Determine which bond in following group will be most polar. C-H, Si-H, or Sn-H. And why?
546448
this polar nature depends on electronegativity. \nIn a bond, say C-H, the most EN element gets partial -ve charge and less EN element gets partial +ve charge\nIn C-H, H is less EN and C is more EN. Thus, C gets partial +ve charge and H gets partial -ve charge.\n\nEN of C is 2.550 EN of H is 2.200\nEN of Si is 1.9 EN of Sn is 1.96\n\nIn this group EN of C is high and H is less in C-H thus, C-H is most polar. \n\nAs EN of Si and Sn is less, than H , H gets positive charge but very less +ve charge. but in C-H C is more EN and gets +ve 2.550.\nHope i was clear enough.\nAny doubts you can ask further.
929337
a) Tetrahedral shape. bond angle : 109.5\nNo lone pair on central carbon\n\nb) Linear shape. Bond angle: 180\nNo lone pair on central Be\n\nc) Trigonal pyramdial shape. Bond angle less than 109.5\na lone pair on Cl. 2 oxgen double-bonded and 1 oxygen single-bonded to Cl. 0 formal charge on Cl and the O=Cl. Formal charge on Cl-O is -1.
554160
A molecule is polar when it has a dipole moment that does not cancel out. For example, consider the molecule CH4 (methane) which is nonpolar. It has a carbon atom bonded to 4 hydrogen atoms. Bonding is the overlap of the electron density. So you have an electron from a hydrogen and an electron from the carbon forming each of the bonds. Carbon is just very slightly more electronegative (tendency of pulling electrons closer to itself) than hydrogen. This means that the electrons are pulled by carbon away from hydrogen, but not by much (the difference is very very small). This shifting of electron density allows for the carbon to be slightly more negatively charged and hydrogen slightly electron deficient. This makes the C-H bond just slightly polar. However, because there are 4 hydrogen atoms that are bonded to the carbon, this bond polarity cancels each other out...meaning that the bonds are slightly polar, but the overall molecule is nonpolar. \n\nTo determine if the bonds are polar you just have to know the electronegativity differences. To know if a molecule is polar, you must know the correct geometry. Since electrons are negatively charged and repel each other, they want to be as far away from each other as possible. If you have a carbon bonded to 4 hydrogens, you will have a tetrahedral geometry with the hydrogens 109.5 degrees apart from each other creating the best possible arrangement. Also, if you have a lone pair of electrons, then the molecule will always be polar because of that charge separation.
843792
Be very careful:\nFirst, look at the electronegativity differences, the presence of groups like OH- or cations (like Na+ of soap)\nThen, you must know molecular geometries, it is highly important:\nFor BF3 this is a  trigonal planar compound, all differences of electronegativity B<F are canceled due to the position of the F atoms (every 120º on the same plane). It is not polar.\n[SnCl3]- this is a trigonal pyramid compound.\nCl\n I\n:Sn-Cl\n I\nCl\nAS you can see, Sn has a lone pair of electrons, making Cl atoms accommodate in the base corners of an imaginary pyramid. Even if Cl cancel each others dipolar momentum on the sides, they all point ↓. This is a polar compound.
734468
A molecule is said to be polar when the atoms are different in its electronegativity and its magnitude does not cancel off each other. Water itself is made up of Oxygen & Hydrogen atom. The electronegativity of o2 and h2 is the same. Thus, there are no dipole movement in the oxygen-hydrogen bond. However, the magnitude of the 2 hydrogen atom that is attached on the oxygen bond is in a 45 degrees manner. Thus, when the magnitude of the whole molecule is considered, the molecule is deemed to be polar.
838159
Sigma corresponds with "s" orbitals.  Pi corresponds with "p" orbitals.  We know that 2 electrons fill up the s orbitals.  We know that it takes 6 electrons to fill up the p orbitals.  Carbon has four bonding sites, so it could theoretically have four bonds.  That is 1 sigma bond and 3 pi bonds.  Bottom line is to remember that it corresponds with orbitals.  Hope this helps!
803112
the amino group of NH2-R, attacks the carbon of -N=C=S group using the lone pair on nitrogen. The amino group acts as nucleophile. \n        -                              +\n-N-C(NH2-R)=S\nThen the N of isothiocyanate will be protonated to give\n(both - and + are on N's)\n-NH-C(NH-R)=S\n\nHere the H of NH2 group is migrated on to the N of the isothiocyanate group\n\nu can go through the following link\n\nAll the best
819193
If you look at how the -CN is bonded [C≡N], you can see that the oxidation number of N is -3.  From that it will follow, as you note, that C is +2.
950410
Mainly H-bonding (intermolecular forces)
809957
The most polar bond will be between the atoms with the greatest difference in electronegativity.\n\nF2 has bond that is not polar since the electrons are equally shared between the two identical atoms. The electronegativity series is As<S<Br<F so the biggest difference is between As and F and the most polar bond is in AsF3.\n\nAs belongs to the same family as N. Thus in AsF3 it forms 3 single bonds and also has 1 lone pair. So the electron-group arrangement is tetrahedral but since there are only 3 identical bonds there will be a net dipole moment.\n\nI already explained why F2 has no dipole moment\n\nBrF3 has trigonal bipyramid electron group arrangment (3 bonds+2 lone pairs) but not corners are occupied by identical atoms (2 have lone pairs of Br) so the dipole moments do not cancel out and the molecule has a net dipole moment\n\nSF4 has again trigonal bipyramid electron group arrangement(4 bonds + 1 lone pair) with one lone pair at one of the corners so it has a net dipole moment\n\nSF6 has octahedral electron group arrangement (6 single bonds)\nand since all corners have F atoms the dipole moments of the bonds cancel each other and the molecule has no net dipole moment.
803448
This is an SN2 reaction because you have a secondary haloalkane and a polar aprotic solvent so inversion of conformation occurs. The answer is B
561562
polar.\n\nThe best way to think of it is in terms of water solubility.  Non polar organic molecules usually have lots of hydrogens, and usually no oxygens.
749482
1-C Hydroboration puts -OH on least substituted C; anti-Markownikov\n2-B Correct\n3-A Correct\n4-C Br goes on the center C; product has two identical CH3CH2- groups on center C, therefore symmetrical and achiral\n5-A Correct\n6-A Correct\n7-B Correct\n8-A Correct\n9-C Markownikov addition puts H on C next to C-CH3 and OH onto HO-C-CH3
423221
Ethane is nonpolar.  Ethanol has a polar hydroxyl group which can form hydrogen bonds with water.
956449
1.) ClO2, because the O=Cl=O angle is 90deg whereas O=S=O is linear, 180deg\n2.) HCl, because Cl is more electronegative than Br\n3.) SCl2, because the Cl-S-Cl angle is 90deg, whereas Cl-Be-Cl is linear, 180deg\n4.) AsF3, because As sits at the top of a tripod of F atoms, whereas AsF5 is symmetrical.
839410
This compound is an amine.  You can identify it by the NH2 (amine group) at the end.  You did a fair job with the structure considering the limits of the text here.  I am sure you realize that two of your carbons do not have the correct number of bonds.
789677
3 is incorrect.  Three of the carbon electrons are located in the three sp2 hybrid orbitals and the remaining electron sits in the lone p orbital (orthogonal/perpendicular) to the molecular plane.  It is this electron in the p orbital which "interacts" with the oxygen lone electron in it's unhybidized p orbital to form the pi bond.\n\nThere are no electrons left in the original carbon 2s orbital because these orbitals have been hybridized to the sp2 orbitals in order to allow for the three sigma bonds (one to each hydrogen and one to the oxygen atom)