This chemistry question involves key chemical concepts and calculations. The detailed solution below walks through each step, from identifying the reaction type to computing the final answer.
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Here are the solutions to your chemistry questions: 14. Beryllium (Z=4) a) The Beryllium atom (Z=4) has 4 electrons. Step 1: Fill the lowest energy orbital, 1s, with 2 electrons. Step 2: Fill the next lowest energy orbital, 2s, with the remaining 2 electrons. The s, p configuration for the Beryllium atom is: 1s^2 2s^2 b) Beryllium forms a Be^2+ ion by losing 2 electrons. Step 1: The neutral Beryllium atom has 4 electrons (1s^2 2s^2). Step 2: The Be^2+ ion has 4 - 2 = 2 electrons. Step 3: The electron configuration of Be^2+ is 1s^2. Step 4: The Noble Gas with 2 electrons and an electron configuration of 1s^2 is Helium. The Noble Gas with the same electronic configuration as Be^2+ is: Helium (He) c) Beryllium prefers to lose 2 electrons rather than gain 6 to fill its outer shell because losing 2 electrons requires significantly less energy. By losing its two valence electrons from the 2s orbital, Beryllium achieves a stable electron configuration identical to that of the Noble Gas Helium (1s^2). Gaining 6 electrons would be energetically unfavorable due to the large energy input required and increased electron-electron repulsion. 15. Aluminum Phosphate (AlPO_4) a) Aluminum (Al) has an atomic number Z=13. It typically forms an Al^3+ ion by losing its 3 valence electrons. Step 1: A neutral Aluminum atom has 13 protons and 13 electrons. Step 2: The Al^3+ ion has 13 protons (in the nucleus) and 13 - 3 = 10 electrons. Step 3: These 10 electrons are arranged in shells. The first shell (K-shell) holds 2 electrons, and the second shell (L-shell) holds the remaining 8 electrons. The structure for the Aluminum ion (Al^3+) can be described as: A nucleus containing 13 protons*. An inner electron shell (K-shell) containing 2 electrons*. An outer electron shell (L-shell) containing 8 electrons*. (Note: Neutrons are not included as the mass number is not provided.) b) The formula for Aluminum Phosphate is AlPO_4 and not Al_3(PO_4)_3 due to the principle of charge balance and the convention of writing ionic formulas in their simplest whole-number ratio. Step 1: Aluminum (Al) is in Group 13 and typically forms an ion with a charge of +3, so it is Al^3+. Step 2: The Phosphate radical (PO_4) has a charge of -3, so it is PO_4^3-. Step 3: For a neutral ionic compound, the total positive charge must balance the total negative charge. (Charge of Al^3+) × (Number of Al^3+ ions) + (Charge of PO_4^3-) × (Number of PO_4^3- ions) = 0 (+3) × (1) + (-3) × (1) = 0 Step 4: A 1:1 ratio of Al^3+ ions to PO_4^3- ions results in a neutral compound. Step 5: Ionic compound formulas are always written using the simplest whole-number ratio of the ions. Therefore, one Al^3+ ion combines with one PO_4^3- ion to form AlPO_4. While Al_3(PO_4)_3 represents the same 1:1 ratio, it is not the simplest formula and thus not the correct chemical formula. The formula is AlPO_4 because one Al^3+ ion balances the charge of one PO_4^3- ion, and the formula represents the simplest whole-number ratio of the ions.