Consider the following representation based on long form of periodic table. [SEE DIAGRAM] Value of all four quantum number for last electron of element 'X' in their ground state is n = 4, l = 1, m = 1

Question

Consider the following representation based on long form of periodic table.
[SEE DIAGRAM]
Value of all four quantum number for last electron of element 'X' in their 
ground state is n = 4, l = 1, m = 1 and s = -1/2 and spin multiplicity of 
element 'X' in their ground state is 4.

Prashant Kotian · Accepted Answer

Correct answer: C. Step 1: Understand that electropositive character (metallic character) increases moving down a group and decreases moving from left to right across a period. Step 2: Alkali metals (Group 1) are always more electropositive than alkaline earth metals (Group 2) within the same period due to lower ionization energies. Step 3: Among the choices, [He] 2s1 (Lithium) and [Xe] 6s1 (Cesium) are alkali metals. Since Cesium is much further down Group 1 than Lithium, its outermost electron is most easily lost, making [Xe] 6s1 the configuration of the most electropositive element.

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