This makes graphite slippery, so it is useful as a lubricant. Explain why diamond does not conduct electricity and why graphite does conduct electricity. Diamond does not conduct electricity because it has no charged particles that are free to move. Graphite does conduct electricity because it has delocalised electrons which move between the layers.
Diamond and graphite Diamond and graphite are different forms of the element carbon. Diamond Structure and bonding Diamond is a giant covalent structure in which: each carbon atom is joined to four other carbon atoms by strong covalent bonds the carbon atoms form a regular tetrahedral network structure there are no free electrons Carbon atoms in diamond form a tetrahedral arrangement Properties and uses The rigid network of carbon atoms, held together by strong covalent bonds, makes diamond very hard.
Graphite Structure and bonding Graphite has a giant covalent structure in which: each carbon atom forms three covalent bonds with other carbon atoms the carbon atoms form layers of hexagonal rings there are no covalent bonds between the layers there is one non-bonded - or delocalised - electron from each atom Dotted lines represent the weak forces between the layers in graphite Properties and uses Graphite has delocalised electrons, just like metals.
Why two carbon atoms don't form more than triple bond with each other? Please tell me simply as I am a student of 10th class. Unfortunately, an answer to this requires a quick dip into orbital theory. The process by which these orbitals form bonds is simply overlap: If you have two orbitals that can be moved together and these two orbitals overlap, you generate a bonding and an antibonding orbital from these two. This is a mathematical operation known as linear combination; plugging two orbitals into a linear combination means that exactly two orbitals are mathematically required to come out of the linear combination.
Image taken from jahschem. And carbon does not have any additional accessable orbitals that it could use for further bonds. Thus, it is not possible for carbon to form quadruple bonds. The next orbital available would be a d orbital; the transition metals are those in which the d orbitals are valence orbitals and take part in bonding.
To the best of my knowledge, quadruple bonds have not been shown to exist between main group elements though. When you say four bonds with each other, it means 1 hybrid orbital for sure which will be the sigma bond inbetween the two carbon atoms. Now inbetween two atoms you can never have more than 1 sigma bond , the reason is for sigma bond we need head one overlapping of hybrid orbitals.
If hybrid orbital of one carbon atom is already in head on overlap with the hybrid orbital of the other carbon atom, the other 3 hybrid orbitals of the carbon atom cannot do any further head on overlapping because angle between the hybrid orbitals would be degrees approx.
So basically they cant reach the hybrid orbital of the other C atom. So you can't have more than 1 sigma bond. As we need one hybrid orbital for sigma bond, we would be creating only one hybrid orbital then, hence sp hybridisation.
If we talk about four bonds inbetween the carbon atom, then the other three has to be pie bonds that will be from 3 pure p orbitals, but 3 pure p orbitals is not possible.
In the case of water, we know that the O-H covalent bond is polar, due to the different electronegativities of hydrogen and oxygen. Since there are two O-H bonds in water, their bond dipoles will interact and may result in a molecular dipole which can be measured.
The following diagram shows four possible orientations of the O-H bonds. The bond dipoles are colored magenta and the resulting molecular dipole is colored blue. In a similar manner the configurations of methane CH 4 and carbon dioxide CO 2 may be deduced from their zero molecular dipole moments.
Since the bond dipoles have canceled, the configurations of these molecules must be tetrahedral or square-planar and linear respectively. The case of methane provides insight to other arguments that have been used to confirm its tetrahedral configuration. For purposes of discussion we shall consider three other configurations for CH 4 , square-planar, square-pyramidal and triangular-pyramidal.
Models of these possibilities may be examined by. Substitution of one hydrogen by a chlorine atom gives a CH 3 Cl compound. Since the tetrahedral, square-planar and square-pyramidal configurations have structurally equivalent hydrogen atoms, they would each give a single substitution product. However, in the trigonal-pyramidal configuration one hydrogen the apex is structurally different from the other three the pyramid base.
Substitution in this case should give two different CH 3 Cl compounds if all the hydrogens react. In the case of disubstitution, the tetrahedral configuration of methane would lead to a single CH 2 Cl 2 product, but the other configurations would give two different CH 2 Cl 2 compounds. These substitution possibilities are shown in the above insert. Structural Formulas It is necessary to draw structural formulas for organic compounds because in most cases a molecular formula does not uniquely represent a single compound.
Different compounds having the same molecular formula are called isomers , and the prevalence of organic isomers reflects the extraordinary versatility of carbon in forming strong bonds to itself and to other elements. When the group of atoms that make up the molecules of different isomers are bonded together in fundamentally different ways, we refer to such compounds as constitutional isomers.
There are seven constitutional isomers of C 4 H 10 O, and structural formulas for these are drawn in the following table. These formulas represent all known and possible C 4 H 10 O compounds, and display a common structural feature. There are no double or triple bonds and no rings in any of these structures. Note that each of the carbon atoms is bonded to four other atoms, and is saturated with bonding partners.
Simplification of structural formulas may be achieved without any loss of the information they convey. In condensed structural formulas the bonds to each carbon are omitted, but each distinct structural unit group is written with subscript numbers designating multiple substituents, including the hydrogens. Shorthand line formulas omit the symbols for carbon and hydrogen entirely. Each straight line segment represents a bond, the ends and intersections of the lines are carbon atoms, and the correct number of hydrogens is calculated from the tetravalency of carbon.
Non-bonding valence shell electrons are omitted in these formulas. Developing the ability to visualize a three-dimensional structure from two-dimensional formulas requires practice, and in most cases the aid of molecular models. As noted earlier, many kinds of model kits are available to students and professional chemists, and the beginning student is encouraged to obtain one.
Constitutional isomers have the same molecular formula, but their physical and chemical properties may be very different. For an example Click Here. Distinguishing Carbon Atoms When discussing structural formulas, it is often useful to distinguish different groups of carbon atoms by their structural characteristics.
The three C 5 H 12 isomers shown below illustrate these terms. Structural differences may occur within these four groups, depending on the molecular constitution. A consideration of molecular symmetry helps to distinguish structurally equivalent from nonequivalent atoms and groups. The ability to distinguish structural differences of this kind is an essential part of mastering organic chemistry. It will come with practice and experience.
Our ability to draw structural formulas for molecules is remarkable. To see how this is done Click Here. Formula Analysis. Although structural formulas are essential to the unique description of organic compounds, it is interesting and instructive to evaluate the information that may be obtained from a molecular formula alone. Three useful rules may be listed: The number of hydrogen atoms that can be bonded to a given number of carbon atoms is limited by the valence of carbon.
The origin of this formula is evident by considering a hydrocarbon made up of a chain of carbon atoms. Here the middle carbons will each have two hydrogens and the two end carbons have three hydrogens each. Thus, when even-valenced atoms such as carbon and oxygen are bonded together in any number and in any manner, the number of remaining unoccupied bonding sites must be even.
If these sites are occupied by univalent atoms such as H, F, Cl, etc. If the four carbon atoms form a ring, two hydrogens must be lost. Similarly, the introduction of a double bond entails the loss of two hydrogens, and a triple bond the loss of four hydrogens. By rule 2 m must be an even number, so if m The presence of one or more nitrogen atoms or halogen substituents requires a modified analysis. The above formula may be extended to such compounds by a few simple principles: The presence of oxygen does not alter the relationship.
All halogens present in the molecular formula must be replaced by hydrogen. Each nitrogen in the formula must be replaced by a CH moiety. However, the structures of some compounds and ions cannot be represented by a single formula. For clarity the two ambiguous bonds to oxygen are given different colors in these formulas.
If only one formula for sulfur dioxide was correct and accurate, then the double bond to oxygen would be shorter and stronger than the single bond. This averaging of electron distribution over two or more hypothetical contributing structures canonical forms to produce a hybrid electronic structure is called resonance.
0コメント