Write the electron configuration for Mg2+ and Cl- ions and the valency of the hypochlorite ion in Ca(ClO)2.

Here are the solutions to the questions: Question 11: a) Write the resulting configuration of an Mg^2+ ion. Magnesium (Mg) has an atomic number of 12. A neutral Mg atom has 12 electrons. The electron configuration for neutral Mg is 1s^2 2s^2 2p^6 3s^2. When Mg forms an Mg^2+ ion, it loses 2 electrons from its outermost shell (3s^2). The resulting configuration for Mg^2+ is: 1s^2 2s^2 2p^6 b) Identify the valency of the hypochlorite anion (ClO^-) in this compound. The compound is Calcium Hypochlorite (Ca(ClO)_2). Calcium (Ca) is in Group 2 and forms a Ca^2+ ion. Since the compound is electrically neutral, the total negative charge from the hypochlorite anions must balance the +2 charge from the calcium ion. There are two ClO anions for every one Ca^2+ ion. Let the charge of the hypochlorite anion be x. (+2) + 2(x) = 0 2x = -2 x = -1 The valency of the hypochlorite anion (ClO^-) is: 1 c) If calcium (Z=20) reacts with chlorine gas (Cl_2), it forms calcium chloride (CaCl_2). i) Write the s,p configuration for the Cl^- ion (Z=17). Chlorine (Cl) has an atomic number of 17. A neutral Cl atom has 17 electrons. The electron configuration for neutral Cl is 1s^2 2s^2 2p^6 3s^2 3p^5. When Cl forms a Cl^- ion, it gains 1 electron to complete its outermost p-orbital. The s,p configuration for the Cl^- ion is: 1s^2 2s^2 2p^6 3s^2 3p^6 ii) Write a balanced equation and include state symbols. Calcium (Ca) is a solid metal. Chlorine gas (Cl_2) is a gas. Calcium chloride (CaCl_2) is an ionic solid. The balanced chemical equation is: Ca(s) + Cl_2(g) CaCl_2(s) Question 12: A chemist is synthesizing a drug that requires Ammonium Carbonate. a) Write the formula for the Ammonium radical and the Carbonate radical including their charges. • Ammonium radical: NH_4^+ • Carbonate radical: CO_3^2- b) Write out the final chemical formula for Ammonium Carbonate. To form a neutral compound, two ammonium ions (NH_4^+) are needed to balance the charge of one carbonate ion (CO_3^2-). The final chemical formula for Ammonium Carbonate is: (NH_4)_2CO_3 Question 13: A forensic chemist in Nairobi is analyzing a gold bar suspected to be a fake alloy. The bar contains a rare element Gallium (Ga) which has 2 isotopes: ^69Ga and ^71Ga. The reported relative atomic mass (Ar) of the certificate is 69.72. a) If the abundance of ^69Ga is approximately 60%, write an equation for the abundance of ^71Ga and calculate the exact percentage abundance of the isotopes. Let x be the percentage abundance of ^69Ga and y be the percentage abundance of ^71Ga. The sum of the abundances must be 100%: x + y = 100 The equation for the abundance of ^71Ga is: y = 100 - x The relative atomic mass (Ar) is calculated using the formula: Ar = (mass of isotope 1 × abundance 1) + (mass of isotope 2 × abundance 2)100 Substitute the given values: 69.72 = ((69 × x) + (71 × y))/(100) Substitute y = 100 - x into the equation: 69.72 = ((69x) + (71(100 - x)))/(100) Multiply both sides by 100: 6972 = 69x + 7100 - 71x 6972 = 7100 - 2x Rearrange to solve for x: 2x = 7100 - 6972 2x = 128 x = (128)/(2) x = 64\% Now calculate y: y = 100 - x = 100 - 64 y = 36\% b) Calculate the exact percentage abundance of the isotopes. The exact percentage abundance of the isotopes are: • ^69Ga: 64\% • ^71Ga: 36\% Question 14: a) Write the s,p configuration for the Beryllium atom (1s^2 2s^2). Beryllium (Be) has an atomic number of 4. The s,p configuration for the Beryllium atom is: 1s^2 2s^2 b) Beryllium often reacts to form a Be^2+ ion which always has the same electronic configuration as the Helium atom. c) Why does Beryllium prefer to lose 2 electrons rather than gain 6 electrons to fill its outer shell? Beryllium prefers to lose 2 electrons because it has 2 valence electrons in its outermost 2s orbital. Losing these 2 electrons results in a stable electron configuration identical to that of Helium (1s^2), which is a noble gas. This process requires significantly less energy than gaining 6 electrons to fill its 2p orbital and achieve a Neon-like configuration (1s^2 2s^2 2p^6). Atoms tend to react in ways that achieve stability with the minimum energy input. Question 15: A new solid state battery uses Aluminum Phosphate (AlPO_4) as an electrolyte. a) Draw the structure of Aluminum ion (Z=13). Aluminum (Al) has an atomic number of 13. A neutral Al atom has 13 electrons. Its electron configuration is 1s^2 2s^2 2p^6 3s^2 3p^1. When Aluminum forms an ion, it typically loses its 3 valence electrons to become Al^3+. The electron configuration for Al^3+ is 1s^2 2s^2 2p^6. This means the Al^3+ ion has 2 electrons in the first shell and 8 electrons in the second shell. Here is a Bohr model representation of the Aluminum ion (Al^3+): (Nucleus: 13 protons)