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Factors Influencing the Formation of Ionic Bond

Following are the factors that favour the formation and stability of ionic bonds:
  1. Low Ionisation Energy
    Ionisation energy of an element is the amount of energy required to remove an electron from the outermost shell of an isolated neutral gaseous atom to form gaseous cation. Lesser the ionization energy, easier is the removal of an electron. Ionisation energy of alkali metals (Na, K etc.) is low thus; they have more tendencies to form positive ions. The ionization energy of alkaline earth metals (Mg, Ca etc.) is higher and they form cations not so easily. For example, ionization energy of sodium is 495 kJ mol-1 while that of Magnesium is 743 kJ mol-1
    Na(g) Na+(g) +e ;   Mg(g) Mg+(g)+ e

General trends in ionization energy with reference to the periodic table

The energy required to remove second electron is very high and thus, the formation of Mg2+ ion is relatively difficult. In case of Aluminium, the formation of Al3+ ion requires 3158 kJ mol-1. This much energy is generally not available and hence Aluminium does not form ionic bond. For ionic bond formation, the Ionisation energy of the metal should be low.
  1. Electron gain Enthalpy
    It is defined as the amount of energy released when an extra electron is added to an isolated gaseous atom to form gaseous anion.
    Higher the electron affinity more is the quantity of energy released and easier is the formation of ionic bond. The electron affinity of chlorine is the highest in the periodic table. It is 348 kJ mol-1. Group 16 elements (oxygen, sulphur etc.) do form divalent anions and in these cases, the formation of divalent anion from monovalent anion requires energy. Thus group 16 elements form ionic bonds with lesser ease compared to group 17 elements. For ionic bond formation, the electron affinity of an element should be high.
  2. Atomic Radii
    The size of an atom can be visualized from its atomic radius, which is defined as the distance between the nucleus and the outermost shell of electrons in an atom. To estimate the atomic radius, the distance between the nuclei of two covalently bonded atoms is measured spectroscopically. The bond distance is then divided between the bonded atoms to calculate the atomic radius. For example, the bond distance in hydrogen molecule (H2) is found to be 74 pm. Therefore, the atomic radius of hydrogen atom is equal to 74 pm/2 =37 pm. In metals, atomic radius is taken as half the internuclear distance separating the metal ions in the crystal.

General trends in atomic radius with reference to periodic table.

  1. Lattice Energy
    For ionic bond formation, the oppositely charged ions combine with the large decrease in energy.
    The energy released when the oppositely charged ions combine to form one mole of the ionic compound is called Lattice energy.
    The higher the value of lattice energy of the ionic compound, the greater is the stability of the compound and also easier is its formation.
    The force of attraction between the oppositely charged ions is directly proportional to the magnitude of charges and varies inversely to the square of the distance between them.
    Force of attraction =
    where q1,q2 are oppositely charged ions and D is the distance between them. Hence, the value of Lattice energy of an ionic compound depends upon the following factors:
    1. Charge on the Ions
      The higher the magnitude of the charge of the ions, greater is the force of attraction and higher is the lattice energy (or amount of energy released).
    2. Size of the Ions
      We know that highly charged ions are rare. Smaller the sizes of the ions, lesser will be the internuclear distance and greater is the force of attraction.

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