Question

which of the following ions has the same number of electrons as br^-?
a k^+
b sr^{2+}
c i^-
d cl^-

Explanation:

Step1: Determine the number of electrons in \(Br^-\)

Bromine (\(Br\)) has an atomic number of \(35\). When it gains one electron to form \(Br^-\), the number of electrons is \(35 + 1=36\).

Step2: Analyze each ion:

• For \(K^+\): Potassium (\(K\)) has an atomic number of \(19\). When it loses one electron to form \(K^+\), the number of electrons is \(19- 1 = 18\).
• For \(Sr^{2+}\): Strontium (\(Sr\)) has an atomic number of \(38\). When it loses two electrons to form \(Sr^{2+}\), the number of electrons is \(38-2 = 36\).
• For \(I^-\): Iodine (\(I\)) has an atomic number of \(53\). When it gains one electron to form \(I^-\), the number of electrons is \(53 + 1=54\).
• For \(Cl^-\): Chlorine (\(Cl\)) has an atomic number of \(17\). When it gains one electron to form \(Cl^-\), the number of electrons is \(17+1 = 18\).

Answer:

B. \(Sr^{2+}\)