Question

i have five more protons than the heaviest metalloid in the fourth period. who am i? view periodic table tap on an option below to select or deselect an answer selenium (se) cesium (cs) zirconium (zr) rubidium (rb) yttrium (y) niobium (nb) strontium (sr) barium (ba) iodine (i) bromine (br)

Explanation:

Step1: Identify the heaviest metalloid in the fourth period

The heaviest metalloid in the fourth period is germanium (Ge) with atomic number \(Z = 32\) (since the atomic number is equal to the number of protons).

Step2: Calculate the number of protons of the unknown element

The unknown element has five more protons than germanium. So the number of protons \(n=32 + 5=37\).

Step3: Identify the element with 37 protons

Looking at the periodic table, the element with atomic number \(Z = 37\) is rubidium (Rb).

Answer:

Rubidium (Rb)