some chemistry help if possible

NiamhDunk

could anyone help me out with these questions....

just some notes aswell,

i can answer question 56, but not 57, so i just put in 56 because 57 follows on from it.

question 57, i understand that both carbons are sp2 hybridized, but dunno how to figure out what hybridization the nitrogen atom has.

then question 61, i know methane is sp3 hybridized, but dunno what hybrdization ammonium ion has.

then q65, i understand it all and know that the carbon atom is sp hybrid, but cant figure out what hybrid the oxygen atoms have.

thanks very much

Marvinthefish

Q57 NaOH is a well known base right? So acids (eg Q56) should react easily with it. Of course,the problem is how many of the Q56 acids react. They tell you that the pKa of water is 15.7 because sodium hydroxide dissociates to give OH(-) ions, the conjugate base of water. So all the acids on one side of 15.7 will react, the ones on the other side won't. (I know which is which, but I'm not telling:P)

Q60 (not Q57 again I presume) The N atom has a double bond to carbon, a single bond to C and a lone pair (count this as a group). If a C atom had a double bond, a single bond and a single bond (analogous to the N's lp) what hybridization would the C atom have? I think it would be the same. Not sure on this. Hybrid orbitals aren't my strong point.

Q61 Ammonia (NH3) looks like :. That is, its orbitals are really similar to methane (CH4), because the lone pair on nitrogen in ammonia is counted as a group (see above). When you have NH4(+), the shape of the molecule is the same and the hybridisation of N is the same.

Q65 (phew!) You're not asked about the hybridisation of oxygen in that Q. I hope it's not a case of...

Anyway, the oxygen atoms have a double bond to C and two lone pairs not involved in the double bond. There has to be one unhybridised p orbital on oxgen (in order to pi bond with C). This slide has a picture of the same sort of O atom. Note the sigma bond is made by overlappping lobes of the hybrid orbital and the pi bond is overlapping (unhybridised) p orbitals.

NiamhDunk

First off, your a legend thanks very much for your help.

ok the only one i still have trouble with is

Marvinthefish wrote: »

Q57 NaOH is a well known base right? So acids (eg Q56) should react easily with it. Of course,the problem is how many of the Q56 acids react. They tell you that the pKa of water is 15.7 because sodium hydroxide dissociates to give OH(-) ions, the conjugate base of water. So all the acids on one side of 15.7 will react, the ones on the other side won't. (I know which is which, but I'm not telling:P)

so any value greater than 15.7 is a weaker acid/stronger base and wont react with water..........any value smaller than 15.7 is a stronger acid/weaker base and hence will react, right? i understand all that.


the only part i have difficulty with is how sodium hydroxide dissociates to make OH-.....is it because the Arhennius definition of bases says that the majority of bases dissociate in water to form OH-? where does the Na go?


thanks again very much for your assistance

Marvinthefish wrote: »

Q60 (not Q57 again I presume) The N atom has a double bond to carbon, a single bond to C and a lone pair (count this as a group). If a C atom had a double bond, a single bond and a single bond (analogous to the N's lp) what hybridization would the C atom have? I think it would be the same. Not sure on this. Hybrid orbitals aren't my strong point.

oh and a carbon atom under these circumstances would be sp2 hybridized

Marvinthefish

NiamhDunk wrote: »

First off, your a legend thanks very much for your help.

You're welcome!

so any value greater than 15.7 is a weaker acid/stronger base and wont react with water

Any value greater than 15.7 won't react with hydroxide ion. The acids will be giving away a proton and, as long as hydroxide ion is around, water (H2O) will not accept a proton.

Right. Sodium hydroxide dissociates in water to give Na(+) and OH(-) ions. The Na(+) ions become solvated. That is, water molecules crowd around the positively charged sodium atom with the lone pairs on oxygen pointing towards it. This effectively leaves the OH(-) ion swimming about on its own looking for any acid to react with! (BTW if OH- meets a water molecule it might react with it, but it will just give another molecule of water and an OH- ion again!):

NiamhDunk

legend! thanks

SOL

Almost right about that pKa thing, infact it is a logarithmic measure of acidity, so it is a sliding scale hence the word almost in the question. So if a base and an acid differ by 1 unit of pka then 1 in 10 would be protonated... so really 1 in 2000 ish acetones will be deprotonated... leading to aldol reactions...