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Grade 10CBSE

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

🔑Concepts

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Reactivity of elements is explained as their tendency to attain a completely filled valence shell, also known as the noble gas configuration or the octet rule.

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Metals lose electrons from their valence shell to form positive ions called cations. For example, Sodium (NaNa) loses one electron to become Na+Na^+.

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Non-metals gain electrons into their valence shell to form negative ions called anions. For example, Chlorine (ClCl) gains one electron to become Cl−Cl^-.

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Ionic Compounds (or Electrovalent Compounds) are formed by the transfer of electrons from a metal to a non-metal, resulting in strong electrostatic forces of attraction between the oppositely charged ions.

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Physical nature: Ionic compounds are solids and are somewhat hard because of the strong force of attraction between the positive and negative ions.

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Melting and Boiling Points: Ionic compounds have high melting and boiling points because a considerable amount of energy is required to break the strong inter-ionic attraction.

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Solubility: Electrovalent compounds are generally soluble in water and insoluble in solvents such as kerosene, petrol, etc.

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Conduction of Electricity: Ionic compounds conduct electricity in the molten state or in an aqueous solution where ions are free to move, but they do not conduct electricity in the solid state because the movement of ions is restricted.

📐Formulae

Na→Na++e−Na \rightarrow Na^+ + e^- (Formation of Sodium Cation)

Cl+e−→Cl−Cl + e^- \rightarrow Cl^- (Formation of Chloride Anion)

Na++Cl−→NaClNa^+ + Cl^- \rightarrow NaCl (Formation of Sodium Chloride)

Mg→Mg2++2e−Mg \rightarrow Mg^{2+} + 2e^- (Formation of Magnesium Cation)

2Cl+2e−→2Cl−2Cl + 2e^- \rightarrow 2Cl^- (Formation of two Chloride Anions)

Mg2++2Cl−→MgCl2Mg^{2+} + 2Cl^- \rightarrow MgCl_2 (Formation of Magnesium Chloride)

💡Examples

Problem 1:

Explain the formation of Sodium Chloride (NaClNaCl) using electron-dot structures.

Solution:

Sodium (Z=11Z=11) has the electronic configuration 2,8,12, 8, 1. It loses its valence electron to achieve an octet: Na→Na+(2,8)+e−Na \rightarrow Na^+ (2, 8) + e^-. Chlorine (Z=17Z=17) has the configuration 2,8,72, 8, 7. It needs one electron to complete its octet: Cl+e−→Cl−(2,8,8)Cl + e^- \rightarrow Cl^- (2, 8, 8). The transfer of one electron from NaNa to ClCl results in the formation of Na+Na^+ and Cl−Cl^-, which are held together by electrostatic forces to form NaClNaCl.

Explanation:

The metal atom loses electrons to become a cation, and the non-metal atom gains those electrons to become an anion. The resulting compound is neutral.

Problem 2:

Why do ionic compounds have high melting points?

Solution:

Ionic compounds consist of a giant lattice structure of oppositely charged ions held together by very strong electrostatic forces of attraction. To overcome these strong forces and move the ions out of their fixed positions, a large amount of thermal energy is required.

Explanation:

High melting point is a direct consequence of the strength of the ionic bond (F∝q1q2r2F \propto \frac{q_1 q_2}{r^2}).

Problem 3:

Show the formation of Magnesium Oxide (MgOMgO) by the transfer of electrons.

Solution:

Magnesium (Z=12Z=12) configuration is 2,8,22, 8, 2. It loses two electrons: Mg→Mg2++2e−Mg \rightarrow Mg^{2+} + 2e^-. Oxygen (Z=8Z=8) configuration is 2,62, 6. It gains two electrons: O+2e−→O2−O + 2e^- \rightarrow O^{2-}. These ions combine: Mg2++O2−→MgOMg^{2+} + O^{2-} \rightarrow MgO.

Explanation:

In MgOMgO, two electrons are transferred from one MgMg atom to one OO atom to satisfy the octet requirements of both.