The answer is B NaCI solid
The Rutherford–Bohr model of the hydrogen atom (Z = 1) or a hydrogen-like ion (Z > 1). In this model it is an essential feature that the photon energy (or frequency) of the electromagnetic radiation emitted (shown) when an electron jumps from one orbital to another, be proportional to the mathematical square of atomic charge (Z2). Experimental measurement by Henry Moseley of this radiation for many elements (from Z = 13 to 92) showed the results as predicted by Bohr. Both the concept of atomic number and the Bohr model were thereby given scientific credence. The atomic number is the number of _z_ an atom.
Answer is: a. Rubidium (Rb) is more reactive than strontium (Sr) because strontium atoms must lose more electrons.
The ionization energy (Ei) is the minimum amount of energy required to remove the valence electron, when element lose electrons, oxidation number of element grows (oxidation process).
Alkaline metals (group 1), in this example rubidium, have lowest ionizations energy and easy remove valence electrons (one electron), they are most reactive metals.
Earth alkaline metals (group 2), in this example strontium, have higher ionization energy than alkaline metals, because they have two valence electrons, they are less reactive.
Rubidium electron configuration: ₃₇Rb 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s¹; one valence electron is 5s¹ orbital.
Strontium electron configuration: ₃₈Sr 1s²2s²2p⁶3s²3p⁶3d¹⁰4s²4p⁶5s²; two valence electrons is 5s² orbital.
