Question

The atomic number of the element from the following with lowest $ 1^{\text{st}} $ ionization enthalpy is:

• A. 87
• B. 19
• C. 32
• D. 35

Hint:

- Ionization energy \(\propto \frac{1}{\text{atomic size}}\) - Alkali metals have lowest ionization energies in their periods - Within Group 1, ionization energy decreases down the group - Francium has the lowest known ionization energy (380 kJ/mol)

Answer 1

The Correct Option is A

The objective is to identify the element with the lowest first ionization enthalpy from the provided atomic numbers. Ionization enthalpy signifies the energy required to detach an electron from a gaseous atom or ion. Generally, elements with lower ionization enthalpy can release electrons more readily.

Step-by-step Explanation:

1. The given atomic numbers are: 87, 19, 32, 35.
2. Corresponding elements from the periodic table are:
   • Atomic number 87: Francium (Fr)
   • Atomic number 19: Potassium (K)
   • Atomic number 32: Germanium (Ge)
   • Atomic number 35: Bromine (Br)
3. Ionization enthalpy decreases down a group in the periodic table due to increased distance of valence electrons from the nucleus and greater shielding by inner electrons.
4. Among these elements:
   • Francium (Fr), an alkali metal in Group 1, possesses low ionization energies.
   • Potassium (K) is also in Group 1 but is positioned above Francium.
   • Germanium (Ge) and Bromine (Br) are located further right in the periodic table and exhibit higher ionization enthalpies than Group 1 elements.
5. Consequently, Francium, being at the bottom of Group 1, exhibits the lowest ionization enthalpy among the listed elements.

Conclusion: Among the given options, Francium (Atomic number 87) has the lowest first ionization enthalpy.