Question

Rubidium (Rb) has an atomic number of 37. It can lose 1 electron to form an ion. What would its charge be?

a. 1–
b. 1
c. 0
d. 2+

Answer 1

Rubidium (Rb) with an atomic number of 37 forms a cation with a +1 charge upon losing one electron. The resulting rubidium ion is denoted as Rb+. The correct option is B.

Step-by-step explanation:

When rubidium (Rb) with an atomic number of 37 loses one electron to form an ion, it obtains a charge of +1. This is due to the fact that when an atom loses an electron, it has one more proton than electrons, which results in a positive charge.

Therefore, the charge of the rubidium ion would be +1, corresponding to option (c) Rb: +1 from the provided selections. The abbreviated electron configuration for Rb, which has 37 electrons, would be [Kr]5s1.

For example, just like potassium (K), which also loses one electron to form a cation (K+), rubidium follows the same trend due to it being in the same group (alkali metals) on the periodic table. Thus, the resulting rubidium ion is symbolized as Rb+. The correct option is B.