(a) For the electron arrangements of the ions X³⁻ (2, 8) and Y²⁻ (2, 8, 8), we can find the corresponding neutral elements X and Y by removing the electrons that are added to form the ions.
- Ion X³⁻ has an electron arrangement of 2, 8. The neutral atom X originally had 18 electrons (2 + 8 + 8) but gained 3 electrons to form the ion; therefore, it would have had 15 electrons before gaining 3, giving it an electronic arrangement of 2, 8, 5.
- Ion Y²⁻ has an electron arrangement of 2, 8, 8. The neutral atom Y originally had 10 electrons (2 + 8) but gained 2 electrons to form the ion; therefore, it would have had 8 electrons before gaining 2, giving it an electron arrangement of 2, 6.
- Thus:
- Element X: 2, 8, 5 (Phosphorus, atomic number 15)
- Element Y: 2, 6 (Oxygen, atomic number 8)
(b) The compound formed between X (P³⁻) and Y (O²⁻) would be written as: \( X_2Y_3 \) or \( P_2O_3 \) (which corresponds to phosphorous oxide).
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Hydrogen can be placed in either Group I or Group VII of the periodic table due to its electronic configuration (1s¹). It has one electron in its outer shell (like alkali metals in Group I), which it can lose to form a positive ion (H⁺) and it can gain one more electron to complete its outer shell, like the halogens in Group VII, thereby forming a negative ion (H⁻). Its ability to either donate or accept an electron gives it the unique versatility to be aligned with either group.
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The diagram for the phosphorus ion \( ^{31}_{15}P^{3-} \) represents 15 protons and 16 neutrons in its nucleus (31 - 15 = 16). The electron arrangement for \( P^{3-} \) is 2, 8, 8.
Diagram (not drawn here but described):
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Nucleus:
- 15 Protons (represented by +)
- 16 Neutrons (represented by n)
\[ + + + + + + + + + + + + + + + + + + (15 protons) \] \[ n n n n n n n n n n n n n n n n (16 neutrons) \]
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Electron Shells:
- 1st shell (2 electrons)
- 2nd shell (8 electrons)
- 3rd shell (8 electrons)
(a) (i) The element which forms a double-charged cation is A, as it has 6 valence electrons (2, 8, 6) and will lose 6 electrons to become \( A^{6+} \).
(ii) The element which forms a soluble carbonate is C because it has 1 valence electron (2, 8, 1) and will form a soluble sodium or potassium carbonate.
(b) The element with the shortest atomic radius is K, as it is found in the same period as L and has the smallest number of electron shells.
(a) The electronic configuration of calcium (atomic number 20) is: \[ 1s^2 2s^2 2p^6 3s^2 3p^6 4s^2 \] For beryllium (atomic number 4): \[ 1s^2 2s^2 \]
(b) Calcium is more reactive than beryllium because it has more energy levels and a larger atomic radius, making it easier for the outermost electrons to be lost, facilitating chemical reactions.
(a) Elements K and L have similar properties. Both are metals but K has a lower atomic radius than L and has more metallic characteristics while they both have similar properties related to size and bonding.
(b) The most likely formula of the oxide of L (which has 7 valence electrons in its outer most shell) is \( L_2O_7 \) or \( L_2O \) since it would need to gain electrons to achieve stability.
(c) Element M (2.8.8.1) is a non-metal. It has more valence electrons that make it likely to gain electrons to complete its outer shell.
(d) The strongest reducing agent is element K, as it has the lowest electronegativity and is able to lose electrons more readily.
(e) The ionic radius of N is less than that of M as N has a full outer octet causing increased effective nuclear charge that pulls electrons closer, reducing radius.
(f) The ionic radius of L is bigger than its atomic radius because when L ionizes, it gains an electron, resulting in increased repulsion among the existing electrons, thus enlarging the ionic radius compared to the neutral atomic radius.
2 answers