Midterm Review

Midterm Review - Total = 26e-FC- O ( double) = 0 S=...

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Midterm Review # 2 (part 2) Answers 1. a) Cas, b) SrO, c) BaS, d) MgO 2. a) H—O(4 e-)—C –(DOUBLE) O(4 E-) I O(4 E-) I H b) H I H—C—S(4E-)—H I H 3. a) O(4 E-) II H—O(4E-) – Br—(double)O ( 4e-) II O ( 4E-) What does it mean all important resonance form? b) (2E-)N—(TRIPLE)N—(DOUBLE, 2E-)N (2E-) c) (4E-)O – (DOUBLE)—N(DOUBLE) O ( 4 E-) I
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F (6E-) 4. a) Fc- I= 7-(1/2(10)+2) =0 F=7-(1/2(2)+6) = 0 I has 2 e- and 5 F’s around it with 6 e- each b) N=5-(1/2(4) + 2) = -1 C= 4-(1/2 (4) + 2) = 0 C has 2e- and a double bond with N and N has 4 e- around it c) N= 5-(1/2(6) + 2)= 0 Cl=7 -(1/2(2)+6)=0 O (double) = 0 Cl has 6 e-, a single bond with N, N has 2 e- and a double bond with O and O has 4 e- 5. a) All three O’s have a double bond with S and each O has 4 e- around them.
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Unformatted text preview: Total = 26e-FC- O ( double) = 0 S= 6-(1/2(12))=0 b) One of the Os ( 4e-) has a double bond with As. The other 3 Os have a single bond with 6e- each. 6. CF is more polar. The electro negativity determines how polar something is. The farther away, the more polar. 7. a) H2O ionic b) SrCl2ionic c) H2metallic d) AgClionic 8. O2, N2, CsI, NaBr, HBr 9. HBr, NaBr, CsI, N2, O2 10. a)-278 b) -498 c) 368 11. a) C(triple)CH Trigonal Plane 120 I H b) H H Triggonal Plane 120 I I HC(double) N I I H H c) This one I wasnt sure about how to do and how to determine the shape...
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Midterm Review - Total = 26e-FC- O ( double) = 0 S=...

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