criss-cross rule
lt: i can use the criss - cross rule for ionic compounds.
oxidation states
a fancy name for \charges\ in chemistry
the oxidation state or the charge of the ion will tell you the charge that
the atom has due to either a loss or gain of electrons
the literal charge (+ or -) will tell you if electrons have been lost or
gained, and the number will tell you how many electrons were
involved.
criss crossing
if you criss cross the numbers, you will get the same result as
drawing!
ionic salts make crystal lattices. this winds up looking like a 3d checker board
reducing
when the subscripts are reducible, reduce them. this will show you the smallest units
the crystal lattice.
ex: mg+o
show what you know
part 1: criss cross the following to determine the
formula. dont forget to reduce when necessary!
1) ba + br
2) aluminum + carbon
3) na+o
4) calcium + fluorine
5) li + n
6) lithium + sulfur
7) sr + se
8) strontium + nitrogen
9) k + o
10) magnesium + phosphorus
part 2: uncriss - cross to determine
the charges of each of the ions.
1) cacl₂
2) k₃p
3) mgs
4) cas
5) k₂se
6) li₂o
7) sro
8) naf
9) cabr₂
10) bas

Question

Accepted Answer

# Explanation:
## Part 1: Criss - Cross to find the formula
### 1) For $Ba + Br$
- **Step1: Determine oxidation states**
  - Barium ($Ba$) has an oxidation state of $+ 2$ (it is in group 2 of the periodic table and loses 2 electrons). Bromine ($Br$) has an oxidation state of $-1$ (it is in group 17 and gains 1 electron).
- **Step2: Apply criss - cross rule**
  - Criss - cross the numbers (the magnitude of the oxidation states). We get $BaBr_{2}$.
### 2) For Aluminum ($Al$) + Carbon ($C$)
- **Step1: Determine oxidation states**
  - Aluminum ($Al$) has an oxidation state of $+3$ (it is in group 13 and loses 3 electrons). Carbon ($C$) in ionic compounds (when it forms an anion) has an oxidation state of $-4$ (it gains 4 electrons to complete its octet).
- **Step2: Apply criss - cross rule**
  - Criss - cross the numbers. We get $Al_{4}C_{3}$.
### 3) For $Na+O$
- **Step1: Determine oxidation states**
  - Sodium ($Na$) has an oxidation state of $+1$ (it is in group 1 and loses 1 electron). Oxygen ($O$) has an oxidation state of $-2$ (it is in group 16 and gains 2 electrons).
- **Step2: Apply criss - cross rule**
  - Criss - cross the numbers. We get $Na_{2}O$.
### 4) For Calcium ($Ca$) + Fluorine ($F$)
- **Step1: Determine oxidation states**
  - Calcium ($Ca$) has an oxidation state of $+2$ (group 2, loses 2 electrons). Fluorine ($F$) has an oxidation state of $-1$ (group 17, gains 1 electron).
- **Step2: Apply criss - cross rule**
  - Criss - cross the numbers. We get $CaF_{2}$.
### 5) For $Li + N$
- **Step1: Determine oxidation states**
  - Lithium ($Li$) has an oxidation state of $+1$ (group 1, loses 1 electron). Nitrogen ($N$) has an oxidation state of $-3$ (group 15, gains 3 electrons).
- **Step2: Apply criss - cross rule**
  - Criss - cross the numbers. We get $Li_{3}N$.
### 6) For Lithium ($Li$) + Sulfur ($S$)
- **Step1: Determine oxidation states**
  - Lithium ($Li$) has an oxidation state of $+1$. Sulfur ($S$) has an oxidation state of $-2$ (group 16, gains 2 electrons).
- **Step2: Apply criss - cross rule**
  - Criss - cross the numbers. We get $Li_{2}S$.
### 7) For $Sr+Se$
- **Step1: Determine oxidation states**
  - Strontium ($Sr$) has an oxidation state of $+2$ (group 2, loses 2 electrons). Selenium ($Se$) has an oxidation state of $-2$ (group 16, gains 2 electrons).
- **Step2: Apply criss - cross rule**
  - Criss - cross the numbers. Since $2$ and $2$ can be reduced (divide both by 2), we get $SrSe$.
### 8) For Strontium ($Sr$) + Nitrogen ($N$)
- **Step1: Determine oxidation states**
  - Strontium ($Sr$) has an oxidation state of $+2$. Nitrogen ($N$) has an oxidation state of $-3$.
- **Step2: Apply criss - cross rule**
  - Criss - cross the numbers. We get $Sr_{3}N_{2}$.
### 9) For $K + O$
- **Step1: Determine oxidation states**
  - Potassium ($K$) has an oxidation state of $+1$ (group 1, loses 1 electron). Oxygen ($O$) has an oxidation state of $-2$.
- **Step2: Apply criss - cross rule**
  - Criss - cross the numbers. We get $K_{2}O$.
### 10) For Magnesium ($Mg$) + Phosphorus ($P$)
- **Step1: Determine oxidation states**
  - Magnesium ($Mg$) has an oxidation state of $+2$ (group 2, loses 2 electrons). Phosphorus ($P$) has an oxidation state of $-3$ (group 15, gains 3 electrons).
- **Step2: Apply criss - cross rule**
  - Criss - cross the numbers. We get $Mg_{3}P_{2}$.

## Part 2: Uncriss - cross to find the charges
### 1) For $CaCl_{2}$
- **Step1: Let the charge of $Ca=x$ and $Cl = y$**
  - Using the formula $CaCl_{2}$, and the rule that in a neutral compound $x + 2y=0$.
  - We know that for halides like $Cl$, the common charge is $-1$. Substitute $y=-1$ into $x + 2y = 0$. Then $x-2=0$, so $x = + 2$. The charge of $Ca^{2+}$ and $Cl^{-}$.
### 2) For $K_{3}P$
- **Step1: Let the charge of $K=x$ and $P = y$**
  - Using the formula $K_{3}P$, and the rule $3x + y=0$.
  - We know that for alkali metals like $K$, the common charge is $+1$. Substitute $x = + 1$ into $3x + y=0$. Then $3 + y=0$, so $y=-3$. The charge of $K^{+}$ and $P^{3 -}$.
### 3) For $MgS$
- **Step1: Let the charge of $Mg=x$ and $S = y$**
  - Using the formula $MgS$, and the rule $x + y=0$.
  - We know that for group 2 elements like $Mg$, the common charge is $+2$. Substitute $x = + 2$ into $x + y=0$. Then $2 + y=0$, so $y=-2$. The charge of $Mg^{2+}$ and $S^{2 -}$.
### 4) For $CaS$
- **Step1: Let the charge of $Ca=x$ and $S = y$**
  - Using the formula $CaS$, and the rule $x + y=0$.
  - We know that for group 2 elements like $Ca$, the common charge is $+2$. Substitute $x = + 2$ into $x + y=0$. Then $2 + y=0$, so $y=-2$. The charge of $Ca^{2+}$ and $S^{2 -}$.
### 5) For $K_{2}Se$
- **Step1: Let the charge of $K=x$ and $Se = y$**
  - Using the formula $K_{2}Se$, and the rule $2x + y=0$.
  - We know that for alkali metals like $K$, the common charge is $+1$. Substitute $x = + 1$ into $2x + y=0$. Then $2 + y=0$, so $y=-2$. The charge of $K^{+}$ and $Se^{2 -}$.
### 6) For $Li_{2}O$
- **Step1: Let the charge of $Li=x$ and $O = y$**
  - Using the formula $Li_{2}O$, and the rule $2x + y=0$.
  - We know that for alkali metals like $Li$, the common charge is $+1$. Substitute $x = + 1$ into $2x + y=0$. Then $2 + y=0$, so $y=-2$. The charge of $Li^{+}$ and $O^{2 -}$.
### 7) For $SrO$
- **Step1: Let the charge of $Sr=x$ and $O = y$**
  - Using the formula $SrO$, and the rule $x + y=0$.
  - We know that for group 2 elements like $Sr$, the common charge is $+2$. Substitute $x = + 2$ into $x + y=0$. Then $2 + y=0$, so $y=-2$. The charge of $Sr^{2+}$ and $O^{2 -}$.
### 8) For $NaF$
- **Step1: Let the charge of $Na=x$ and $F = y$**
  - Using the formula $NaF$, and the rule $x + y=0$.
  - We know that for alkali metals like $Na$, the common charge is $+1$. We know that for halides like $F$, the common charge is $-1$. Substitute $x = + 1$ and $y=-1$ into $x + y=0$, which holds true. The charge of $Na^{+}$ and $F^{-}$.
### 9) For $CaBr_{2}$
- **Step1: Let the charge of $Ca=x$ and $Br = y$**
  - Using the formula $CaBr_{2}$, and the rule $x + 2y=0$.
  - We know that for group 2 elements like $Ca$, the common charge is $+2$. We know that for halides like $Br$, the common charge is $-1$. Substitute $x = + 2$ and $y=-1$ into $x + 2y=0$ ($2+2\times(-1)=0$), which holds true. The charge of $Ca^{2+}$ and $Br^{-}$.
### 10) For $BaS$
- **Step1: Let the charge of $Ba=x$ and $S = y$**
  - Using the formula $BaS$, and the rule $x + y=0$.
  - We know that for group 2 elements like $Ba$, the common charge is $+2$. Substitute $x = + 2$ into $x + y=0$. Then $2 + y=0$, so $y=-2$. The charge of $Ba^{2+}$ and $S^{2 -}$.

# Answer:
## Part 1:
1. $BaBr_{2}$
2. $Al_{4}C_{3}$
3. $Na_{2}O$
4. $CaF_{2}$
5. $Li_{3}N$
6. $Li_{2}S$
7. $SrSe$
8. $Sr_{3}N_{2}$
9. $K_{2}O$
10. $Mg_{3}P_{2}$

## Part 2:
1. $Ca^{2+}$ and $Cl^{-}$
2. $K^{+}$ and $P^{3 -}$
3. $Mg^{2+}$ and $S^{2 -}$
4. $Ca^{2+}$ and $S^{2 -}$
5. $K^{+}$ and $Se^{2 -}$
6. $Li^{+}$ and $O^{2 -}$
7. $Sr^{2+}$ and $O^{2 -}$
8. $Na^{+}$ and $F^{-}$
9. $Ca^{2+}$ and $Br^{-}$
10. $Ba^{2+}$ and $S^{2 -}$

Sovi.AI - AI Math Tutor

QUESTION IMAGE

Question

Explanation:

Part 1: Criss - Cross to find the formula

1) For \(Ba + Br\)

2) For Aluminum (\(Al\)) + Carbon (\(C\))

3) For \(Na+O\)

4) For Calcium (\(Ca\)) + Fluorine (\(F\))

5) For \(Li + N\)

6) For Lithium (\(Li\)) + Sulfur (\(S\))

7) For \(Sr+Se\)

8) For Strontium (\(Sr\)) + Nitrogen (\(N\))

9) For \(K + O\)

10) For Magnesium (\(Mg\)) + Phosphorus (\(P\))

Part 2: Uncriss - cross to find the charges

1) For \(CaCl_{2}\)

Answer:

Part 1:

Part 2: