(a)
Interpretation:
Interpret formula of Copper(i) iodide.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The formula of Binary ion contains the formula of cation first then proceeded the formula of anion. The formula of cation in Binary ion remain same to that of metal while for anion its formula ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state formula is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
Answer to Problem 26CR
The formula of given compound is CuI.
Explanation of Solution
Copper(i) is denoted as Cu+ while iodide denoted as I- and thus formula of given compound is CuI.
(b)
Interpretation:
Interpret formula of Cobaltous chloride.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The formula of Binary ion contains the formula of cation first then proceeded the formula of anion. The formula of cation in Binary ion remain same to that of metal while for anion its formula ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state formula is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
Answer to Problem 26CR
The formula of given compound is CoCl2.
Explanation of Solution
Cobalt is present in lower oxidation state such as + 2 and thus formula of given compound is CoCl2.
(c)
Interpretation:
Interpret formula of Silver sulfide.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The formula of Binary ion contains the formula of cation first then proceeded the formula of anion. The formula of cation in Binary ion remain same to that of metal while for anion its formula ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state formula is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
Answer to Problem 26CR
The formula of given compound is AgF.
Explanation of Solution
Ag is present in lower oxidation state such as + 1 and thus formula of given compound is AgF.
(d)
Interpretation:
Interpret formula of Mercurous bromide.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The formula of Binary ion contains the formula of cation first then proceeded the formula of anion. The formula of cation in Binary ion remain same to that of metal while for anion its formula ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state formula is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
Answer to Problem 26CR
The formula of given compound is HgBr2.
Explanation of Solution
Hg is present in lower oxidation state such as + 2 and thus formula of given compound is HgBr2.
(e)
Interpretation:
Interpret formula of Mercuric oxide.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The formula of Binary ion contains the formula of cation first then proceeded the formula of anion. The formula of cation in Binary ion remain same to that of metal while for anion its formula ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state formula is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
Answer to Problem 26CR
The formula of given compound is Hg2 O3.
Explanation of Solution
Hg is present in higher oxidation state such as + 3 and thus formula of given compound is Hg2 O3.
(f)
Interpretation:
Interpret formula of Chromium (iii) sulfide.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The formula of Binary ion contains the formula of cation first then proceeded the formula of anion. The formula of cation in Binary ion remain same to that of metal while for anion its formula ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state formula is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
Answer to Problem 26CR
The formula of given compound is Cr2 S3.
Explanation of Solution
Cr is present in higher oxidation state such as + 3 and thus formula of given compound is Cr2 S3.
(g)
Interpretation:
Interpret formula of Plumbic oxide.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The formula of Binary ion contains the formula of cation first then proceeded the formula of anion. The formula of cation in Binary ion remain same to that of metal while for anion its formula ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state formula is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
Answer to Problem 26CR
The formula of given compound is PbO2.
Explanation of Solution
Pb is present in higher oxidation state such as + 4 and thus formula of given compound is PbO2.
(h)
Interpretation:
Interpret formula of Potassium nitride.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The formula of Binary ion contains the formula of cation first then proceeded the formula of anion. The formula of cation in Binary ion remain same to that of metal while for anion its formula ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state formula is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
Answer to Problem 26CR
The formula of given compound is K3 N.
Explanation of Solution
N is present in oxidation state such as -3 and thus formula of given compound is K3 N.
(i)
Interpretation:
Interpret formula of Stannous fluoride.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The formula of Binary ion contains the formula of cation first then proceeded the formula of anion. The formula of cation in Binary ion remain same to that of metal while for anion its formula ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state formula is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
Answer to Problem 26CR
The formula of given compound is SnF2.
Explanation of Solution
Sn2 + is present in lower oxidation state such as + 2 and thus formula of given compound is SnF2.
(j)
Interpretation:
Interpret formula of Ferric oxide.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The formula of Binary ion contains the formula of cation first then proceeded the formula of anion. The formula of cation in Binary ion remain same to that of metal while for anion its formula ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state formula is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
Answer to Problem 26CR
The formula of given compound is Fe2 O3.
Explanation of Solution
Fe is present in higher oxidation state such as + 3 and thus formula of given compound is Fe2 O3.
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Chapter 5 Solutions
EBK INTRODUCTORY CHEMISTRY
- Relative Intensity Part VI. consider the multi-step reaction below for compounds A, B, and C. These compounds were subjected to mass spectrometric analysis and the following spectra for A, B, and C was obtained. Draw the structure of B and C and match all three compounds to the correct spectra. Relative Intensity Relative Intensity 100 HS-NJ-0547 80 60 31 20 S1 84 M+ absent 10 30 40 50 60 70 80 90 100 100- MS2016-05353CM 80- 60 40 20 135 137 S2 164 166 0-m 25 50 75 100 125 150 m/z 60 100 MS-NJ-09-43 40 20 20 80 45 S3 25 50 75 100 125 150 175 m/zarrow_forwardDon't used hand raiting and don't used Ai solutionarrow_forwardPredicting the pro Predict the major products of this organic reaction. Explanation Check m ☐ + 5 1.03 Click and drag t drawing a stru 2. (CH₂)₂S 3 2 © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacy Centerarrow_forward
- starting material target If so, draw a synthesis below. If no synthesis using reagents ALEKS recognizes is possible, check the box under the drawing area. Be sure you follow the standard ALEKS rules for submitting syntheses. + More... X Explanation Check C टे Br T Add/Remove step ☐ Br Br © 2025 McGraw Hill LLC. All Rights Reserved. Terms of Use | Privacarrow_forwardDon't used hand raitingarrow_forwardRelative Intensity Part VI. consider the multi-step reaction below for compounds A, B, and C. These compounds were subjected to mass spectrometric analysis and the following spectra for A, B, and C was obtained. Draw the structure of B and C and match all three compounds to the correct spectra. Relative Intensity Relative Intensity 100 HS-NJ-0547 80 60 31 20 S1 84 M+ absent 10 30 40 50 60 70 80 90 100 100- MS2016-05353CM 80- 60 40 20 135 137 S2 164 166 0-m 25 50 75 100 125 150 m/z 60 100 MS-NJ-09-43 40 20 20 80 45 S3 25 50 75 100 125 150 175 m/zarrow_forward
- Part II. Given two isomers: 2-methylpentane (A) and 2,2-dimethyl butane (B) answer the following: (a) match structures of isomers given their mass spectra below (spectra A and spectra B) (b) Draw the fragments given the following prominent peaks from each spectrum: Spectra A m/2 =43 and 1/2-57 spectra B m/2 = 43 (c) why is 1/2=57 peak in spectrum A more intense compared to the same peak in spectrum B. Relative abundance Relative abundance 100 A 50 29 29 0 10 -0 -0 100 B 50 720 30 41 43 57 71 4-0 40 50 60 70 m/z 43 57 8-0 m/z = 86 M 90 100 71 m/z = 86 M -O 0 10 20 30 40 50 60 70 80 -88 m/z 90 100arrow_forwardPart IV. C6H5 CH2CH2OH is an aromatic compound which was subjected to Electron Ionization - mass spectrometry (El-MS) analysis. Prominent m/2 values: m/2 = 104 and m/2 = 9) was obtained. Draw the structures of these fragments.arrow_forwardFor each reaction shown below follow the curved arrows to complete each equationby showing the structure of the products. Identify the acid, the base, the conjugated acid andconjugated base. Consutl the pKa table and choose the direciton theequilibrium goes. However show the curved arrows. Please explain if possible.arrow_forward
- A molecule shows peaks at 1379, 1327, 1249, 739 cm-1. Draw a diagram of the energy levels for such a molecule. Draw arrows for the possible transitions that could occur for the molecule. In the diagram imagine exciting an electron, what are its various options for getting back to the ground state? What process would promote radiation less decay? What do you expect for the lifetime of an electron in the T1 state? Why is phosphorescence emission weak in most substances? What could you do to a sample to enhance the likelihood that phosphorescence would occur over radiationless decay?arrow_forwardRank the indicated C—C bonds in increasing order of bond length. Explain as why to the difference.arrow_forwardUse IUPAC rules to name the following alkanearrow_forward
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