3. consider mgf₂. this formula tells us that one mg²⁺ ion bonds with two f⁻ ions. what is the overall charge on mgf₂?
4. what is the overall charge on any ionic compound?
5. why is calcium nitride written like ca₃n₂ and not something like can₂ or ca₂n₃? in other words why do exactly three calcium ions bond with exactly two nitride ions?
6. the formula ca₃n₂ can never be written as n₂ca₃. to find out why, take note of each of the four example formulas given above.
a) in terms of charge, what do the first ions named all have in common?
b) in terms of charge, what do the second ions named all have in common?
c) now, why can’t ca₃n₂ ever be written like n₂ca₃?
7. there are two rules to follow when writing formulas for ionic compounds. one has to do with charges (see questions 4 and 5) and the other has to do with which atom to write first and which one to write second (see question 6). what are these two rules?
8. what is wrong with the following formulas?
a) al₂s
b) pna₃
c) mg₂s₂
9. write the formula and name for the compound that forms when the following atoms form ionic compounds. the first is done for you.
a) nitrogen and sodium
na₃n
sodium nitride
b) barium and sulfur
c) magnesium and iodine
d) oxygen and aluminum
e) calcium and phosphorus
f) sodium and sulfur
10. given the following compounds, determine the charge on the unknown ion “x”.
a) x₂s
b) mgx
c) x₃p₂
copyright 2002 - 2004 by jason neil. all rights reserved.
to make copies permission must be obtained from www.chemistryinquiry.com

Question

3. consider mgf₂. this formula tells us that one mg²⁺ ion bonds with two f⁻ ions. what is the overall charge on mgf₂?
4. what is the overall charge on any ionic compound?
5. why is calcium nitride written like ca₃n₂ and not something like can₂ or ca₂n₃? in other words why do exactly three calcium ions bond with exactly two nitride ions?
6. the formula ca₃n₂ can never be written as n₂ca₃. to find out why, take note of each of the four example formulas given above.
   a) in terms of charge, what do the first ions named all have in common?
   b) in terms of charge, what do the second ions named all have in common?
   c) now, why can’t ca₃n₂ ever be written like n₂ca₃?
7. there are two rules to follow when writing formulas for ionic compounds. one has to do with charges (see questions 4 and 5) and the other has to do with which atom to write first and which one to write second (see question 6). what are these two rules?
8. what is wrong with the following formulas?
   a) al₂s
   b) pna₃
   c) mg₂s₂
9. write the formula and name for the compound that forms when the following atoms form ionic compounds. the first is done for you.
   a) nitrogen and sodium
      na₃n
      sodium nitride
   b) barium and sulfur
   c) magnesium and iodine
   d) oxygen and aluminum
   e) calcium and phosphorus
   f) sodium and sulfur
10. given the following compounds, determine the charge on the unknown ion “x”.
    a) x₂s
    b) mgx
    c) x₃p₂
copyright 2002 - 2004 by jason neil. all rights reserved.
to make copies permission must be obtained from www.chemistryinquiry.com

Accepted Answer

### Question 3
# Explanation:
## Step1: Identify ion charges
$Mg^{2+}$ has a charge of $+2$, each $F^-$ (fluoride ion) has a charge of $-1$. There are 2 $F^-$ ions.
## Step2: Calculate total charge
Total charge = Charge of $Mg^{2+}$ + (Charge of $F^-$ × number of $F^-$ ions) = $+2 + (2 × -1)$ = $+2 - 2$ = $0$.
# Answer: 0

### Question 4
# Explanation:
Ionic compounds are formed by the electrostatic attraction between positive and negative ions. The positive and negative charges must balance out, so the overall charge of any ionic compound is neutral (zero).
# Answer: 0 (neutral)

### Question 5
# Explanation:
## Step1: Determine ion charges
Calcium ($Ca$) forms $Ca^{2+}$ (charge $+2$), nitride ion is $N^{3-}$ (charge $-3$).
## Step2: Find the least common multiple (LCM) of 2 and 3
LCM of 2 and 3 is 6. To balance the charges: number of $Ca^{2+}$ ions = $\frac{6}{2} = 3$, number of $N^{3-}$ ions = $\frac{6}{3} = 2$. So the formula is $Ca_3N_2$ to balance $3×(+2) + 2×(-3) = 6 - 6 = 0$. Formulas like $CaN_2$ (1×(+2) + 2×(-3) = -4, unbalanced) or $Ca_2N_3$ (2×(+2) + 3×(-3) = -5, unbalanced) don't balance charges.
# Answer: To balance charges: $Ca^{2+}$ (3 ions: $3×+2 = +6$) and $N^{3-}$ (2 ions: $2×-3 = -6$) balance to zero, so $Ca_3N_2$. Other formulas don't balance charges.

### Question 6a
# Explanation:
Looking at ionic compound formulas (e.g., $MgF_2$, $Ca_3N_2$, $Na_3N$), the first ions (Mg, Ca, Na) are metal ions (cations) which are positively charged (they lose electrons to form positive ions).
# Answer: The first ions (cations) are all positively charged (metal cations).

### Question 6b
# Explanation:
The second ions (F, N, N in the examples) are non - metal ions (anions) which are negatively charged (they gain electrons to form negative ions).
# Answer: The second ions (anions) are all negatively charged (non - metal anions).

### Question 6c
# Explanation:
Ionic compound formulas follow the rule that the cation (positive ion, usually metal) is written first, and the anion (negative ion, usually non - metal) is written second. In $Ca_3N_2$, $Ca^{2+}$ is the cation (metal) and $N^{3-}$ is the anion (non - metal). Writing $N_2Ca_3$ would reverse the order of cation and anion, which goes against the naming convention for ionic compounds where cation comes first.
# Answer: Ionic compound formulas have the cation (metal, positive ion) first and anion (non - metal, negative ion) second. $Ca^{2+}$ (cation) should come before $N^{3-}$ (anion), so $Ca_3N_2$ can't be $N_2Ca_3$.

### Question 7
# Explanation:
1. Charge balance rule: The sum of the positive charges from cations and negative charges from anions in the formula unit must be zero (overall charge of ionic compound is zero).
2. Order rule: In the formula of an ionic compound, the cation (usually a metal ion or a positive polyatomic ion) is written first, and the anion (usually a non - metal ion or a negative polyatomic ion) is written second.
# Answer:
1. The sum of positive and negative charges in the formula unit is zero (charge balance).
2. Cation (positive, usually metal) is written first, anion (negative, usually non - metal) second.

### Question 8a
# Explanation:
## Step1: Identify ion charges
$Al^{3+}$ (aluminum ion, charge $+3$), $S^{2-}$ (sulfide ion, charge $-2$).
## Step2: Check charge balance
To balance charges, find LCM of 3 and 2, which is 6. Number of $Al^{3+}$ ions = $\frac{6}{3}=2$, number of $S^{2-}$ ions = $\frac{6}{2} = 3$. So the correct formula should be $Al_2S_3$, but the given formula is $Al_2S$ which has unbalanced charges ($2×+3+1×-2 = +4
eq0$).
# Answer: Unbalanced charges. $Al^{3+}$ (2 ions: $+6$) and $S^{2-}$ (1 ion: $-2$) don't balance. Correct formula should be $Al_2S_3$.

### Question 8b
# Explanation:
## Step1: Identify ion charges
$Na^+$ (sodium ion, charge $+1$), $P^{3-}$ (phosphide ion, charge $-3$).
## Step2: Check order and balance
Ionic formula should have cation (Na$^+$) first and anion (P$^{3-}$) second. Also, to balance charges: number of $Na^+$ ions = 3 (since $3×+1$ to balance $-3$ from P$^{3-}$). The given formula $PNa_3$ reverses the order of cation and anion (anion P$^{3-}$ first, cation Na$^+$ second) and also, although the charge balance is there (3×+1 + (-3)=0), the order is wrong.
# Answer: The order is wrong. Ionic formula should have cation (Na$^+$) first, anion (P$^{3-}$) second. Correct formula: $Na_3P$.

### Question 8c
# Explanation:
## Step1: Identify ion charges
$Mg^{2+}$ (magnesium ion, charge $+2$), $S^{2-}$ (sulfide ion, charge $-2$).
## Step2: Check formula
For $Mg^{2+}$ and $S^{2-}$, the ratio of ions should be 1:1 (since $+2$ and $-2$ balance with 1 each). The given formula $Mg_2S_2$ can be simplified to $MgS$ (dividing subscripts by 2), and having $Mg_2S_2$ is incorrect as we use the simplest whole - number ratio for ionic formulas.
# Answer: Ionic formula should use the simplest whole - number ratio. $Mg^{2+}$ and $S^{2-}$ have a 1:1 ratio, so formula should be $MgS$, not $Mg_2S_2$.

### Question 9b
# Explanation:
## Step1: Identify ion charges
Barium ($Ba$) forms $Ba^{2+}$ (charge $+2$), sulfur ($S$) forms $S^{2-}$ (charge $-2$).
## Step2: Determine formula and name
Since the charges are equal in magnitude (2+ and 2 -), the ratio of $Ba^{2+}$ to $S^{2-}$ is 1:1. The formula is $BaS$. The name is barium sulfide (cation name first, anion name second, sulfide for $S^{2-}$).
# Answer: Formula: $BaS$; Name: barium sulfide

### Question 9c
# Explanation:
## Step1: Identify ion charges
Magnesium ($Mg$) forms $Mg^{2+}$ (charge $+2$), iodine ($I$) forms $I^-$ (charge $-1$).
## Step2: Determine formula and name
To balance charges: number of $Mg^{2+}$ = 1, number of $I^-$ = 2 (since $+2+(2× - 1)=0$). The formula is $MgI_2$. The name is magnesium iodide (cation name first, anion name (iodide for $I^-$) second).
# Answer: Formula: $MgI_2$; Name: magnesium iodide

### Question 9d
# Explanation:
## Step1: Identify ion charges
Aluminum ($Al$) forms $Al^{3+}$ (charge $+3$), oxygen ($O$) forms $O^{2-}$ (charge $-2$).
## Step2: Find the ratio of ions
Find LCM of 3 and 2, which is 6. Number of $Al^{3+}$ ions = $\frac{6}{3}=2$, number of $O^{2-}$ ions = $\frac{6}{2}=3$. So the formula is $Al_2O_3$. The name is aluminum oxide (cation name first, anion name (oxide for $O^{2-}$) second).
# Answer: Formula: $Al_2O_3$; Name: aluminum oxide

### Question 9e
# Explanation:
## Step1: Identify ion charges
Calcium ($Ca$) forms $Ca^{2+}$ (charge $+2$), phosphorus ($P$) forms $P^{3-}$ (charge $-3$).
## Step2: Find the ratio of ions
LCM of 2 and 3 is 6. Number of $Ca^{2+}$ ions = $\frac{6}{2}=3$, number of $P^{3-}$ ions = $\frac{6}{3}=2$. So the formula is $Ca_3P_2$. The name is calcium phosphide (cation name first, anion name (phosphide for $P^{3-}$) second).
# Answer: Formula: $Ca_3P_2$; Name: calcium phosphide

### Question 9f
# Explanation:
## Step1: Identify ion charges
Sodium ($Na$) forms $Na^+$ (charge $+1$), sulfur ($S$) forms $S^{2-}$ (charge $-2$).
## Step2: Find the ratio of ions
To balance $-2$ charge of $S^{2-}$, we need 2 $Na^+$ ions (since $2×+1 = +2$). So the formula is $Na_2S$. The name is sodium sulfide (cation name first, anion name (sulfide for $S^{2-}$) second).
# Answer: Formula: $Na_2S$; Name: sodium sulfide

### Question 10a
# Explanation:
## Step1: Identify known ion charge
$S^{2-}$ (sulfide ion) has a charge of $-2$. Let the charge of $X$ be $x$. There are 2 $X$ ions.
## Step2: Set up charge balance equation
Total charge = 0, so $2x+(-2)=0$. Solving for $x$: $2x = 2$, so $x = +1$.
# Answer: +1

### Question 10b
# Explanation:
## Step1: Identify known ion charge
$Mg^{2+}$ (magnesium ion) has a charge of $+2$. Let the charge of $X$ be $x$.
## Step2: Set up charge balance equation
Total charge = 0, so $+2 + x=0$. Solving for $x$: $x=-2$.
# Answer: -2

### Question 10c
# Explanation:
## Step1: Identify known ion charge
$P^{3-}$ (phosphide ion) has a charge of $-3$. There are 2 $P^{3-}$ ions, so total negative charge = $2×-3=-6$. Let the charge of $X$ be $x$. There are 3 $X$ ions.
## Step2: Set up charge balance equation
Total charge = 0, so $3x+(-6)=0$. Solving for $x$: $3x = 6$, so $x = +2$.
# Answer: +2

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QUESTION IMAGE

Question

Explanation:

Question 3

Step1: Identify ion charges

Step2: Calculate total charge

Step1: Determine ion charges

Step2: Find the least common multiple (LCM) of 2 and 3

Answer:

Question 4