why a relatively larger amount of energy is required to break the bond symmetrically than assymetrically?
829392
just a piont of view, but normally to divide something symmetrically you have to do it at the widest piont along the centre line. where as assymetricaly is off the centre line at a narrower piont and therefore not a symmetry,this only applies to poloygons and symmetric shapes.  hope this helps
138726
Mg more reactive than Al cuz it has lesser valence electrons to release,making it easier to achieve stable noble gas configuration
950381
It takes energy not only to break up the solid solute into its individual particles, but also to force the solvent molecules apart.  Now, if the solute can offer the solvent some very attractive forces (like sodium ions and chloride ions do from table salt) the whole process will work. The energy required to break up chalk is just too great.
577456
basically all things tend to have minimum potential energy. this can be achieved when the charges stay as far away as possible from each other. \n\nhence when charges are given to a conductor the are distributed uniformly,as in the case of a symmetrical conductor, on the outer surface of the conductor to minimise potential energy. \n\nif the conductor is asymmetrical, then the charge distribution is inversely proportional to the radius of curvature.
901447
Note that we're not taking energy differences into account.  In cases where the difference in orbital energy is greater than 15eV you get negligible overlap.  Now onto the symmetry-allowed overlaps.\n\n1s and 2s are both spherical and overlap to form a sigma bond.\n\n1s and 2p(x) form a sigma bond.  You can tell because if you draw the 2p(x) lobe you'll see it overlap with the corresponding 1s.\n\nThe 2p(x) and the 2p(y) do not overlap.  There is no possible bond.  Draw the 2p(x) extending toward the other atom as positive or shaded (remember: positive does not denote charge).  The other atom has its 2p(y) pointing up and down (perpendicular) and there is equal positive and negative overlap, thus canceling both of them out.  No net overlap.\n\n3p(y)-3p(y) form a pi bond.  Once again, draw these out and see it for yourself.  Draw the 3p(y) orbitals as rather fat so you can see how they overlap.  Since they are not cylindrically symmetric they are not a sigma bond.  Pretend you were to rotate one of the atoms about the x-axis... you would change orientation and there'd be less overlap.  For the sigma bonds above, rotation does not affect the overlap.\n\n2p(x)-2p(x) have positive overlap.  Draw them and see why.  They point toward one another.  Moreover, if you rotate one of the 2p(x) orbitals about the x-axis you keep the same structure/orientation since the 2p(x) is along the x-axis.  Thus, you can tell that this gives a sigma bond.
540888
My understanding is that the shear stretches the molecules out, allowing more interactive sites to be exposed.  In a linear form, the molecule becomes more able to participate in bonding with its neighbor.
184098
has to do with intermolecular bonds. The bonds that hold a solid together are more ridgid than that of a liquid.
1137289
Covalent bonds with a bond strength of 50-150 kcal/mol are a lot stronger than ionic bonds (less than 10 kcal/mol)
933000
I'm vaguely guessing here but it has something to do with the law of conservation in thermodynamics. Matter (or atoms) cannot be created or destroyed hence the number of atoms on the left hand side must equal the no. of atoms on the right hand side of the equation. Hope that helps!
378789
From my understanding, it should not be, unless subjected to stress, such as bending, which forces molecular orientation.
943097
I think it's false. If the ionisation energy is increased, it's more difficult achieve ionisation, therefore less likely to be reactive. This is why those at the top of the group are more reactive.
423186
Its a smaller molecule, with weaker bonds, so is broken down more quickly supplying the sportsmen with quick energy supply! I think
296542
when a bond is formed, some electrons don't participate in that because of hybrid levels, thats why there are some electrons left known as lone pair, they are never single, they are a pair of opposite spin.
15546
Interesting scenario first of all i would like to comment. Well actually the bond is due to electronic forces. Hence mass has litle effect on the bond strength. Thus i thikn the effect of mass change should be a small one!! unless you change the charge of proton too!!
893177
Its all about mobility.  \n\nHow does the energy (in the form of heat) get transported over the structure?  Lattice vibrations, right?\n\nThe bonds have to deform in order to vibrate, and the higher electronegativity, associated with a higher bond strength, is more resistant to the imposed deformations, and therefore to transporting the heat.\n\nExcept for radiant transport (and thats a factor of T^4) the best conductive insulator or convective insulator is... a vacuum.  No mass means no bucket in which to ship the energy.
4764
You can't balance it because you don't know what it yields.\n\n\nP.S. This needs to be in the chemistry category. You really must not know much about chemistry and physics.
784817
actually it is not really double bonds.these are between single & double bonds.even i will say \nthat it is a cloud of elecrons present over these\nbonds.this elecron cloud is not localised and such type of bonding is called delocalised\nbonding.  and phenomenon is said to be\nresonance.