Cu+ : 1s2 2s2 2p6 3s2 3p6 3d10 4s0 Cu2+ : 1s2 2s2 2p6 3s2 3p6 3d9 4s0 Cu2+ has an incomplete d subshell whereas Cu+ doesn't. The incomplete d subshell is allows transition metal ions to have colour. When a photon of light is absorbed by Cu2+, electrons in the lower energy levels of the 3d subshell jump up to the higher energy levels, and gain energy of a specific wavelength which corresponds to a colour. This colour is absorbed and therefore not seen. The colours that aren't absorbed are seen. The electrons of Cu+ can't jump up that energy level because the 3d subshell is full, there's no space to move up to. Therefore no energy is absorbed, all the colours are reflected and seen as white light. You don't need to know that, however, in the question they probably just want you to link the incomplete d subshell to the ability to form coloured complexes.
Can someone please explain if hydrogen bonds form between C=O groups, or just -OH groups? And also, if I have -COOH group, is the proton attached to the oxygen a singlet? Because it's not directly attached to the carbon, so won't be split?
If we consider an ester, hydrogen bonds can't form between molecules of the ester because although there is lone pairs on the oxygen atom there is no partially positive hydrogen to be attracted to them. Hydrogen bonds can form between an ester and water, however. Just remember that to have a hydrogen bond, there has to be nitrogen, oxygen, or fluorine in the molecule and a partially positive hydrogen atom. As for the second question, that's right
Cu+ : 1s2 2s2 2p6 3s2 3p6 3d10 4s0 Cu2+ : 1s2 2s2 2p6 3s2 3p6 3d9 4s0 Cu2+ has an incomplete d subshell whereas Cu+ doesn't. The incomplete d subshell is allows transition metal ions to have colour. When a photon of light is absorbed by Cu2+, electrons in the lower energy levels of the 3d subshell jump up to the higher energy levels, and gain energy of a specific wavelength which corresponds to a colour. This colour is absorbed and therefore not seen. The colours that aren't absorbed are seen. The electrons of Cu+ can't jump up that energy level because the 3d subshell is full, there's no space to move up to. Therefore no energy is absorbed, all the colours are reflected and seen as white light. You don't need to know that, however, in the question they probably just want you to link the incomplete d subshell to the ability to form coloured complexes.
Could someone also help with this sketching titration question. Not sure how to get the marks. Where would the equivalence point? Where would the end Ph be?
Could someone explain to me why benzene will only undergo substitution reactions but not addition reactions? I've read the explanation in the book and it didn't help
Could someone explain to me why benzene will only undergo substitution reactions but not addition reactions? I've read the explanation in the book and it didn't help
because benzene is more stable structure than if something were to be added, cause addition reactions it disrupts the delocalised pi electron, idk how correct this is though to be honest
Could someone also help with this sketching titration question. Not sure how to get the marks. Where would the equivalence point? Where would the end Ph be?
Hey, the equivalence point is the volume (of the solution being added) at which the number of moles of both solutions are the same.
To work out the end pH: it says that 0.1 moldm-3 of NaOH is added (obviously in excess) so the end pH will be the pH of the NaOH solution.
is the nitration of phenol the same as the nitration of benzene?
No but we do not need to know bout it. There's an application question in a past paper which gives the different conditions for the nitration of phenol and then you have to compare them and say why phenol does not need concentrated acids or a h2so4 catalyst and can occur at lower temperatures but we do not need to know this but somehow I do
Okay so we have added an equal volume of water, so let's say, with 1 dm3 of the acid, we added 1 dm3 of water. The total volume of the solution has doubled but the number of moles of acid remains the same
in qs like thse i neevr know if to for example: make all NH2 into NH3+ or just the ones who were involved in the bonds? if that makes sense same goes for zwitter ions do i make all cooh coo- and all nh2 nh3+, or with alkali or acid hydrolysis i just get confused. help!? i hope i made sense.
in qs like thse i neevr know if to for example: make all NH2 into NH3+ or just the ones who were involved in the bonds? if that makes sense same goes for zwitter ions do i make all cooh coo- and all nh2 nh3+, or with alkali or acid hydrolysis i just get confused. help!? i hope i made sense.
All amines (if they can, i.e primary or secondary amines) accept a H+ ion in acid hydrolysis (lone pair of e- on N atom) as there is an abundance of H+ ions, supplied by the acid
Similarly, all carboxylic acid groups become carboxylate ions and form a salt in alkali hydrolysis. So all COOHs turn into COO-Na+ with NaOH
Hey, the equivalence point is the volume (of the solution being added) at which the number of moles of both solutions are the same.
To work out the end pH: it says that 0.1 moldm-3 of NaOH is added (obviously in excess) so the end pH will be the pH of the NaOH solution.
Hint pH = 14 - pOH
Strong bases, i.e NaOH, dissociate fully in water
I learnt that the end pH would be different as the alkali will be diluted as it has been added to the acid. Don't really know though. To work it out I was told to make the volume equal to the acid plus the alkali added. And use the normal nber of moles. Sub this into moles=CV/1000 and solve for concentration which you then put into -log(ans) and then do 14 minus whatever you got.
Could someone also help with this sketching titration question. Not sure how to get the marks. Where would the equivalence point? Where would the end Ph be?
Hi What paper is this from if you don't mind me asking? Thanks
If we consider an ester, hydrogen bonds can't form between molecules of the ester because although there is lone pairs on the oxygen atom there is no partially positive hydrogen to be attracted to them. Hydrogen bonds can form between an ester and water, however. Just remember that to have a hydrogen bond, there has to be nitrogen, oxygen, or fluorine in the molecule and a partially positive hydrogen atom. As for the second question, that's right
I learnt that the end pH would be different as the alkali will be diluted as it has been added to the acid. Don't really know though. To work it out I was told to make the volume equal to the acid plus the alkali added. And use the normal nber of moles. Sub this into moles=CV/1000 and solve for concentration which you then put into -log(ans) and then do 14 minus whatever you got.
Oh yeah, you're right! Sorry about that it was a little too late for me I think