1. When a potassium atom reacts with a bromine atom, the

bromine atom will
a) lose 1 electron and form a positive ion
b) lose 2 electrons and form a positive ion
c) gain 1 electron and form a negative ion
d) gain 2 electrons and form a negative ion

I know its either C/D but im not sure which one.

To determine the correct answer, we need to look at the electron configurations of potassium (K) and bromine (Br) atoms.

A potassium atom has an electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s¹. It has one valence electron in the 4s orbital.

A bromine atom has an electron configuration of 1s² 2s² 2p⁶ 3s² 3p⁶ 4s² 3d¹⁰ 4p⁵. It has seven valence electrons in the 4s and 4p orbitals.

Based on the concept of electron configuration and the octet rule, atoms tend to gain or lose electrons to achieve a stable electron configuration similar to the noble gases.

In the case of potassium (K), it has a single valence electron in the 4s orbital. To achieve a stable electron configuration similar to argon (Ar) with 18 electrons, it is easier for potassium to lose one electron. This would result in a neutral potassium ion with a +1 charge, denoted as K+.

On the other hand, bromine (Br) has seven valence electrons in the 4s and 4p orbitals. To achieve a stable electron configuration similar to krypton (Kr) with 36 electrons, it is easier for bromine to gain one electron to complete its outer energy level. This would result in a negative bromide ion, denoted as Br-.

Therefore, the correct answer is option (c) gain 1 electron and form a negative ion.

When a potassium atom (K) reacts with a bromine atom (Br), the bromine atom will gain one electron and form a negative ion. Therefore, the correct answer is option c) gain 1 electron and form a negative ion.

Br is in group VII (or 17 depending upon the system your prof is using; either system, however, tells you there are 7 electrons in the outside shell. That means 1 electron is missing to make an outside shell of 8; therefore, it gains the 1 electron becoming a -1 charge in the process.