name: date: ionic bonding practice worksheet what would be the compounds formed by the following elements after they underwent ionic bonding? a. potassium + chlorine → b. magnesium + brominee → c. sodium + fluorine → d. sodium + bromine → e. magnesium + chlorine → f. aluminum + bromine → g. strontium + iodine → h. sodium + oxygen → i. sodium + chlorine → j. calcium + fluorine → k. aluminum + chlorine → l. potassium + iodine → m. magnesium + oxygen→ n. rubidium + oxygen → o. selenium + chlorine → p. potassium + sulfur → q. gallium + chloride → r. sodium + sulfur →

Question

Accepted Answer

### Part a: Potassium + Chlorine
# Explanation:
## Step1: Determine ions
Potassium (K) is in Group 1, so it forms $ K^+ $. Chlorine (Cl) is in Group 17, forms $ Cl^- $.
## Step2: Balance charges
The charges $ +1 $ and $ -1 $ balance, so the formula is $ KCl $.
# Answer: $ KCl $

### Part b: Magnesium + Bromine (Note: "Brominee" is likely a typo, should be Bromine)
# Explanation:
## Step1: Determine ions
Magnesium (Mg) is in Group 2, forms $ Mg^{2+} $. Bromine (Br) is in Group 17, forms $ Br^- $.
## Step2: Balance charges
To balance $ +2 $ and $ -1 $, we need 2 $ Br^- $ for 1 $ Mg^{2+} $. Formula: $ MgBr_2 $.
# Answer: $ MgBr_2 $

### Part c: Sodium + Fluorine
# Explanation:
## Step1: Determine ions
Sodium (Na) is Group 1, $ Na^+ $. Fluorine (F) is Group 17, $ F^- $.
## Step2: Balance charges
$ +1 $ and $ -1 $ balance, formula: $ NaF $.
# Answer: $ NaF $

### Part d: Sodium + Bromine
# Explanation:
## Step1: Determine ions
Sodium (Na): $ Na^+ $, Bromine (Br): $ Br^- $.
## Step2: Balance charges
Charges balance ($ +1 $ and $ -1 $), formula: $ NaBr $.
# Answer: $ NaBr $

### Part e: Magnesium + Chlorine
# Explanation:
## Step1: Determine ions
Magnesium (Mg): $ Mg^{2+} $, Chlorine (Cl): $ Cl^- $.
## Step2: Balance charges
Need 2 $ Cl^- $ for $ Mg^{2+} $, formula: $ MgCl_2 $.
# Answer: $ MgCl_2 $

### Part f: Aluminum + Bromine
# Explanation:
## Step1: Determine ions
Aluminum (Al) is Group 13, forms $ Al^{3+} $. Bromine (Br): $ Br^- $.
## Step2: Balance charges
To balance $ +3 $ and $ -1 $, need 3 $ Br^- $. Formula: $ AlBr_3 $.
# Answer: $ AlBr_3 $

### Part g: Strontium + Iodine
# Explanation:
## Step1: Determine ions
Strontium (Sr) is Group 2, $ Sr^{2+} $. Iodine (I) is Group 17, $ I^- $.
## Step2: Balance charges
Need 2 $ I^- $ for $ Sr^{2+} $, formula: $ SrI_2 $.
# Answer: $ SrI_2 $

### Part h: Sodium + Oxygen
# Explanation:
## Step1: Determine ions
Sodium (Na): $ Na^+ $, Oxygen (O) is Group 16, forms $ O^{2-} $.
## Step2: Balance charges
Need 2 $ Na^+ $ for $ O^{2-} $, formula: $ Na_2O $.
# Answer: $ Na_2O $

### Part i: Sodium + Chlorine
# Explanation:
## Step1: Determine ions
Sodium (Na): $ Na^+ $, Chlorine (Cl): $ Cl^- $.
## Step2: Balance charges
Charges balance, formula: $ NaCl $.
# Answer: $ NaCl $

### Part j: Calcium + Fluorine
# Explanation:
## Step1: Determine ions
Calcium (Ca) is Group 2, $ Ca^{2+} $. Fluorine (F): $ F^- $.
## Step2: Balance charges
Need 2 $ F^- $ for $ Ca^{2+} $, formula: $ CaF_2 $.
# Answer: $ CaF_2 $

### Part k: Aluminum + Chlorine
# Explanation:
## Step1: Determine ions
Aluminum (Al): $ Al^{3+} $, Chlorine (Cl): $ Cl^- $.
## Step2: Balance charges
Need 3 $ Cl^- $ for $ Al^{3+} $, formula: $ AlCl_3 $.
# Answer: $ AlCl_3 $

### Part l: Potassium + Iodine
# Explanation:
## Step1: Determine ions
Potassium (K): $ K^+ $, Iodine (I): $ I^- $.
## Step2: Balance charges
Charges balance, formula: $ KI $.
# Answer: $ KI $

### Part m: Magnesium + Oxygen
# Explanation:
## Step1: Determine ions
Magnesium (Mg): $ Mg^{2+} $, Oxygen (O): $ O^{2-} $.
## Step2: Balance charges
Charges balance ($ +2 $ and $ -2 $), formula: $ MgO $.
# Answer: $ MgO $

### Part n: Rubidium + Oxygen
# Explanation:
## Step1: Determine ions
Rubidium (Rb) is Group 1, $ Rb^+ $. Oxygen (O): $ O^{2-} $.
## Step2: Balance charges
Need 2 $ Rb^+ $ for $ O^{2-} $, formula: $ Rb_2O $.
# Answer: $ Rb_2O $

### Part o: Selenium + Chlorine (Note: Selenium is non - metal, Chlorine is non - metal, but for ionic bonding context, Selenium (Se) in Group 16 forms $ Se^{2-} $, Chlorine (Cl) forms $ Cl^- $. Wait, actually, Selenium and Chlorine form a covalent compound, but if we consider ionic (maybe a typo, or considering Se as a metal - like in some contexts, but more accurately, Se is non - metal. However, following the worksheet's ionic bonding premise:
# Explanation:
## Step1: Determine ions (assuming ionic)
Selenium (Se): $ Se^{2-} $, Chlorine (Cl): $ Cl^- $. Wait, no, to form ionic, Se would need to lose electrons, but Se is a non - metal. Maybe the worksheet expects us to use oxidation states. Se can have $ -2 $, Cl $ -1 $. But for ionic, we need a metal and non - metal. Maybe a typo, Selenium should be Selenium metal (but Se is a non - metal). Alternatively, maybe the question has an error. But if we proceed with Se as $ Se^{2-} $ and Cl as $ Cl^- $, to balance $ -2 $ and $ -1 $? No, that's not ionic. Wait, maybe Selenium is a metal here (incorrect, but for the worksheet):
Wait, no, Selenium is a non - metal, Chlorine is a non - metal. They form a covalent compound $ SeCl_2 $ or $ SeCl_4 $ etc. But the worksheet is about ionic bonding, so maybe it's a typo, and the first element is a metal. If we assume it's a metal (e.g., Selenium is a typo for a metal, but since we have to answer, maybe the intended is a metal and Chlorine. Alternatively, maybe the question is wrong. But if we proceed with Se as $ Se^{2-} $ and Cl as $ Cl^- $, we can't form an ionic compound with two non - metals. So there might be a typo. If we assume the first element is a metal (e.g., Strontium instead of Selenium), but as per the given, we'll try. Wait, maybe the user meant Selenium as a cation? No, Se is a non - metal. So this question might have an error. But if we force ionic, maybe Se is $ Se^{2+} $ (unlikely, but for the sake of answering):
## Step1: Assume Se is $ Se^{2+} $, Cl is $ Cl^- $
## Step2: Balance charges: need 2 $ Cl^- $ for $ Se^{2+} $, formula $ SeCl_2 $
# Answer: $ SeCl_2 $ (Note: This is a covalent compound in reality, but following the worksheet's ionic bonding premise)

### Part p: Potassium + Sulfur
# Explanation:
## Step1: Determine ions
Potassium (K): $ K^+ $, Sulfur (S) is Group 16, forms $ S^{2-} $.
## Step2: Balance charges
Need 2 $ K^+ $ for $ S^{2-} $, formula: $ K_2S $.
# Answer: $ K_2S $

### Part q: Gallium + Chloride (Note: "Chloride" is $ Cl^- $, Gallium (Ga) is Group 13, forms $ Ga^{3+} $)
# Explanation:
## Step1: Determine ions
Gallium (Ga): $ Ga^{3+} $, Chloride ($ Cl^- $): $ Cl^- $.
## Step2: Balance charges
Need 3 $ Cl^- $ for $ Ga^{3+} $, formula: $ GaCl_3 $.
# Answer: $ GaCl_3 $

### Part r: Sodium + Sulfur
# Explanation:
## Step1: Determine ions
Sodium (Na): $ Na^+ $, Sulfur (S): $ S^{2-} $.
## Step2: Balance charges
Need 2 $ Na^+ $ for $ S^{2-} $, formula: $ Na_2S $.
# Answer: $ Na_2S $

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QUESTION IMAGE

Question

Explanation:

Part a: Potassium + Chlorine

Step1: Determine ions

Step2: Balance charges

Step1: Determine ions

Step2: Balance charges

Step1: Determine ions

Step2: Balance charges

Answer:

Part b: Magnesium + Bromine (Note: "Brominee" is likely a typo, should be Bromine)