Question

Consider the following Lewis structure. What is the formal charge on the nitrogen atom in this structure? :0: N :O: 19. Draw the Lewis structure for CIO,, making sure that all of the atoms obey the octet rule. What is the formal charge about the central atom, and how many lone pairs of electrons does it have? a) -1, one lone pair b) +1, one lone pair c) -1, zero lone pairs d) -1, two lone pairs e) +1, two lone pairs Table of Bond Energies (in kJ/mol) H-H 432 C-H 413 O-H 467 H-F 565 C-C 347 O-O (in O?) 498 H-Cl 427 C-O 358 C=O (in 799 CO?) H-Br 363 C-Cl 339 C-Cl 243 H-I 295 C-Br 276 Br-Br 193 C-C 614 20. Use the Table of Bond Energies to determine the molar enthalpy of combustion of ethene, C?H?, in kJ/mol (The Lewis structure of C?H? is shown.) H H a) -1228 kJ/mol b) -1802 kJ/mol c) -1304 kJ/mol d) -3454 kJ/mol e) -2956 kJ/mol 

          Consider the following Lewis structure. What is the formal charge on the nitrogen atom in this structure?
:0:
N
:O:
19. Draw the Lewis structure for CIO,, making sure that all of the atoms obey the octet rule. What is the formal charge about the
central atom, and how many lone pairs of electrons does it have?
a) -1, one lone pair
b) +1, one lone pair
c) -1, zero lone pairs
d) -1, two lone pairs
e) +1, two lone pairs
Table of Bond Energies (in kJ/mol)
H-H	432	C-H	413	O-H	467
H-F	565	C-C	347	O-O (in O?)	498
H-Cl	427	C-O	358	C=O (in	799
				CO?)
H-Br	363	C-Cl	339	C-Cl	243
H-I	295	C-Br	276	Br-Br	193
C-C	614
20. Use the Table of Bond Energies to determine the molar enthalpy of combustion of ethene, C?H?, in kJ/mol
(The Lewis structure of C?H? is shown.)
H
H
a) -1228 kJ/mol
b) -1802 kJ/mol
c) -1304 kJ/mol
d) -3454 kJ/mol
e) -2956 kJ/mol
Show more…
Consider the following Lewis structure. What is the formal charge on the nitrogen atom in this structure? :0: N :O: 19. Draw the Lewis structure for CIO,, making sure that all of the atoms obey the octet rule. What is the formal charge about the central atom, and how many lone pairs of electrons does it have? a) -1, one lone pair b) +1, one lone pair c) -1, zero lone pairs d) -1, two lone pairs e) +1, two lone pairs Table of Bond Energies (in kJ/mol) H-H 432 C-H 413 O-H 467 H-F 565 C-C 347 O-O (in O?) 498 H-Cl 427 C-O 358 C=O (in 799 CO?) H-Br 363 C-Cl 339 C-Cl 243 H-I 295 C-Br 276 Br-Br 193 C-C 614 20. Use the Table of Bond Energies to determine the molar enthalpy of combustion of ethene, C?H?, in kJ/mol (The Lewis structure of C?H? is shown.) H H a) -1228 kJ/mol b) -1802 kJ/mol c) -1304 kJ/mol d) -3454 kJ/mol e) -2956 kJ/mol

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Chemistry: Structure and Properties
Chemistry: Structure and Properties
Nivaldo Tro 2nd Edition
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Text: Atom in this structure :0 1+2 '61 Draw the Lewis structure for ClOy, making sure that all of the central atom, and how many lone pairs of electrons does it have? Si a) one lone pair b) one lone pair c) zero lone pairs d) two lone pairs +1.two lone pairs Table of Bond Energies in kJ/mol: 432 CH 413 OH 565 CC 347 OOinO 427 Co 358 C=O(in CO) 363 15-0 339 CI-CI 295 C-Br 276 C-C 614 H-H H-F H-Cl 467 498 799 H-Br H-I 243 193 Use the Table of Bond Energies to determine the molar enthalpy of combustion of ethene, CH2, in kJ/mol (The Lewis structure of CH2 is shown.) H TH a) -1228 kJ/mol b) 1802 kJ/mol c) 1304 kJ/mol d) 3454 kJ/mol -2956 kJ/mol
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B. Draw Lewis Structures for the following molecules (include all lone pair electrons). For each molecule, follow these steps: 1. Count the total number of valence electrons for each atom. 2. Predict the general connectivity of the molecule. 3. Use any remaining electrons as lone pairs to fill up unsatisfied octets. a. CH4 b. CH2Cl2 c. HI d. Br2 e. BF3 f. H2S g. N2H4 h. CH5N Typical number of bonds per charge-neutral atom: - Group 17 (F, Cl, Br, I) + Hydrogen: 1 - Group 16 (O, S, Se): 2 - Group 15 (N, P): 3 - Group 14 (C, Si): 4 Some molecules contain double/triple bonds, which require an additional rule: 4. If all electrons have been placed as bonds/lone pairs, but some atoms remain unfilled (subvalent), remove a lone pair from an adjacent atom and use it to form a double bond with the subvalent atom. If some atoms are still subvalent, remove additional lone pairs to make a triple bond or additional double bonds. a. O2 b. N2H2 c. CH2O d. C2H2 e. N2 f. C2Cl4 g. CO2 h. HCN 5. If the molecule has a charge, add electrons (for a negative charge) or subtract electrons (for a positive charge) from the total electron count in step 1. Some atoms will not have their typical number of bonds (as in the chart above). a. CH3CO2- b. CO3 2- c. NH4+ d. OH

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Draw the Lewis structure for SeS2 and answer the following questions: How many valence electrons are present in this compound? 18 How many bonding electrons are present in this compound? 6 How many lone pair (non-bonding) electrons are present in this compound? 12 What, if anything, is wrong with each of the following structural representations of SeS2? (Only the lone pairs on the central atom are shown. Assume the double bonded atoms have two lone pairs and the single bonded atoms have three lone pairs.) too many electrons used nothing is wrong too many electrons used What are the approximate bond angles made by the atoms in this structure? Select all that apply. 60° 90° 109° 120° 180°

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a. Draw the Lewis structure of $\mathrm{H}_{2} \mathrm{CO}$. b. Describe the orbitals used by the carbon atom in bonding and indicate the approximate bond angles. SOLUTION TO 31 a. Our first attempt at a Lewis structure (drawing the atoms with the hydrogens on the outside of the molecule) shows that carbon is the only atom that does not form the needed number of bonds. If we place a double bond between carbon and oxygen and move the $\mathrm{H}$ from $\mathrm{O}$ to $\mathrm{C}$ (which still keeps it on the outside of the molecule), then all the atoms end up with the correct number of bonds. Lone-pair electrons are used to give oxygen a filled outer shell. When we check to see if any atom needs to be assigned a formal charge, we find that none of them does. SOLUTION TO 31 b. Because carbon forms a double bond, we know that it uses $s p^{2}$ orbitals (as it does in ethene) to bond to the two hydrogens and the oxygen. It uses its "leftover" $p$ orbital to form the second bond to oxygen. Because carbon is $s p^{2}$ hybridized, the bond angles are approximately $120^{\circ}$.

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Transcript

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00:01 Hello, so here in this question we are provided with eight compounds and we need to find the total valency of each of this compound, then the bond connectivity and also the number of lawn pairs.
00:22 Okay, so let's see the answer for this question.
00:26 Here we are having carbon and hydrogen.
00:29 So carbon, the valency is four and hydrogen we have the valency as one.
00:35 Okay.
00:35 So we can have the bond connectivity as we're having one carbon atom and four hydrogen atom and each of this hydrogen will give one electron to this carbon atom.
00:46 Okay.
00:47 So they will have two electrons in a single bond like this.
00:53 So the connectivity will be carbon, hydrogen, hydrogen, hydrogen and hydrogen.
01:01 Okay.
01:01 So our total valency will be one, two, three, four, five, six, six, 7 and 8.
01:08 So the total valency is 8 and the number of loan pair is 0.
01:15 Okay.
01:16 Now for the next question what we have is we have carbon, hydrogen and chlorine.
01:23 Chlorin the valency is 7.
01:26 Okay.
01:26 So we can draw the connectivity as carbon with two hydrogen atoms and two chlorine atoms.
01:35 As each of this chlorine atom is donating one electron to the carbon atom, the rest of it will remain as lawn pair.
01:42 So the number of lawn pairs will be 1, 2, 3, 4, 5 and 6.
01:50 Okay, so the total valency will be 1, 2, 3, 4, 5, 6 and 8.
01:57 And number of lawn pairs will be 1, 2, 3, 4, 5 and 6.
02:01 So there are 6 long pair of electrons.
02:04 Now for the next one, what we have is hydrogen iodide, okay? so for hydrogen iodide as we have earlier said, hydrogen has a valency of one and iodine, as it is one of the halogens, the valency will be seven.
02:17 Okay, so we can draw it as hydrogen and iodine will form a single bond by donating one electron and the rest of the electrons will form lawn pairs for this iodine.
02:28 Okay? so the general connectivity will be h, i and the others will remain as lawn pair.
02:35 So there are three lawn pair and a bond pair.
02:37 So the total valency is 2 and the number of lawn pairs is 3...
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