## Chemistry (7th Edition)

Published by Pearson

# Chapter 2 - Atoms, Molecules, and Ions - Section Problems - Page 74: 124

#### Answer

(a) $Be^{2+}$ has 4 protons and 2 electrons. (b) $Rb^+$ 37 protons and 36 electrons. (c) $Se^{2-}$ 34 protons and 36 electrons. (d) $Au^{3+}$ 79 protons and 76 electrons,

#### Work Step by Step

The number of protons is equal to the atomic number (Z), which can be found in a periodic table, next to the symbol of the element. To calculate the number of electrons, since each one represents a negative charge, and each proton a postive charge, we can use this formula: $(charge) = Z - n^{\circ}e^-$ $n^{\circ}e^- = Z - (charge)$ (a) Atomic number for $Be$: $4$ $n^{\circ}e^- = 4 - (2+) = 4 - 2 = 2e^-$ (b) Atomic number for $Rb$: $37$ $n^{\circ}e^- = 37 - (1+) = 37 - 1 = 36e^-$ (c) Atomic number for $Se$: $34$ $n^{\circ}e^- = 34 - (-2) = 34 + 2 = 36e^-$ (d) Atomic number for $Au$: $79$ $n^{\circ}e^- = 79 - (3+) = 79 - 3 = 76e^-$

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