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Atomic Structure - Properties of Metals and Non-metals

Grade 9IB

Review the key concepts, formulae, and examples before starting your quiz.

🔑Concepts

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Metals are elements that lose electrons to form positive ions called cations. This is represented as M→Mn++ne−M \rightarrow M^{n+} + ne^-.

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Non-metals are elements that gain electrons to form negative ions called anions, such as X+ne−→Xn−X + ne^- \rightarrow X^{n-}, or share electrons to form covalent bonds.

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The tendency of metals to lose electrons is known as electropositivity, while the tendency of non-metals to gain electrons is known as electronegativity.

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Atomic structure determines properties: Metals typically have 1,2,1, 2, or 33 valence electrons, whereas non-metals typically have 4,5,6,4, 5, 6, or 77 valence electrons.

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Metals form basic oxides (e.g., Na2O+H2O→2NaOHNa_2O + H_2O \rightarrow 2NaOH), while non-metals form acidic oxides (e.g., SO2+H2O→H2SO3SO_2 + H_2O \rightarrow H_2SO_3).

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Physical properties: Metals are malleable (can be hammered into sheets) and ductile (can be drawn into wires). Non-metals are generally brittle in solid form.

📐Formulae

n=A−Zn = A - Z

Max electrons in shell n=2n2\text{Max electrons in shell } n = 2n^2

4M+O2→2M2O (Metal oxide)4M + O_2 \rightarrow 2M_2O \text{ (Metal oxide)}

S+O2→SO2 (Non-metal oxide)S + O_2 \rightarrow SO_2 \text{ (Non-metal oxide)}

💡Examples

Problem 1:

An element XX has an atomic number of 1313. Determine its electronic configuration, identify if it is a metal or non-metal, and predict the charge of its ion.

Solution:

The atomic number Z=13Z = 13. The electronic configuration is 2,8,32, 8, 3. Since it has 33 electrons in its outermost shell, it is a metal. It loses 33 electrons to achieve a stable octet: X→X3++3e−X \rightarrow X^{3+} + 3e^-. The ion formed is X3+X^{3+}.

Explanation:

Elements with 1,2,1, 2, or 33 valence electrons are generally metals because they have low ionization energy and tend to lose electrons to achieve the stable noble gas configuration of 2,82, 8.

Problem 2:

Compare the nature of oxides formed by Magnesium (Z=12Z=12) and Sulfur (Z=16Z=16). Write the chemical equations for their reaction with oxygen.

Solution:

Magnesium (Metal): 2Mg+O2→2MgO2Mg + O_2 \rightarrow 2MgO MgOMgO is a basic oxide.

Sulfur (Non-metal): S+O2→SO2S + O_2 \rightarrow SO_2 SO2SO_2 is an acidic oxide.

Explanation:

Metallic oxides react with water to form hydroxides (bases), whereas non-metallic oxides react with water to form acids. This distinction is a key chemical property used to classify elements.

Problem 3:

Calculate the number of neutrons in an atom of Aluminum given its mass number is 2727 and atomic number is 1313.

Solution:

Using the formula: n=A−Zn = A - Z n=27−13n = 27 - 13 n=14n = 14 There are 1414 neutrons in the Aluminum atom.

Explanation:

The mass number (AA) is the sum of protons and neutrons, while the atomic number (ZZ) represents the number of protons.