how many pie-bond we find in an alkane ?
455047
r u really serious !!! how can one find a pie bond in an alkane????
1032656
C4H10\n\nCH3-CH2-CH2-CH3
562353
pure butane is neutral for sure. And butane, no matter isobutane or just butane, is alkane. Alkane does not dissolve in water since no ionic bond exists in the compound, so measuring its pH does not make much sense.\n\nCarbon and hydrogen have very similar electronegativity, so it's very difficult to remove hydrogen from the carbon chain. It's pka value is around 50.
1112022
An alkene, or olefine, in organic chemistry is an unsaturated hydrocarbon containing at least one carbon to carbon double bond. The simplest alkenes, with only one double bond, form a homologous series with general formula CnH2n.\n\nThe simplest alkene is ethylene (C2H4), which has the International Union of Pure and Applied Chemistry (IUPAC) name ethene. Alkenes are also called olefins (an archaic synonym, widely used in the petrochemical industry) or vinyl compounds.\n\nShape of alkenes\n\nAs predicted by the VSEPR model of electron pair replusion, the molecular geometry of alkenes includes bond angles about each carbon in a double bond of about 120°. The angle may vary because of steric strain introduced by nonbonded interactions created by functional groups attached to the carbons of the double bond. For example, the C-C-C bond angle in propylene is 123.9°. The alkene double bond is stronger than a single covalent bond and also shorter with an average bond length of 133 picometres.\n\n\nMolecular geometry\n\nLike single covalent bonds, double bonds can be described in terms of overlapping atomic orbitals, except that unlike a single bond (which consists of a single sigma bond), a carbon-carbon double bond consists of one sigma bond and one pi bond.\n\nEach carbon of the double bond uses its three sp2 hybrid orbitals to form sigma bonds to three atoms. The unhybridized 2p atomic orbitals, which lie perpendicular to the plane created by the axes of the three sp2 hybrid orbitals, combine to form the pi bond.\ncis-2-Butylene\nEnlarge\ncis-2-Butylene\ntrans-2-Butylene\nEnlarge\ntrans-2-Butylene\n\nBecause it requires a large amount of energy to break a pi bond (264 kJ/mol in ethylene), rotation about the carbon-carbon double bond is very difficult and therefore severely restricted. As a consequence substituted alkenes may exist as one of two isomers called a cis isomer and a trans isomer. For example, in cis-2-butylene the two methyl substituents face the same side of the double bond and in trans-2-butylene they face the opposite side.\n\nIt is certainly not impossible to twist a double bond. In fact, a 90° twist requires an energy approximately equal to half the strength of a pi bond. The misalignment of the p orbitals is less than expected because pyridalization takes place. trans-Cyclooctene is a stable strained alkene and the orbital misalignment is only 19° with a dihedral angle of 137° (normal 120°) and a degree of pyramidalization of 18°. This explains the dipole moment of 0.8 D for this compound (cis-isomer 0.4 D) where a value of zero is expected.[1] The trans isomer of cycloheptene is only stable at low temperatures.
423134
CH4 does not have any non-bonding electron pairs.  It has 4 sp3 hybrid orbitals which are all involved in bonding.  SF2 has 4 electron domains according to it's Lewis structure.  This makes the hybridization sp3.  two of these hybrid orbitals are involved with bonding while the other two contain the non-bonding pairs of electrons.  To find the non-bonding pairs you must draw the correct Lewis structure.  a single atom in the molecule will only have a single hybridization.  S in SF2 can not be sp2 and sp3 hybrid.  it is only sp3.  Therefore the non-bonding electrons exist in the sp3 hybrid orbital.
712605
#10.\n\n3,4 - dimethylhept - 3 - ene\n\nI cant really draw the other ones.\nJust look up your reaction mechanisms.\n\nGood Luck!
855406
alkanes are sp3 hybridized and tetrahedral. (there are single bonds between the carbons)\nalkenes are sp2 hybridized and triangular planar. (there is at least one double bond in the carbon chain)\nalkynes are sp hybridized and linear (there is at least one triple bond)\n\nwhether or not they are polar depends on their substituents. if they only have hydrogen atoms attached they will be nonpolar.
318950
Geometrical isomerism (Cis and trans)is not shown on the compound you ask for, if you ask about an alkene(carbon,carbon double bond ) i.e 1,4-dichlorohexene the sturcure is (Cl)CH2-CH2=CH2-CH2(Cl)CH2CH3.
1032046
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!!!
764122
The first bond between two atoms is always a sigma bond.  Additional bonds between the same atoms are always pi bonds.  So carbon dioxide has one sigma bond and one pi bond between both the carbon and each oxygen.  So in total there are two sigma bonds and two pi bonds.
1154659
there are 2 pi  & 1 sigma bond in cn & 1 pi & 1 sigma bond in co
950288
1)  Only the valence orbitals matter, and all the valence orbitals count (more complicated version: all valence orbitals which are of the correct symmetry to hybridize to form MO.  If you don't know what that means then you're in a lower level class and it doesn't matter).  # of atomic orbitals = # of molecular orbitals, and the molecular orbitals may be bonding, nonbonding, or antibonding.  Not depending on the particular class you're taking your teacher may or may not want you to draw the non-valence orbitals, but they don't matter at all.\n\n2)  You start filling all the MOs with electrons, starting with the lowest energy and moving up.\n\n3)  You wouldn't determine that for C2 because that's not correct.  The bonding and antibonding orbitals formed from the 2s atomic orbitals would both be filled (net bond order of zero) the lowest bonding (sigma) MO from the 2p AO would be filled, and each of the higher bonding (pi) MO from the 2p AO would have one electron.  None of the antibonding orbitals from the 2p AO would be filled.\n\nThis predicts that C2 would have 2 unpaired electrons, whereas a Lewis structure would predict a quadruple bond between the carbons.  One guess as to which one is right....
1068778
pure butane is neutral for sure. And butane, no matter isobutane(2-methylpropane) or just butane, is an alkane. \n\npH = -log[concentration of hydronium ions] (hydronium is H3O+), but Alkane does not dissolve in water since no ionic bond exists in the compound, so measuring its pH does not make much sense.\n\nCarbon and hydrogen have very similar electronegativity, so it's very difficult to remove hydrogen from the saturate carbon chain. alkane has pka value around 50, pKa value for water is around 16.
793123
All alkanes exist as two or more isomers is not true\n\nbecause methane, ethane and propane are also alkane, but they have no isomers.
546716
The level of unsaturation\n\nAlkanes = all C-C single bonds\nAlkenes = at least one C=C double bond\nAlkynes = at least one C-C triple bond.
542309
CnHnx2\n\nFor instance, the simplest alkene is ethene, or C2H4.
273309
wat r u ppl talking about....u ppl r absolutely wrong.....its alkynes n i am 100% sure about\n\nthey r more reactive becoz of there triple bonds...\n\n\n2 ppl in a row a row giving a wrong answer can confuse a person completely