Hydroxyl, ether, carbonyl, or amine?
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The allowable combination for the atomic orbital is n=3, l=1 or 2, =+1.
Three quantum numbers specify an atomic orbital:
- The principal quantum number, n, which is a positive integer, describes the relative size of the orbital and its distance from the nucleus.
- l is the angular momentum quantum number that is related to the shape of the orbital; l is an integer from 0 to n-1 (so n limits l ),
- is the magnetic quantum number that prescribes the three-dimensional shape of the orbital around the nucleus;
values are integers from -l to =l(l limits ml)
For n = 3, l can have three values: 0, 1, and 2. Hence, the l value must be lower than 3. Because of = +1, the l value must be higher than 0. Two l values are consistent with n and
values:
l= 1 or 2
Therefore, the allowable condition is n=7, l=1 or 2, =+3.
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For this reaction to proceed, the following bond breaking should occur:
*one C-H bond
* one Cl-Cl bond
After, the following bond formations should occur:
*one C-Cl bond
*one H-Cl bond
Now, add the bond energies for the respective bond energies which can be found in the attached picture. For bond formations, energy is negative. For bond breaking, energy is positive.
ΔHrxn = (1)(413) + (1)(242) + 1(-328) + 1(-431) = <em>-104 kJ</em>
*one C-H bond
* one Cl-Cl bond
After, the following bond formations should occur:
*one C-Cl bond
*one H-Cl bond
Now, add the bond energies for the respective bond energies which can be found in the attached picture. For bond formations, energy is negative. For bond breaking, energy is positive.
ΔHrxn = (1)(413) + (1)(242) + 1(-328) + 1(-431) = <em>-104 kJ</em>