(a)
Interpretation:
Interpret name of CoCl2.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The name of Binary ion contains the name of cation first then proceeded the name of anion. The name of cation in Binary ion remain same to that of metal while for anion its name ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state name is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
(b)
Interpretation:
Interpret name of CrBr3.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The name of Binary ion contains the name of cation first then proceeded the name of anion. The name of cation in Binary ion remain same to that of metal while for anion its name ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state name is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
(c)
Interpretation:
Interpret name of PbO.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The name of Binary ion contains the name of cation first then proceeded the name of anion. The name of cation in Binary ion remain same to that of metal while for anion its name ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state name is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
(d)
Interpretation:
Interpret name of SnO2.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The name of Binary ion contains the name of cation first then proceeded the name of anion. The name of cation in Binary ion remain same to that of metal while for anion its name ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state name is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
(e)
Interpretation:
Interpret name of Co2 O3.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The name of Binary ion contains the name of cation first then proceeded the name of anion. The name of cation in Binary ion remain same to that of metal while for anion its name ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state name is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
(f)
Interpretation:
Interpret name of FeCl3.
Concept Introduction:
Binary ionic compound is the species contains two ions in the compound to form a chemical species.
The name of Binary ion contains the name of cation first then proceeded the name of anion. The name of cation in Binary ion remain same to that of metal while for anion its name ends by suffix ‘ate’ while anion of group 6 and 7 ends with suffix ‘ide’.
For cation with lower oxidation state name is ended with ‘ous’ and with high oxidation state is ended with ‘ic’.
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Chapter 5 Solutions
Introductory Chemistry: Foundation - Text (Looseleaf)
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- III O Organic Chemistry Using wedges and dashes in skeletal structures Draw a skeletal ("line") structure for each of the molecules below. Be sure your structures show the important difference between the molecules. key O O O O O CHON Cl jiii iiiiiiii You can drag the slider to rotate the molecules. Explanation Check Click and drag to start drawing a structure. Q Search X G ©2025 McGraw Hill LLC. All Rights Reserved. Terms of Use F 3 W C 3/5arrow_forward3. Use Kapustinskii's equation and data from Table 4.10 in your textbook to calculate lattice energies of Cu(OH)2 and CuCO3 (4 points)arrow_forward2. Copper (II) oxide crystalizes in monoclinic unit cell (included below; blue spheres 2+ represent Cu²+, red - O²-). Use Kapustinski's equation (4.5) to calculate lattice energy for CuO. You will need some data from Resource section of your textbook (p.901). (4 points) CuOarrow_forward
- What is the IUPAC name of the following compound? OH (2S, 4R)-4-chloropentan-2-ol O (2R, 4R)-4-chloropentan-2-ol O (2R, 4S)-4-chloropentan-2-ol O(2S, 4S)-4-chloropentan-2-olarrow_forwardIn the answer box, type the number of maximum stereoisomers possible for the following compound. A H H COH OH = H C Br H.C OH CHarrow_forwardSelect the major product of the following reaction. Br Br₂, light D Br Br Br Brarrow_forward
- Select all molecules which are chiral. Brarrow_forwardUse the reaction coordinate diagram to answer the below questions. Type your answers into the answer box for each question. (Watch your spelling) Energy A B C D Reaction coordinate E A) Is the reaction step going from D to F endothermic or exothermic? A F G B) Does point D represent a reactant, product, intermediate or transition state? A/ C) Which step (step 1 or step 2) is the rate determining step? Aarrow_forward1. Using radii from Resource section 1 (p.901) and Born-Lande equation, calculate the lattice energy for PbS, which crystallizes in the NaCl structure. Then, use the Born-Haber cycle to obtain the value of lattice energy for PbS. You will need the following data following data: AH Pb(g) = 196 kJ/mol; AHƒ PbS = −98 kJ/mol; electron affinities for S(g)→S¯(g) is -201 kJ/mol; S¯(g) (g) is 640kJ/mol. Ionization energies for Pb are listed in Resource section 2, p.903. Remember that enthalpies of formation are calculated beginning with the elements in their standard states (S8 for sulfur). The formation of S2, AHF: S2 (g) = 535 kJ/mol. Compare the two values, and explain the difference. (8 points)arrow_forward
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