Chemistry: Post your doubts here!

[Pratik.Dhakal]

Mine was great! It was really short and quite simple

oh really ? i took a lot of time to draw the graphs !!!

 

[smartangel]

Electronic configuration of Cu is 1s2 2s2 2p6 3s2 3p6 3d10 4s1... its d orbital is completely full so how is it a transition element??

 

[Aarjit]

Electronic configuration of Cu is 1s2 2s2 2p6 3s2 3p6 3d10 4s1... its d orbital is completely full so how is it a transition element??

It's the transition metal ION which needs to have a partially filled d-orbital, not the element in its ground state. Remembering the following definition will help:

"A transition element is an element that forms at least one stable ion with a partially filled d-orbital." [Cu forms Cu+(aq) and Cu2+(aq) ions, Cu2+(stable) having partially filled d-orbitals]

 

[smartangel]

It's the transition metal ION which needs to have a partially filled d-orbital, not the element itself. Remembering the following definition will help:

"A transition element is an element that forms at least one ion with a partially filled d-orbital." [Cu forms Cu+(aq) and Cu2+(aq) ions, both having partially filled d-orbitals]

thanks..could you please explain the spliiting of degenerate orbitals.. as in why it happens.. And why d x2-y2 and dz2 move to a higher energy level..

 

[Aarjit]

thanks..could you please explain the spliiting of degenerate orbitals.. as in why it happens.. And why d x2-y2 and dz2 move to a higher energy level..

The degenerate d-orbitals split because of the existent repulsion between the electrons. This helps to form transition-metal-complexes. While forming a complex with a ligand the d-orbitals interact with the surrounding electron cloud, causing it to become non-degenerate (variable energies). This causes photons to be adsorbed by the electrons (which then undergo state transitions) to emit radiations with corresponding frequencies (f) to the energy (E) of the photons [where E=hf].

P.S. Electrons in d-orbitals are repelled from the electrons in the ligands, based on electrostatic interactions. This causes two of the d-orbitals (dz2 & dx2-y2) [which have a stronger (repulsive) interaction with the ligand] to be at higher energy than the other three (dxy, dyz, dxz).

 

[smartangel]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_4.pdf
Q4 c.i..why is the ans for solution D blue..why not violet?

 

[Aarjit]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_4.pdf
Q4 c.i..why is the ans for solution D blue..why not violet?

(i) The absorption peak (D) is ~ 650 nm which corresponds to the color (emitted) having a wavelength of 450 nm, i.e. blue. If the peak were to be shifted towards 700 nm, you would see a much darker (violet) solution.

(ii) Apply E = h.f =h. (c/λ) [Since E is inversely proportional to λ, the shorter wavelength (450 nm) of C will have a larger ΔE]

 

[arkif rizan]

pls can u tell what is the disappearance of NMR peaks on the addition of D2O means in the NMR spectrum??

 

[arkif rizan]

and the reason why lines in an NMR spectrum can be split into doublets, triplets, quartets and
multiplets ??

 

[Aarjit]

and the reason why lines in an NMR spectrum can be split into doublets, triplets, quartets and
multiplets ??

The splitting of the absorption peaks (n) depends upon the # adjacent hydrogen atoms (h). n = (h +1) peaks

 

[smartangel]

(i) The absorption peak (D) is ~ 650 nm which corresponds to the color (emitted) having a wavelength of 450 nm, i.e. blue. If the peak were to be shifted towards 700 nm, you would see a much darker (violet) solution.

(ii) Apply E = h.f =h. (c/λ) [Since E is inversely proportional to λ, the shorter wavelength (450 nm) of C will have a larger ΔE]

i dont get part i..

 

[Aarjit]

pls can u tell what is the disappearance of NMR peaks on the addition of D2O means in the NMR spectrum??

On adding heavy water (D2O), peaks (Eg. -OH) disappear because the deuterium nucleus has a nuclear spin of +/- 1 rather than +/- .5, so it does not absorb the same frequency range as H1.

See how the -OH is removed from ethanol after adding heavy water:

CH3CH2OH + D2O <---> CH3CH2OD + HOD

 

[Aarjit]

i dont get part i..

Try to understand that the wavelength of the light absorbed is not equal to the wavelength of the light emitted. [i.e. if yellow light (600 nm) is absorbed, the same yellow light is not emitted! A light with the corresponding wavelength in the ABSORPTION SPECTRUM ~ blue is emitted]

Here, this will clear your dubiety: read the third paragraph;

http://www.dartmouth.edu/~chemlab/chem6/dyes/full_text/chemistry.html

 

[arkif rizan]

On adding heavy water (D2O), peaks (Eg. -OH) disappear because the deuterium nucleus has a nuclear spin of +/- 1 rather than +/- .5, so it does not absorb the same frequency range as H1.

See how the -OH is removed from ethanol after adding heavy water:

CH3CH2OH + D2O <---> CH3CH2OD + HOD

thnx mate

 

[smartangel]

http://www.xtremepapers.com/papers/... AS Level/Chemistry (9701)/9701_w11_qp_41.pdf
Q2a..how is it 1st order with respect to H+..im getting zero :s

 

[smartangel]

Try to understand that the wavelength of the light absorbed is not equal to the wavelength of the light emitted. [i.e. if yellow light (600 nm) is absorbed, the same yellow light is not emitted! A light with the corresponding wavelength in the ABSORPTION SPECTRUM ~ blue is emitted]

Here, this will clear your dubiety: read the third paragraph;

http://www.dartmouth.edu/~chemlab/chem6/dyes/full_text/chemistry.html

hmm..so if talk in terms of the pastpaer question.. C has an absorption peak at 450nm aprrox. frm the table we get it as blue. but the answer is red!

 

[Aarjit]

hmm..so if talk in terms of the pastpaer question.. C has an absorption peak at 450nm aprrox. frm the table we get it as blue. but the answer is red!

Exactly!

C's absorption peak is ~ 450 - 500 nm which corresponds to RED*.

*You need to infer this from the emission spectrum, not from the table. It's best if you memorize the colors and their corresponding frequencies.

 

[Pratik.Dhakal]

Electronic configuration of Cu is 1s2 2s2 2p6 3s2 3p6 3d10 4s1... its d orbital is completely full so how is it a transition element??

To be a transition element it should form at least form an ion with partial filled d-orbital and now if copper loses one electron then it will have a complete d-orbital which means it won't be a transition metal but we must remember that copper is stable in +2 oxidation state so it can be called a transition metal !!!

 

[Pratik.Dhakal]

http://www.xtremepapers.com/papers/CIE/Cambridge International A and AS Level/Chemistry (9701)/9701_s07_qp_4.pdf
Q4 c.i..why is the ans for solution D blue..why not violet?

the question has mentioned you (i) A list of possible colours for these complexes is as follows.


yellow red green blue

so its obviously blue because the question has itself stated that the complex can form colors only mentioned above !!!

 

[smartangel]

Exactly!

C's absorption peak is ~ 450 - 500 nm which corresponds to RED*.

*You need to infer this from the emission spectrum, not from the table. It's best if you memorize the colors and their corresponding frequencies.

OH. so they want us to memorize the spectrum..that sounds weird. but anyway thanks