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BCl3- Emphasize the completion of the octet rather than any minimization of formal charge. Enter your answers as a sequence of numbers (one decimal place) separated by commas (for example, 1.3,1.0,1.5,2.0,3.0).

A) N2
B) PO43-
C) PCl5
D) H2
E) BCl3

Can someone please help me with this question? I have an exam on Friday and I need to know how to work this out :(

2007-06-07 04:37:36 · 3 answers · asked by Confused 1 in Science & Mathematics Chemistry

3 answers

3, 5, 5, 1, 3

2007-06-07 05:09:48 · answer #1 · answered by ag_iitkgp 7 · 0 0

1. a. Others have metallic, network, or hydrogen bonds 2. c. sp2 hybridization 3. a. Have to have a lone pair of electrons on an atom for hydrogen bonding to take place, and the nitrogen in ammonium does not have one. 4. c. Ethanol is a polar molecule 5. Can't answer without ther graphic 6. c. Van der Waals forces allow the sheets of carbon in graphite to "slide," whereas diamond's network bonding won't. 7. d. Non-bonding pairs take up more space (VSEPR Theory) 8. c. Carbon tetrachloride is a non-polar molecule with polar bonds, so there is no permanent dipole in the molecule 9. c. Ammonia will, to a degree, react with water to produce ammonium hydroxide, which will dissociate to ammonium ions and hydroxide ions. 10. d. Trichloromethane is the only polar molecule in the group. 11. c. Others are polar covalent. 12. d. Stronger lattice energy between the divalent species, and then the smaller radii allow for closer proximity and more energy required to break bonds. 13. a. Network bonding explains these properties. 14. Can't answer without the structure 15. sorry, not sure 16. d. Network bonding in SiO2 17. a. BF3 is sp2 hybridized, with 120 degree angles. The others are sp3 hybridized, with angles of 109.5 degrees or less. 18. d. SO3(2-) is trigonal planar 19. No really one absolutely correct answer, but a is the closest. 20. a.

2016-05-19 00:13:11 · answer #2 · answered by ? 3 · 0 0

A) 3.0
B) 1.3 (1.25)
C) 1.0
D) 1.0
E) 1.0

2007-06-07 04:47:41 · answer #3 · answered by Matt C 2 · 0 0

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