The question is about determining which element has the lowest first ionization enthalpy from the given atomic numbers. Ionization enthalpy is the energy required to remove an electron from a gaseous atom or ion. Typically, elements with a lower ionization enthalpy can lose electrons more easily.
Step-by-step Explanation:
Conclusion: The element with the lowest first ionization enthalpy among the options is Francium (Atomic number 87).
Concept:
The first ionization enthalpy decreases:
- Down a group (atomic size increases)
- From right to left across a period (effective nuclear charge decreases)
Analysis of Options:
Option 1 (87): Francium (Fr) - Group 1, Period 7 element - Largest atomic size in periodic table - Lowest effective nuclear charge on valence electron - Lowest ionization energy among given options.
Option 2 (19): Potassium (K) - Group 1, Period 4 - Higher ionization energy than Fr (smaller size).
Option 3 (32): Germanium (Ge) - Group 14, Period 4 - Much higher ionization energy than alkali metals.
Option 4 (35): Bromine (Br) - Group 17, Period 4 - Highest ionization energy among options (high effective nuclear charge).
Periodic Trend:
Ionization Energy Order: Fr (87) < K (19) < Ge (32) < Br (35)
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)

Cobalt chloride when dissolved in water forms pink colored complex $X$ which has octahedral geometry. This solution on treating with cone $HCl$ forms deep blue complex, $\underline{Y}$ which has a $\underline{Z}$ geometry $X, Y$ and $Z$, respectively, are
Given below are two statements:
Statement (I): According to the Law of Octaves, the elements were arranged in the increasing order of their atomic number.
Statement (II): Meyer observed a periodically repeated pattern upon plotting physical properties of certain elements against their respective atomic numbers.
In the light of the above statements, Choose the correct answer from the options given below:
Given below are two statements:
Statement I: $ H_2Se $ is more acidic than $ H_2Te $
Statement II: $ H_2Se $ has higher bond enthalpy for dissociation than $ H_2Te $
In the light of the above statements, choose the correct answer from the options given below.
What will be the equilibrium constant of the given reaction carried out in a \(5 \,L\) vessel and having equilibrium amounts of \(A_2\) and \(A\) as \(0.5\) mole and \(2 \times 10^{-6}\) mole respectively?
The reaction : \(A_2 \rightleftharpoons 2A\)