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Question:For a molecule who's central atom has a sp3d hybridization, which of the following shapes are possible? Linear, "Seesaw", or tetrahedral?

Answers:Both linear and seesaw shapes are possible with dsp3 (sp3d) hybridization. It depends on how many nonbonding pairs you have and how many groups are attached to the central atom. The seesaw shape occurs when the molecule has the form AB4E, where A is the central atom, B represents the atoms attached to the central atom and E is a nonbonding pair of electrons attached to the central atom. The linear shape occurs when the molecule has the form AB2E3, where A, B, and E are as defined as before.

Question:If an atom has sp3d hybridization in a molecule. What is the maximum number of pi bonds that atom can form? & If an atom has sp3 hybridization in a molecule. What is the maximum number of pi bonds that atom can form? Been having a lot of trouble with these question. Any help is greatly appreciated!

Answers:Zero for both. Fully hybridized structures cannot form pi bonds.

Question:i didnt even know you can add a d to that? I thought the only options where sp3, sp2, and sp? Where did they d come from?

Answers:The d orbital comes into play when the central atom has more than 8 valence electrons. That's called an expanded octet and can happen when the central atom is in the third row of the periodic table or lower (as sulfur is). When you draw the Lewis structure for SF4 you see that it has four single sulfur-fluorine bonds, but then you have an additional pair of electrons that must go on the central sulfur atoms as a lone pair. That's five electron groups surrounding the sulfur atom, and sp3 hybridization can only accommodate four electron groups. The fifth hybridized orbital comes from the sulfur atom's potential (but unoccupied in the unbonded atom) 3d sublevel. I hope that helps. Good luck!

Question:1) The hybridization of Cl in ClF2+ is ? A. sp. B. sp2. C. sp3. D. dsp3. E. d2sp3. 2) In the BeF2 molecule the Be valence orbitals are: A. sp hybrids B. sp2 hybrids C. sp3 hybrids D. dsp2 hybrids E. none of these 3) Consider the molecular orbital energy level diagrams for O2 and NO. Which of the following is true? I. Both molecules are paramagnetic. II. The bond strength of O2 is greater than the bond strength of NO. III. NO is an example of a homonuclear diatomic molecule. IV. The ionization energy of NO is smaller than the ionization energy of NO+. A. I only B. I and II C. I and IV D. II and III E. I, II, and IV 4) The fact that O2 is paramagnetic can be explained by A. the Lewis structure of O2. B. resonance. C. a violation of the octet rule. D. the molecular orbital diagram for O2. E. hybridization of atomic orbitals in O2

Answers:1,a 2,a 3,?? 4,a

From Youtube

Sp3d Hybridization and the Trigonal Bipyramidal Arrangement :

Hybridization Geometries & Bond Angles :Understanding the terminology and geometries of hybridized orbitals (part of the valence bond theory). sp = linear = 180 deg sp2= trigonal planar = 120 deg sp3 = tetrahedral = 109.5 deg dsp3 = trigonal bipyramid = 120 and 90 d2sp3 = octahedral = 90 deg I also give a brief explanation of sigma and pi bonds, hybridization, and some of the predictions you can make once you've studied the theory. No, the valence bond theory isn't perfect (consider the paramagnetic properties of O2). It also does a really bad job of predicting some of the real bond angles (measured by crystallography and spectroscopy experimental methods). Overall, it's still a useful explanation. ...It's helped us understand bonding enough to make predictions about unknown molecules that can be verified by experiment. :) I find this fascinating! :) hehe