Predict the formulas of the stable binary compounds that would be formed by the combination of the following p

Exercises · 3.32

Q. Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.

(a) Lithium and oxygen

(b) Magnesium and nitrogen

(d) Silicon and oxygen

(e) Phosphorus and fluorine

(f) Element 71 and fluorine

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Concept understanding — Ionic Compound Prediction

Ionic Compound Prediction: From Intuition to Rule

Imagine you have a bag of positively charged magnets (cations) and negatively charged magnets (anions). If you just dump them together, they'll snap into a neutral clump — but only if the total positive charge exactly cancels the total negative charge. That's the core idea behind ionic compound formation: the compound must be electrically neutral overall.

The Intuition

Sodium ( $\text{Na}$ ) wants to lose one electron to become $\text{Na}^+$ . Chlorine ( $\text{Cl}$ ) wants to gain one electron to become $\text{Cl}^-$ .

If you put one $\text{Na}^+$ and one $\text{Cl}^-$ together, the charges cancel: $+1 + (-1) = 0$ . That's why sodium chloride is $\text{NaCl}$ — one sodium ion for every chloride ion.

But what about magnesium ( $\text{Mg}$ ) and chlorine? Magnesium loses two electrons to become $\text{Mg}^{2+}$ . One $\text{Mg}^{2+}$ needs two $\text{Cl}^-$ ions to balance: $+2 + 2(-1) = 0$ . So the formula is $\text{MgCl}_2$ .

The rule is simple: the total positive charge must equal the total negative charge. You're just finding the smallest whole-number ratio of ions that makes this happen.

The Precise Statement

X^{m+}

In plain language: the subscript on the cation ( $a$ ) times its charge ( $m$ ) must equal the subscript on the anion ( $b$ ) times its charge ( $n$ ).

How to Apply It — Step by Step

Write the ions with their charges.

Example: calcium ( $\text{Ca}^{2+}$ ) and phosphate ( $\text{PO}_4^{3-}$ ).
Find the smallest numbers that balance the charges.

The charges are $+2$ and $-3$ . The least common multiple of 2 and 3 is 6.
- To get $+6$ from $\text{Ca}^{2+}$ , you need 3 calcium ions: $3 \times (+2) = +6$ .
- To get $-6$ from $\text{PO}_4^{3-}$ , you need 2 phosphate ions: $2 \times (-3) = -6$ .
Write the formula with those numbers as subscripts.

$\text{Ca}_3(\text{PO}_4)_2$ — the parentheses around phosphate show it's a polyatomic ion taken as a unit.

Tip

A quick shortcut: swap the charges (without the signs) and use them as subscripts.

For $\text{Ca}^{2+}$ and $\text{PO}_4^{3-}$ , swap 2 and 3 → $\text{Ca}_3(\text{PO}_4)_2$ .

For $\text{Al}^{3+}$ and $\text{O}^{2-}$ , swap 3 and 2 → $\text{Al}_2\text{O}_3$ .

This always works because $a \times m = b \times n$ is exactly the cross-multiplication of the charges.

Common Pitfalls

Watch out

Never change the charge on an ion. The charge is fixed — $\text{Na}$ is always $+1$ , $\text{O}$ is always $-2$ . You only change how many of each ion you use.

Another trap: forgetting to reduce the ratio. If you get $\text{Ca}_2\text{O}_2$ , that's wrong — it should be $\text{CaO}$ (the smallest whole numbers are 1 and 1). Always simplify.

Why This Works …

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The key idea is to use the octet rule and the common oxidation states of each element to predict the simplest neutral formula. The final compounds are: (a) $\text{Li}_2\text{O}$ ,

(b) $\text{Mg}_3\text{N}_2$ ,

(d) $\text{SiO}_2$ , (e) $\text{PF}_3$ or $\text{PF}_5$ , (f) $\text{LuF}_3$ .

This is a classic exercise in predicting ionic (or polar covalent) binary compounds. The logic is straightforward: each element wants to achieve a stable noble gas electron configuration. Metals tend to lose electrons to become positive ions (cations), and non-metals tend to gain electrons to become negative ions (anions). The formula is the smallest whole-number ratio that makes the total positive charge equal the total negative charge.

Let’s go through each pair step by step.

Lithium and oxygen

Lithium (Li) is in Group 1. It has one valence electron and loses it easily to form $\text{Li}^+$ (a $1+$ ion). Oxygen (O) is in Group 16. It needs two electrons to complete its octet, forming $\text{O}^{2-}$ (a $2-$ ion).

To balance charges: two $\text{Li}^+$ ions give $2+$ , which exactly balances one $\text{O}^{2-}$ ion.

The formula is $\text{Li}_2\text{O}$ .
Magnesium and nitrogen

Magnesium (Mg) is in Group 2. It loses two electrons to form $\text{Mg}^{2+}$ . Nitrogen (N) is in Group 15. It needs three electrons to complete its octet, forming $\text{N}^{3-}$ .

The lowest common multiple of the charges $2$ and $3$ is $6$ . So we need three $\text{Mg}^{2+}$ ions (total $6+$ ) and two $\text{N}^{3-}$ ions (total $6-$ ).

The formula is $\text{Mg}_3\text{N}_2$ .
Aluminium and iodine

Aluminium (Al) is in Group 13. It loses three electrons to form $\text{Al}^{3+}$ . Iodine (I) is in Group 17. It gains one electron to form $\text{I}^-$ .

To balance: one $\text{Al}^{3+}$ needs three $\text{I}^-$ ions.

The formula is $\text{AlI}_3$ .
Silicon and oxygen

Silicon (Si) is in Group 14. It is a metalloid. With oxygen, it typically forms covalent bonds, but we can still use the octet rule. Silicon has four valence electrons; it tends to share them to get an octet. Oxygen needs two electrons.

The simplest ratio is one silicon atom sharing with two oxygen atoms: each oxygen forms two bonds, and silicon forms four.

The formula is $\text{SiO}_2$ (silicon dioxide).
Phosphorus and fluorine

Phosphorus (P) is in Group 15. It has five valence electrons. Fluorine (F) is in Group 17 and needs one electron.

Phosphorus can form three single bonds (using three of its electrons) to get an octet, giving $\text{PF}_3$ . However, phosphorus can also expand its octet (it has available $3d$ orbitals) and form five bonds, giving $\text{PF}_5$ . Both are stable compounds. …

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Q.Predict the formulas of the stable binary compounds that would be formed by the combination of the following pairs of elements.

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