Midterm Review

# Midterm Review - Total = 26e-FC O double = 0 S= 6(1/2(12)=0...

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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 O’s ( 4e-) has a double bond with As. The other 3 O’s have a single bond with 6e- each. 6. C—F is more polar. The electro negativity determines how polar something is. The farther away, the more polar. 7. a) H2O – ionic b) SrCl2—ionic c) H2—metallic d) AgCl—ionic 8. O2, N2, CsI, NaBr, HBr 9. HBr, NaBr, CsI, N2, O2 10. a)-278 b) -498 c) 368 11. a) C(triple)C—H Trigonal Plane 120 I H b) H H Triggonal Plane 120 I I H—C—(double) N I I H H c) This one I wasn’t sure about how to do and how to determine the shape...
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## This note was uploaded on 07/24/2009 for the course BIO 80J taught by Professor Zavenelli during the Winter '08 term at UCSC.

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Midterm Review - Total = 26e-FC O double = 0 S= 6(1/2(12)=0...

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