Answers to Problems. VSEPR examples. reported structure: CO pyramid. ClF (7)=42 vse A X 5 E 1. XeO (6)=32 vse A X 4 E 0.
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1 Answers to Problems VEPR examples e tetrahedron e bent e.m.l.t deg. 6+(7)=0 vse A X E 3 - triangle bent 4+3(6)+=4 vse A X 3 E o l.t. 10 deg. Xe 4 tetrahedron Xe deg tetrahedron 8+4(6)=3 vse A X 4 E 0 Xe 5 octahedron square-based pyramid 90 and 180 deg (7)=4 vse A X 5 E 1
2 Br trigonal bipyramid Br 3 1- Br Br 1- linear 7+(7)+1= vse Br bond angle: 180 deg. A X E tetrahedral - trigonal pyramid 3-6+3(6)+=6 vse l.t deg. A X 3 E 1 linear linear 4+(6)=16 vse bond angle: 180 deg. A X E 0 Kr 4 octahedral square Kr 4 8+4(7)=36 vse A X 4 E Kr 90 and 180 deg.
3 4 trigonal bipyramid 4 teeter totter 6+4(7)=34 vse 90, 10, and 180 deg. A X 4 E 1 3 (ozone) bond angle triangle bent l.t. 10 deg. 3 6+(6)=18 vse A X E 1 v. a (7) = 6 vse has 4 bonded atoms and no unshared electron pairs, so n = 4, m = 0 VEPR formula: A X 4 E 0 basis str. tetrahedron reported str: TETRAEDRN angle = deg. are the most EN atoms present. Determine the location of their common center. and are more electropositive. Determine the location of their charge center. Do the centers of negative and positive charges coincide? AN: N o 3 would be expected to have a permanent dipole moment. b. N (1) - 1 = 8 vse N has 4 bonded atoms and no unshared electron pairs, so n = 4, m = 0 VEPR formula: A X 4 E 0 basis str. tetrahedron
4 reported str: TETRAEDRN angle = deg. N is the most EN atom present. o the center of negative charge is the N atom. are the more electropositive atoms present. Determine their common center N Do the centers of negative and positive charges coincide? AN: YE o N 4 1+ would NT be expected to have a permanent dipole moment. c. 3(6) = 18 vse has bonded atoms and one unshared electron pairs, so n =, m = 1 VEPR formula: A X E 1 basis str. triangle reported str. BENT angle = L.T. 10 deg. d (6) + = 4 vse has 3 bonded atoms and no unshared electron pairs, so n = 3, m = 0 VEPR formula: A X 3 E 0 basis str. triangle reported str. TRIANGLE angle = 10 deg
5 e. I (7) = 4 vse I has 5 bonded atoms and one unshared electron pair, so n = 5, m = 1 VEPR formula: A X 5 E 1 basis str. octahedron reported str: QUARE-BAED PYRAMID angles, 90 I vi. a. a TW electron system N electrons in TW electrons in BND RDER = - 0 * anti- = 1 * w /b diamagnetic b. 1- a TREE electron system NE electron in TW electrons in * anti- * BND RDER = w /b paramagnetic = 1/
6 c. 1+ a NE electron system N electrons in NE electron in * anti- * BND RDER = 1-0 = 1/ 1+ w /b paramagnetic d. e a UR electron system TW electrons in TW electrons in BND RDER = - = 0 e has never been observed. * anti- * vii. irst-row-of-eight atoms have and valence shells, which can accommodate maxima of TW and IX electrons respectively, for a total of EIGT per atom. Molecular orbital energy level diagrams for first-row-of-eight diatomic molecules and ions must therefore accommodate up to IXTEEN vse. These m.o. diagrams are just a little more complex than those used above. In fact two diagrams are used for these cases b/c there is a change in m.o. energy levels between B and, and the remaining members N,, and Ne. vii. a. B (3) = 6 vse BND RDER = (4 - ) / = 1 PARAMAGNETI ( unpaired electrons) (Use ig. A) B * *
7 * b. (4) = 8 vse BND RDER = (6 - ) / = B DIAMAGNETI (Use ig. A) * * c. N (5) = 10 vse BND RDER = (8 - ) / = 3 DIAMAGNETI (Use ig. B) B * * d. (6) = 1 vse BND RDER = (8-4) / = PARAMAGNETI ( unpaired electrons) (Use ig. B) B * Refer to TABLE. (on class web site) for a listing of bond orders, B D E, magnetic properties, and bond lengths for remaining diatomic molecules and ions in exercise ix.
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