Question:medium

Observe the following data (\(\Delta_i H_1\), \(\Delta_i H_2\) and \(\Delta_{eg}H\) represent the first, second ionisation enthalpies and electron gain enthalpy respectively).
Using the data identify the most reactive metal.

Show Hint

For identifying the most reactive metal, the primary factor to look for is the lowest first ionization enthalpy (\(\Delta_i H_1\)). For the most reactive non-metal, look for the most negative electron gain enthalpy (\(\Delta_{eg}H\)).
Updated On: Mar 26, 2026
  • II
  • I
  • IV
  • III
Show Solution

The Correct Option is A

Solution and Explanation

To determine the most reactive metal using the provided ionization and electron gain enthalpy data, we need to understand how these values correlate with metallic reactivity:

  1. First Ionisation Enthalpy (\(\Delta_i H_1\)): The energy required to remove the first electron. A lower value indicates the metal can easily lose an electron, suggesting higher reactivity.
  2. Second Ionisation Enthalpy (\(\Delta_i H_2\)): The energy required to remove the second electron. It is important for elements that can form ions with a +2 charge.
  3. Electron Gain Enthalpy (\(\Delta_{eg} H\)): The energy change when an electron is added. This value impacts non-metal reactivity more; however, negative values (indicating a natural tendency to gain electrons) can still influence metal reactivity indirectly.

Let's analyze the given data:

Element\(\Delta_i H_1\) (kJ mol\(^{-1}\))\(\Delta_i H_2\) (kJ mol\(^{-1}\))\(\Delta_{eg}H\) (kJ mol\(^{-1}\))
I5207300-60
II4903051-48
III16813374-328
IV23725251+48

Conclusion:

  • Element II has the lowest first ionisation enthalpy (490 kJ mol\(^{-1}\)), which means it can lose an electron more easily than the others.
  • The second ionisation enthalpy and electron gain enthalpy values of Element II indicate it can reasonably stabilize after losing an electron compared to others.

Therefore, the most reactive metal is Element II. This aligns with the correct answer option.

Was this answer helpful?
0