahl, Susan L. Zumdahl, Donald J. DeCostePublisher:Brooks / Cole / Cengage Learning
![Text book image](https://www.bartleby.com/isbn_cover_images/9781559539418/9781559539418_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781337399425/9781337399425_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781337399692/9781337399692_smallCoverImage.gif)
![Text book image](https://www.bartleby.com/isbn_cover_images/9781133109655/9781133109655_smallCoverImage.jpg)