QUESTION IMAGE
Question
part 2: determine the products formed by each of the following pairs of elements after an ionic reaction.
pair of elements charges ionic formula of the compound formed name of compound
ex k, i k+1 i-1 ki potassium iodide
1 be, f be2+, f- bef2 beryllium fluoride
2 al, br al3+, br- albr3 aluminum bromide
3 na, cl na+, cl- nacl sodium chloride
4 mg, s mg2+, s2- mgs magnesium sulfide
5 al, n al3+, n3- aln aluminum nitride
6 cs, p
7 ba, n
8 al, s
9 al, i
10 rb, n
11 cs, s
12 ca, o
13 al, f
14 fr, f
Step1: Determine ion charges
For metal - non - metal pairs, metals lose electrons to form cations and non - metals gain electrons to form anions. For example, Be (Group 2) forms $Be^{2 + }$ and F (Group 17) forms $F^{-}$.
Step2: Write ionic formula
The ionic formula is written such that the total positive charge equals the total negative charge. For $Be^{2+}$ and $F^{-}$, we need 2 F⁻ ions to balance the + 2 charge of Be²⁺, so the formula is $BeF_{2}$.
Step3: Name the compound
The name of the compound is formed by writing the name of the metal first, followed by the name of the non - metal with the ending changed to “-ide”.
| # | Pair of Elements | Charges | Ionic Formula of the Compound Formed | Name of Compound |
|---|---|---|---|---|
| 2 | Al, Br | $Al^{3 + },Br^{-}$ | $AlBr_{3}$ | Aluminum bromide |
| 3 | Na, Cl | $Na^{+},Cl^{-}$ | $NaCl$ | Sodium chloride |
| 4 | Mg, S | $Mg^{2+},S^{2 - }$ | $MgS$ | Magnesium sulfide |
| 5 | Al, N | $Al^{3+},N^{3 - }$ | $AlN$ | Aluminum nitride |
| 6 | Cs, P | $Cs^{+},P^{3 - }$ | $Cs_{3}P$ | Cesium phosphide |
| 7 | Ba, N | $Ba^{2+},N^{3 - }$ | $Ba_{3}N_{2}$ | Barium nitride |
| 8 | Al, S | $Al^{3+},S^{2 - }$ | $Al_{2}S_{3}$ | Aluminum sulfide |
| 9 | Al, I | $Al^{3+},I^{-}$ | $AlI_{3}$ | Aluminum iodide |
| 10 | Rb, N | $Rb^{+},N^{3 - }$ | $Rb_{3}N$ | Rubidium nitride |
| 11 | Cs, S | $Cs^{+},S^{2 - }$ | $Cs_{2}S$ | Cesium sulfide |
| 12 | Ca, O | $Ca^{2+},O^{2 - }$ | $CaO$ | Calcium oxide |
| 13 | Al, F | $Al^{3+},F^{-}$ | $AlF_{3}$ | Aluminum fluoride |
| 14 | Fr, F | $Fr^{+},F^{-}$ | $FrF$ | Francium fluoride |
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Step1: Determine ion charges
For metal - non - metal pairs, metals lose electrons to form cations and non - metals gain electrons to form anions. For example, Be (Group 2) forms $Be^{2 + }$ and F (Group 17) forms $F^{-}$.
Step2: Write ionic formula
The ionic formula is written such that the total positive charge equals the total negative charge. For $Be^{2+}$ and $F^{-}$, we need 2 F⁻ ions to balance the + 2 charge of Be²⁺, so the formula is $BeF_{2}$.
Step3: Name the compound
The name of the compound is formed by writing the name of the metal first, followed by the name of the non - metal with the ending changed to “-ide”.
| # | Pair of Elements | Charges | Ionic Formula of the Compound Formed | Name of Compound |
|---|---|---|---|---|
| 2 | Al, Br | $Al^{3 + },Br^{-}$ | $AlBr_{3}$ | Aluminum bromide |
| 3 | Na, Cl | $Na^{+},Cl^{-}$ | $NaCl$ | Sodium chloride |
| 4 | Mg, S | $Mg^{2+},S^{2 - }$ | $MgS$ | Magnesium sulfide |
| 5 | Al, N | $Al^{3+},N^{3 - }$ | $AlN$ | Aluminum nitride |
| 6 | Cs, P | $Cs^{+},P^{3 - }$ | $Cs_{3}P$ | Cesium phosphide |
| 7 | Ba, N | $Ba^{2+},N^{3 - }$ | $Ba_{3}N_{2}$ | Barium nitride |
| 8 | Al, S | $Al^{3+},S^{2 - }$ | $Al_{2}S_{3}$ | Aluminum sulfide |
| 9 | Al, I | $Al^{3+},I^{-}$ | $AlI_{3}$ | Aluminum iodide |
| 10 | Rb, N | $Rb^{+},N^{3 - }$ | $Rb_{3}N$ | Rubidium nitride |
| 11 | Cs, S | $Cs^{+},S^{2 - }$ | $Cs_{2}S$ | Cesium sulfide |
| 12 | Ca, O | $Ca^{2+},O^{2 - }$ | $CaO$ | Calcium oxide |
| 13 | Al, F | $Al^{3+},F^{-}$ | $AlF_{3}$ | Aluminum fluoride |
| 14 | Fr, F | $Fr^{+},F^{-}$ | $FrF$ | Francium fluoride |