Question

Which of the following is favourable condition for the formation of ionic bond ?

• A. Low ionization enthalpy of metal and low negative value of electron gain enthalpy of non metal.
• B. Low ionization enthalpy of metal and high negative value of electron gain enthalpy of non metal.
• C. High ionization enthalpy of metal and high negative value of electron gain enthalpy of non metal.
• D. High ionization enthalpy of metal and low negative value of electron gain enthalpy of non metal.

Hint:

An ionic bond forms easily when the metal loses electrons easily and the non-metal gains them with a large release of energy.

Answer 1

The Correct Option is B

Step 1: Understand the concept
An ionic bond forms by complete transfer of electrons from a metal to a non-metal. The metal becomes a cation and the non-metal becomes an anion. These ions then attract each other.

Step 2: Condition for the metal
Forming the cation needs energy equal to the ionization enthalpy. A low ionization enthalpy makes it cheap for the metal to lose its electron.

Step 3: Condition for the non-metal
Forming the anion releases energy equal to the electron gain enthalpy. A highly negative value means more energy is released, so the anion forms readily.

Step 4: Check the options
Option (A) has a poorly negative electron gain enthalpy, so the non-metal is unwilling to accept electrons. Option (C) makes the metal reluctant to lose electrons. Option (D) has both factors unfavourable. Only option (B) has both factors favourable, and a large lattice energy then stabilises the solid.

Final Answer:
Low ionization enthalpy of the metal with high negative electron gain enthalpy of the non-metal favours ionic bonding. This is option (B). \[ \boxed{\text{(B) }\text{Low IE of metal, high negative } \Delta_{eg}H \text{ of non-metal}} \]