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 homework questions. Exercise 5.3 1. Draw the Lewis dot structures for each of the following molecules: a) H₂S Step 1: Calculate total valence electrons. Hydrogen (Group 1) has 1 valence electron. Sulfur (Group 16) has 6 valence electrons. Total valence electrons = (2 × 1) + 6 = 8 electrons. Step 2: Arrange atoms and form single bonds. Sulfur is the central atom. H-S-H (4 electrons used). Step 3: Distribute remaining electrons as lone pairs. Remaining electrons = 8 - 4 = 4 electrons. Place these on the central sulfur atom. H : ..S : H b) CH₃Br Step 1: Calculate total valence electrons. Carbon (Group 14) has 4 valence electrons. Hydrogen (Group 1) has 1 valence electron. Bromine (Group 17) has 7 valence electrons. Total valence electrons = 4 + (3 × 1) + 7 = 14 electrons. Step 2: Arrange atoms and form single bonds. Carbon is the central atom. C-H (3 bonds) and C-Br (1 bond) = 8 electrons used. Step 3: Distribute remaining electrons as lone pairs. Remaining electrons = 14 - 8 = 6 electrons. Place these on the more electronegative bromine atom. & H & \\ & | & \\ H & - & C & - & ..Br : \\ & | & & & .. \\ & H & & & c) HCN Step 1: Calculate total valence electrons. Hydrogen (Group 1) has 1 valence electron. Carbon (Group 14) has 4 valence electrons. Nitrogen (Group 15) has 5 valence electrons. Total valence electrons = 1 + 4 + 5 = 10 electrons. Step 2: Arrange atoms and form single bonds. Carbon is the central atom. H-C-N (4 electrons used). Step 3: Distribute remaining electrons as lone pairs and form multiple bonds to satisfy octets. Remaining electrons = 10 - 4 = 6 electrons. Place these on nitrogen. Carbon needs 4 more electrons, and nitrogen has a full octet with 6 lone pair electrons. Move two lone pairs from nitrogen to form a triple bond with carbon. H - C ..N d) K₂O Step 1: Identify as an ionic compound (metal K and nonmetal O). Potassium (Group 1) forms a K⁺ ion. Oxygen (Group 16) forms an O²⁻ ion. Step 2: Show the electron transfer and resulting ions with their charges and octets. Two K atoms each lose 1 electron, and one O atom gains 2 electrons. K^+ [ ....O ]^2- K^+ e) Al₂O₃ Step 1: Identify as an ionic compound (metal Al and nonmetal O). Aluminum (Group 13) forms an Al³⁺ ion. Oxygen (Group 16) forms an O²⁻ ion. Step 2: Determine the ratio of ions to balance charges. To balance the charges, two Al³⁺ ions (total +6) combine with three O²⁻ ions (total -6). Step 3: Show the resulting ions with their charges and octets. 2 Al^3+ 3 [ ....O ]^2- 2. In the Lewis structures listed below, M and X represent various elements in the third period of the periodic table. Write the formula of each compound using the chemical symbols of each element: The third period elements are Na, Mg, Al, Si, P, S, Cl, Ar. a) M^+ [::X::]^- Step 1: Identify M and X based on their charges and electron configuration. M has a +1 charge, indicating it is an alkali metal (Group 1). In the third period, this is Na (Sodium). X has a -1 charge and an octet, indicating it is a halogen (Group 17). In the third period, this is Cl (Chlorine). Step 2: Write the chemical formula. NaCl b) M^2+ [::X::]^2- Step 1: Identify M and X based on their charges and electron configuration. M has a +2 charge, indicating it is an alkaline earth metal (Group 2). In the third period, this is Mg (Magnesium). X has a -2 charge and an octet, indicating it is a chalcogen (Group 16). In the third period, this is S (Sulfur). Step 2: Write the chemical formula. MgS c) M^3+ [::X::]^2-_2 Step 1: Identify M and X based on their charges and electron configuration. M has a +3 charge, indicating it is from Group 13. In the third period, this is Al (Aluminum). X has a -2 charge and an octet, indicating it is a chalcogen (Group 16). In the third period, this is S (Sulfur). Step 2: Write the chemical formula, ensuring charge neutrality. The given structure implies two X²⁻ ions for one M³⁺ ion, which is incorrect for charge balance. For M³⁺ and X²⁻, the formula should be M₂X₃. Assuming the structure meant to represent the ions that form M₂X₃. Al_2S_3 d) M^2+ [::X::]^-_2 Step 1: Identify M and X based on their charges and electron configuration. M has a +2 charge, indicating it is an alkaline earth metal (Group 2). In the third period, this is Mg (Magnesium). X has a -1 charge and an octet, indicating it is a halogen (Group 17). In the third period, this is Cl (Chlorine). Step 2