AP Chem

use simple structure and bonding models to account for each of the following:
A. the bond length between the two carbon atoms is shorter in C2h4 then in c2h6.

B.the bond lengths in SO3 are all identical and are shorter than a sufur oxygen single bond.

C. the N-O bonds in the NO2- ion are equal in length, whereas they are unequal in HNO2.

D. for sulfur, the fluorides SF2,SF4, and SF6, are known to exist, whereas for oxygen only OF3 is known to exist.

e. the I3- ion is linear.

PLEASE HELP!!

A. Ethane is H3C-CH3.
Ethene is H2C=CH2
The double bond is shorter than a single bond.

B. Draw the electron dot structure of SO3. You will find a S=O bond and two
S-O bonds. You can draw two more resonance structures with the double bond alternating (they don't actually move--these are just the structures we can draw). The actual structure, then, is a resonacne hybrid. The reason the bond lengths are the same is because of resonance (or you can say the hybrid structure is the true structure instead of any one of the resonance structures you drew above.)

Now, I suggest you reread your text and notes about resonance, hybrids, and electron dot strucutures, and try your hand at C, D, and E. We shall be happy to critique your thoughts.

use simple structure and bonding models to account for each of the following:
A. the bond length between the two carbon atoms is shorter in C2h4 then in c2h6.

B.the bond lengths in SO3 are all identical and are shorter than a sufur oxygen single bond.

C. the N-O bonds in the NO2- ion are equal in length, whereas they are unequal in HNO2.

D. for sulfur, the fluorides SF2,SF4, and SF6, are known to exist, whereas for oxygen only OF3 is known to exist.

e. the I3- ion is linear.

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  1. -o-n--c

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  2. c.) The NO2
    -1 ion exhibits resonance, giving bonds of equal length, but in HNO2, the bonding of
    the H atom fixes one oxygen atom’s bond to N as a single bond. The other oxygen atom is
    bonded to the nitrogen with a double bond. There is no resonance in HNO2.

    d) The central atom sulfur can form hybrid orbitals capable of forming equal bonds with fluorine
    in SF4 and SF6, which oxygen cannot because oxygen electrons on n=2 have no d orbitals
    available with which to hybridize. Recall that d orbitals start at n=3.

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