(a)(i) Define the first ionization energy of an element (ii) Consider the following table and...
(a)(i) Define the first ionization energy of an element
(ii) Consider the following table and use it to answer te question that follows
| Element | Li | Be | b | C | N | O | F | Ne |
| Atomic number | 3 | 4 | 5 | 6 | 7 | 8 | 9 | 10 |
| 1st I.E/kj mol\(^{-1}\) | 520 | 900 | 801 | 1086 | 1402 | 1314 | 1681 | 2081 |
Explain briefly why the first ionization energy ofBis less than that ofBedespite the fact that the atomic number ofBis greater than that ofBe.
(b) When Titanium chloride was electrolysed by passing 0.12Acurrent through the solution for 500 seconds, 0.015 g of titanium was deposited. What is the charge on the titanium ion?
[ IF= 96500 C, Ti= 48.0 ]
(c)(i) Aluminium can be obtained by the application of electrolysis. State the electrolyte which yields aluminium on electrolysis.
(ii) Name two major factors which would favour the siting of an aluminium smelter in a country.
(d)(i) Define the term paramagnetism.
(ii) Consider the following ions: 24\(^{Cr ^{2+}}\), 24\(^{Cr^{6+}}\)
(I) Deduce the number of unpaired electrons in each of the ions.
(II) State which of the ions will have a greater power of paramagnetism
(l) Give a reason for the answer stated in (d)(ii)(II)
Explanation
(a)(i) Is the energy required to remove one mole of an electron from a gaseous atom.
OR
Is the minimum energy required to convert one mole of a gas of atom into one mole of a gaseous plus 1 ions
(ii) \(_{4}\)Be: Is\(^{2}\) 2s\(^{2}\)
\(_{5}\)B :Is\(^{2}\) 2s\(^{2}\) 2p\(^{1}\)
The electron to be removed from Be is in the 2s orbital which is closer to the nucleus, nuclear attraction is greater hence first ionization energy is greater but the electron to be removed from B is in the 2p orbital which is farther away from the nucleus, hence nuclear attraction is weaker hence first ionization energy is smaller.
OR
2p has less energy than 2s
So it is easier to remove an electron from 2p as the nuclear charge is weaker in 2p because 2p is further away from the nucleus. Hence, first ionization energy is small.
(b) Q - It
Q = 0.12 x 500
= 60 C
96500 = 1F
= 60 C liberates 0.015g
\(\therefore\) 96500 will liberate \(\frac{96500}{60}\) x 0.015
= 24 g
n((T\(_{1}\)) = \(\frac{m}{M}\)
= \(\frac{24}{48}\) = 0.5mol
1F = 0.5mol
2F = 1mol
hence charge = +2
Alternative A
Q = lt
= 0.12 x 500
= 60 C
0.015 g of Ti is deposited by 60 C .
\(\therefore\) 48 g will be deposited by = \(\frac{60}{0.015}\) x 45
= 1920000
96500 C\(\equiv\) i mole of electron
\(\therefore\) 192000 C \(\frac{1}{96500}\) x 192
= 1.989
\(\approx\) 2.0 moles of electron
\(\therefore\) charge on Ti ion is +2
Alternate B
\(\frac{m}{M} = \frac{It}{nF}\)
n = \(\frac{Mlt}{mF}\)
= \(\frac{48 \times 0.12 \times 500}{0.015 \times 96500}\)
= 1.989 \(\equiv\) 2
\(\therefore\) Change on Ti ion is +2
(c)(i) Alumina (Al\(_{2}\)\(_{3}\)) mixed with molten cryolite (Na\(_{3}\)AIF\(_{6}\))
(i) - abundant deposits of bauxite
- cheap source of electricity
(d)(i) Is the ability of a substance to be attracted strongly into a magnetic field.
(ii) I. \(_{24}\)Cr\(^{2+}\) - Is\(^{2}\) 2P\(^{6}\) 3S\(^{2}\) 3P\(^{6}\) 3d\(^{4}\)
unpaired electrons = 4
\(_{24}\)Cr\(^{6}\) - Is\(^{2}\) 2s\(^{2}\) 2p\(^{6}\) 3s\(^{2}\) 3d\(^{6}\)
unpaired electrons = 0
(II) \(_{24}\)Cr\(^{2+}\)
III. The greater the number of unpaired electrons in the 3d - orbital, the| greater-the paramagnetism.