What shape is a molecule of CF4?
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CF4 or carbon tetrafluoride has a tetrahedral structure\nthat is the carbon atom at the center of a regular tetrahedron and the F atoms at the corners\nthis is because the carbon atom is sp3 hybridised and has these four sp3 hybridised orbitals bonded with 4 F atoms which orient themselves symmetrically in space so as to balance the forces of repulsion between them and reduce energy thus givin a tetrahedral molecule
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tetrahedral\n\nmethane, CH4\nCH4 is composed of 5 atoms, 1 atom of carbon and 4 atoms of hydrogen covalently bonded.\nCarbon, C, is the central atom in the molecule.\nCarbon has 4 valence electrons (electrons that can be used in bonding).\nEach of carbon's 4 valence electrons will form a bonding pair with 1 of hydrogen's electrons.\nThe central carbon atom will therefore have no lone pairs of electrons and 4 bonding pairs of electrons,\nThe bonding pairs of electrons will repel each other as much as possible, so the molecule will be tetrahedral.\n\nCH4 is tetrahedral in shape
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In all cases the arragngement is tetrahedral and thus the ideal bond angle would be 109.5 degrees. In detail:\n\na) O has 2 unpaired e and 2 lone pairs. In H3O+, the 2 e participate in two single bonds with H and one lone pair participates in a dative bond with H+. Thus you have 3 single bonds + 1 lone pair so the arrangement is tetrahedral. lone pair-bond electron repulsion is stronger than bond electron-bond electron and thus the actual bond angles will be smaller.\nThe shape of the molecule will be trigonal pyramid (It is actually tetrahedral with O in the center and on one corner you have the lone pair but the lone pair is not "visible" in the molecular level)\n\nb) Si belongs to the same group as C. It forms 4 identical single  bonds in SiCl4 and has no lone pairs. Thus both the molecular shape and the electron arrangement are normal tetrahedral with bond angles 109.5\n\nc)N has 3 unpaired e and one lone pair. In NF3 it forms 3 single bonds through the unpaired e and so it has 3 bonds + 1 lone pair. Thus it is similar to H3O+ (tetrahedral electron arrangement, ideal bond angle 109.5, true bond angle smaller, molecular shape trigonal pyramid)
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Fullerene.  C60.  Picture a soccer ball made of carbon.  For your blank in the question above, I think the answer is 60.
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Molecules exist in different shapes. Many of the physical & chemical prpoerties of molecules arise due to the different shapes.\n\nSome of the common geometrical shapes are trigonal, planar, tetra hedral etc.The VSEPR theory provides a simple treatment for predicting the shapes of polyatomic molecules. The theory was originally proposed by Sigdwick and Powell in 1940. \n\nIn case of molecules with central atom having one or more lone pairs VSEPR treatment is as follows: In these type of molecules, both lone and bonded pairs of electrons are present. The lone pairs are localised on the central atom, & bonded pairs are shared between the two atoms. Consequently, the lone pair of electrons occupy more space as compared to the bonding pair. This causes great repulsion between lone pair of electrons.\n\nThese repulsion effectscause deviations in the predicted bond angles in molecules. \n\nThe descending order of repulsion interaction is lp - lp > lp - bp > bp - bp\n\nlp - lone pair\nbp - bonded pair
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No, we don't use the electronegativity chart; instead we draw the full structure of the molecule and then see which 'pulling of bonds' cancel each other. \nIn the molecule AsF3, since As has 5 outer most electrons(in group 5), each F is bonded to As with a single covalent bond. Hence, the shape of the molecule becomes trigonal pyramidal, with 3 bond pairs (with 3 F atoms) and 1 lone pair. The 3 bond pairs pull the electrons towards themselves in the lower plane, while the 1 lone pair pulls at the top. The pulling effect of the 2 F atoms at the 2 sides cancel out  because the pulling effect is the same for both F atoms; however, the pulling effect of the lone pair at the top and the 1 F bond pair at the bottom will not cancel, because it is not from the same atom/species. Hence the overall dipole moment( dipole moment = charge x distance) does not cancel out, so the molecule is polar, with some charge distribution remaining due to the unequal pulling effect.\n\nIn CF3Cl, the shape is tetrahedral, with the 3 Cl atoms and 1 F atom bonded by single covalent bonds to the central C atom. The bond pairs spread out with the 3 Cl atoms pulling at the bottom and the 1 F atom pulling at the top. The pulling effect of the 2 Cl atoms at the sides cancels out, but that of the F atom at the top and the Cl atom at the bottom does not, as the pullin effect of 2 different atoms is different. Hence, as before, overall dipole does not cancel and the molecule is hence polar.
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its C2F6,it actually contains two carbon atoms in a molecule and each one of them has three fluorine atoms attached to it,both these carbon atoms are bonded by a covalent bond!!!
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Molecules are made of atoms
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i have 4 for you along with their formulas:\nXenon hexafluoride F6Xe\nXenon difluoride F2Xe\nXenon tetrafluoride F4Xe \nXenon tetrafluoride oxide F4OXe
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Carbon (C), Hydrogen (H), and Oxygen (O)
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In the proposed structure for C60(fullerenes), each vertex of the truncated icosahedron is occupied by a carbon atom, and each carbon is three-connected to other carbon atoms by one double bond and two single bonds. Carbon atoms with this kind of connectivity are usually referred to as "sp carbons" because the orbitals used to sigma-bond the three adjacent carbons are hybrids of the 2s orbital and the two 2p orbitals (2p and 2p). The remaining 2p orbital (2p) is responsible for the pi-bond.\n\nUsually, an sp carbon and its three neighboring atoms are all coplanar. This is the case, for example, in graphite where there are infinite planar sheets of sp carbons arranged in edge-sharing hexagons. The p orbitals all lie parallel to each other and perpendicular to the graphite plane, generating a sea of - electron density above and below the plane.\n\nIn graphite, the C-C bonds are all equal in length and intermediate between normal single and double bonds. This "delocalization" can be explained by noting that graphite has many different but equivalent "resonance structures". (The resonance structures differ only in the placement of the double bonds.) \nthe proposed structure for C60 is not planar! The angle between a p axis and a C-C bond vector is 101.6° (as compared to 90° in planar graphite) . The bowl-shape or concavity at each sp carbon center introduces some strain into the molecule. However, the high symmetry distributes that strain evenly across the entire structure.\n\n# Other Fullerenes\n\nThe truncated icosahedron is not the only possible fullerene-type structure. There are, in fact, many other hollow cage structures that can be constructed using only pentagons and hexagons (Kroto, 1987). Interestingly, each of these structures contains exactly twelve pentagons, while the number of hexagons is arbitrary. The pentagons are necessary for closure. (Recall that graphite, which consists only of fused hexagons, is essentially a planar sheet. While this sheet may warp and bend, it can never close.) The number of vertices in closed fullerene-type structures is necessarily even.\n\nThe smallest possible fullerene would be C20, containing twelve pentagons and zero hexagons. However, such a structure would possess a great deal of strain, because the local topology at each carbon center would be highly non-planar. Other possible fullerenes include C28, C32, C50, and C70. Because the molecular strain tends to be concentrated in the five-membered rings that are responsible for closure, structures that avoid contiguous (edge-sharing) pentagons are particularly stable. It turns out that C and C are the smallest carbon clusters for which this can be achieved.\n\n# Other Possible Structures for C60\n\nWhile the truncated icosahedron structure for C60 is elegant and aesthetically-pleasing, it is important to bear in mind that when it was proposed, there was only one piece of supporting experimental evidence - an unusually large peak at 720 in the mass spectrum. Other C60 cluster structures, including the following three, were possible.\n\n   1. Planar graphite fragments.\n      The problem with this type of structure is that it contains many unsatisfied valences, and it is not clear why C60 should be any more stable than any other cluster size.\n   2. Cyclic polyalkynes.\n      In this structure, there are no unsatisfied valences but the stability of C60 vs. other cluster sizes is not easily rationalized.\n   3. Other fullerenes, i.e., other hollow cage structures with different atom arrangements.\n      These would necessarily be less highly symmetric than the truncated icosahedron and would less effectively distribute strain evenly around the molecule. \n\nGiven these possible, albeit less attractive, alternatives, it was crucial to obtain additional experimental evidence for the structure of C60.
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The space between the molecules
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Here is a model image of a carbon molecule:\nhttp://www.lloydminsterheavyoil.com/carbon.jpg
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Ozone is trigonal planar.  The lone pairs on the terminal oxygens give the trigonal structure, and the molecule is planar (2 dimensional) mainly due to the double bonds in the most stable resonance structure of ozone.\n\nH2Se should be a bent molecule.  The lone pairs take up maximum available space forcing the tiny hydrogen atoms closer together than they's like to be at an angle of about 91 deg from the Se atom.
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XeF4 - square planar (Octahedral geometric family)\nTeF4 - see-saw (Trigonal Bipyramidal geometric family)\nSeCl2 - linear (linear geometric family)
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The molecular geometry is basically the three dimensional arrangement of the atoms that make up a molecule. The study of their geometry gives us the ability to determine several properties such as the reactivity, polarity, phase of matter, color, magnetism, and biological activity.
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It is trigonal bipyramidal in shape.                                                    try this & you would understand better                                            take a cardboard or a  paper.cut it in da shape of a triangle. suspend threads from da vertices of da triangle both upward & downward.                                                                                       now catch-hold of da threads.now 1 hand holds da threads from da top. The triangle remains in da mid position & the other hand holds da other 3 threads at the bottom                                            NOW the 2 hands are Cl atoms .THE 3 vertices of da triangle r also Cl atoms                                                                                   Phosphorous is at centroid of the triangle