Question

how many electrons are in the following ions?
mg²+
al³+
s²-
na+
cl-
cu²+
n³-
cs+
br-

Explanation:

Step1: Recall the atomic number concept

The atomic number of an element gives the number of protons and for a neutral atom, the number of electrons. For ions, if it's a cation (positive ion), electrons are lost (number of electrons = atomic number - charge). If it's an anion (negative ion), electrons are gained (number of electrons = atomic number+ charge).

• For \(Mg^{2 +}\):

Atomic number of \(Mg\) is \(12\). Since it has a \(+ 2\) charge, number of electrons \(=12 - 2=10\)

• For \(Al^{3+}\):

Atomic number of \(Al\) is \(13\). Since it has a \(+3\) charge, number of electrons \(=13 - 3 = 10\)

• For \(S^{2-}\):

Atomic number of \(S\) is \(16\). Since it has a \(-2\) charge, number of electrons \(=16+2 = 18\)

• For \(Na^{+}\):

Atomic number of \(Na\) is \(11\). Since it has a \(+1\) charge, number of electrons \(=11 - 1=10\)

• For \(Cl^{-}\):

Atomic number of \(Cl\) is \(17\). Since it has a \(-1\) charge, number of electrons \(=17 + 1=18\)

• For \(Cu^{2+}\):

Atomic number of \(Cu\) is \(29\). Since it has a \(+2\) charge, number of electrons \(=29-2 = 27\)

• For \(N^{3-}\):

Atomic number of \(N\) is \(7\). Since it has a \(-3\) charge, number of electrons \(=7+3 = 10\)

• For \(Cs^{+}\):

Atomic number of \(Cs\) is \(55\). Since it has a \(+1\) charge, number of electrons \(=55 - 1=54\)

• For \(Br^{-}\):

Atomic number of \(Br\) is \(35\). Since it has a \(-1\) charge, number of electrons \(=35+1 = 36\)

Answer:

\(Mg^{2+}:10\), \(Al^{3+}:10\), \(S^{2-}:18\), \(Na^{+}:10\), \(Cl^{-}:18\), \(Cu^{2+}:27\), \(N^{3-}:10\), \(Cs^{+}:54\), \(Br^{-}:36\)