(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
Introductory Chemistry: Foundation - Text (Looseleaf)
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- A mixture of C7H12O2, C9H9OCl, biphenyl and acetone was put together in a gas chromatography tube. Please decide from the GC resutls which correspond to the peak for C7,C9 and biphenyl and explain the reasoning based on GC results. Eliminate unnecessary peaks from Gas Chromatography results.arrow_forwardIs the molecule chiral, meso, or achiral? CI .CH3 H₂C CIarrow_forwardPLEASE HELP ! URGENT!arrow_forward
- Identify priority of the substituents: CH3arrow_forwardHow many chiral carbons are in the molecule? OH F CI Brarrow_forwardA mixture of three compounds Phen-A, Acet-B and Rin-C was analyzed using TLC with 1:9 ethanol: hexane as the mobile phase. The TLC plate showed three spots of R, 0.1 and 0.2 and 0.3. Which of the three compounds (Phen-A; Acet-B or Rin-C) would have the highest (Blank 1), middle (Blank 2) and lowest (Blank 3) spot respectively? 0 CH: 0 CH, 0 H.C OH H.CN OH Acet-B Rin-C phen-A A A <arrow_forward
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