Chapter 2 - Atoms, Molecules, and Ions - Additional Exercises - Page 78: 97

Element X $\Longrightarrow$ Radium (Ra) Number of neutrons $\Longrightarrow$ 142

Since the formula of the compound is $XBr_{2}$ , and Br has 7 electrons in its valence (last) shell, which means Bromine's ion has a net negative charge = -1. Thus ion is $Br^{-}$ Since the bromine ion is $Br^{-}$ - or a single negative charge, and the compound $XBr_{2}$ has two Bromine atoms and only one atom of element "X". This means element X must have a net positive charge = +2, to balance the two single negative charges of the Bromine ions. Thus ion of X is $X^{2+}$ Since the ion of X has 86 electrons, then the neutral atom should have 2 more electrons, since the positively charged ion has a deficit of two electrons ( and hence is positive) Thus number of electrons of neutral atom of X = 86 + 2 = 88. Since X is a neutral atom , number of electrons = number of protons. Thus number of protons = 88. Also, number of protons = atomic number of X. Thus atomic number of X is 88. So element X is Radium. Atomic number = Z = 88 Mass number = A = 230 Thus number of neutrons = A - Z = 230 - 88 = 142