ALL CIE CHEMISTRY DOUBTS HERE !!

[Adzel]

yes please...

and take your time.

[Deadly_king]

Can someone plz answer these questions for me  ...

It would surely have been quicker if you specified your problems. I mean, you would not gain anything if we do everything for you.

Anyway I'll be starting by the last number. I guess we'll be meeting midway Garfield

[Adzel]

OK do Q3(all), Q4(all) and Q6(b) ONLY...

[nid404]

Question 4

a) The first ionization energy is the energy required to remove one electron from each atom in one mole of gaseous atoms of an element.

b)
                 
              kJ/mol       log₁₀I.E
1st I.En    801                 2.9
2nd I.E     2427               3.4
3rd I.E      3660               3.6
4th I.E     25026              4.4
5th I.E     32808              4.5

Boron has electronic configuration 1s²2s²2p¹

c)As you can see (graph attached), there is a general increasing trend of I.Es. It is easier to remove the further most electron and hence a low First I.E. Once the p electron is removed, shielding effect decreases and electrons are more closely held by the nucleus and hence an increase in the 2nd and 3rd I.E. Following,there are only two electrons in the first shell and they are closest to the nucleus. Hence a huge change between the 3rd and 4th I.E.

(since they haven't provided the marks, I have explained what I think you would ought to know)

d) i) Sulphur



    ii)Potassium

[nid404]

On question 3....in a while.

[Adzel]

Thanks ALOT!!!! 

[Deadly_king]

Question 6

I guess Ari must have already helped with part (a)

But anyway here's a link which might help:
http://www.s-cool.co.uk/alevel/chemistry/transition-metals/electronic-configuration.html

(b)
(i) The 4s orbital is at a lower energy level than 3d. Hence we fill the 4s orbital before the 3d one.

(ii) Chromium is a transition element and has only 4 electrons int its 3d sub-shell while 2 electrons are present in the 4s sub-shell. The latter suffer repulsion such that one of them moves into one of the empty sub-shells of the 3d orbital. This results in greater stability of the element.

(iii) Copper experiences greater stability with a filled 3d-orbital rather than a filled 4s orbital. Hence it contains only one electron in 4s.

[nid404]

Question 3

Since they don't tell you how much you need to answer I will just say whatever is required


a)The Atomic Spectra of an element are the lines which are present at characteristic wavelengths in the light emitted from an electric arc between electrodes of that element. Each line in the spectrum corresponds to the energy difference between two orbitals in the atom. According to the quantum theory, the electrons in an atom can only reside in specific orbitals, each of which has a fixed energy level, and one photon is emitted for each electronic transition between orbitals.

When an electron drops from an orbital of high energy to one of lower energy, a photon (i.e. a packet of light) is emitted corresponding to the energy difference between the orbitals.

b) If electrons possessed energy, it would be a continuous spectrum

The graph I'm not sure if it's plotted in terms of frequency or wavelength, so I cannot be of much help.

http://www.chemguide.co.uk/atoms/properties/hspectrum.html

Both possibilities are mentioned here....should help 

[moon]

pls help can sio2 react with NaOH? If yes pls write the equation. Thanx in advance.

[Freaked12]

SiO2 is acidic oxide, so yes it can.
SiO2+2NaOH->Na2SiO3+H2O

[Deadly_king]

Good job Garfield 

Rightly said Requiem

[Dania]

What happens when you add dilute sulfuric acid to a cyanohidrin? 

E.g. OHCH(CN)CH(CN)OH + H₂SO₄ = ?

[nid404]

dilute or conc?

[Dania]

Oops. Forgot to mention it. The question says dilute. I know concenrated would dehydrate it.

[Deadly_king]

What happens when you add dilute sulfuric acid to a cyanohidrin? 

E.g. OHCH(CN)CH(CN)OH + H₂SO₄ = ?

Hmm.......the CN groups will get converted to CO₂H

In other words the nitrile is oxidised to a carbolxylic group

Therefore resulting compound will be OHC(CO₂H)HC(CO₂H)HOH

Hope it helps