Question

One of the ions of tin is tin(IV). a. What is the symbol for this ion? b. How many protons and electrons are in the ion? c. What is the formula of tin(IV) oxide? d. What is the formula of tin(IV) phosphate?

Short answer

a. \( \text{Sn}^{4+} \); b. 50 protons, 46 electrons; c. \( \text{SnO}_2 \); d. \( \text{Sn}_3(\text{PO}_4)_4 \)

Step-by-step solution

Determine the symbol for tin(IV) ion

Tin(IV) indicates that the tin ion has a +4 charge. The element symbol for tin is Sn, so the symbol for tin(IV) ion is written as \( \text{Sn}^{4+} \).

Find the number of protons and electrons

Tin has an atomic number of 50, meaning it has 50 protons. For the tin(IV) ion \( \text{Sn}^{4+} \), it loses 4 electrons. Therefore, it has 50 protons and 46 electrons.

Determine the formula of tin(IV) oxide

Tin(IV) ion is \( \text{Sn}^{4+} \) and oxide ion is \( \text{O}^{2-} \). To balance the charges, we need two \( \text{O}^{2-} \) ions for every \( \text{Sn}^{4+} \) ion. The formula is \( \text{SnO}_2 \).

Determine the formula of tin(IV) phosphate

Tin(IV) ion is \( \text{Sn}^{4+} \) and phosphate ion is \( \text{PO}_4^{3-} \). To balance the charges, we need three \( \text{Sn}^{4+} \) ions for every four \( \text{PO}_4^{3-} \) ions. The formula is \( \text{Sn}_3(\text{PO}_4)_4 \).

Key concepts

Ion charge determination

Understanding ion charges is crucial in chemistry. Ions are atoms or molecules that have lost or gained electrons, giving them a positive or negative charge. For tin(IV), the Roman numeral IV indicates a +4 charge.

When an atom loses electrons, it becomes positively charged (cation). Conversely, when it gains electrons, it becomes negatively charged (anion). For the tin(IV) ion, tin has lost 4 electrons. Therefore, the ion charge is +4, and its symbol is written as \( \text{Sn}^{4+} \).

Knowing the charge helps in determining how the ion will combine with others to form compounds.

Atomic structure

The atomic structure of an element involves understanding its protons, neutrons, and electrons. The atomic number of an element equals the number of protons in its nucleus. For example, tin (Sn) has an atomic number of 50, which means it has 50 protons.

The number of electrons in a neutral atom equals the number of protons. However, when an atom becomes an ion, the number of electrons changes. For tin(IV), the ion loses 4 electrons. Therefore, \( \text{Sn}^{4+} \) has:

• 50 protons
• 46 electrons (since 50 - 4 = 46)

This loss of electrons changes the charge and affects how tin interacts with other elements.

Formula calculation

To write the formula of a compound, the charges on the ions must balance out to zero. Consider tin(IV) oxide. The tin(IV) ion has a charge of +4 (\( \text{Sn}^{4+} \)). The oxide ion has a charge of -2 (\( \text{O}^{2-} \)).

To balance these charges, you need two oxide ions for every tin(IV) ion:

• 1 \( \text{Sn}^{4+} \)
• 2 \( \text{O}^{2-} \)

So, the formula for tin(IV) oxide is \( \text{SnO}_2 \).

For tin(IV) phosphate, the phosphate ion has a charge of -3 (\( \text{PO}_4^{3-} \)). Balancing the charges means you need three \( \text{Sn}^{4+} \) ions and four \( \text{PO}_4^{3-} \) ions:

• 3 \( \text{Sn}^{4+} \)
• 4 \( \text{PO}_4^{3-} \)

Hence, the formula is \( \text{Sn}_3(\text{PO}_4)_4 \).

Compound naming

Naming ionic compounds follows specific rules. First, you name the cation (positive ion) followed by the anion (negative ion). For naming an ionic compound:

• The cation retains the name of the element.
• The anion's name is derived from its element but ends in -ide for simple ions, like oxide (\( \text{O}^{2-} \)), or retains a specific polyatomic name, like phosphate (\( \text{PO}_4^{3-} \)).

For example, \( \text{SnO}_2 \) is called tin(IV) oxide because we name the cation (tin(IV)) followed by the anion (oxide).

For \( \text{Sn}_3(\text{PO}_4)_4 \), the name is tin(IV) phosphate following the same rule. This systematic approach helps in accurately identifying and naming compounds in chemistry.