On the basis of their electron configurations, predict the formula of the simple binary ionic compounds likely to form when the following pairs of elements react with each other.
aluminum, and sulfur
radium, and oxygen
calcium, and fluorine
cesium, and nitrogen
The correct answer and explanation is :
When elements form ionic compounds, they do so by transferring electrons to achieve a stable electron configuration, typically a noble gas configuration. To predict the formulas of binary ionic compounds, we first consider the electron configurations of the involved elements and their tendency to either lose or gain electrons. Let’s break down each pair:
1. Aluminum (Al) and Sulfur (S):
- Aluminum (Al) has an electron configuration of ( [Ne]3s^2 3p^1 ). It has 3 valence electrons and tends to lose these 3 electrons to achieve a stable configuration, resembling the noble gas neon (Ne).
- Sulfur (S) has an electron configuration of ( [Ne]3s^2 3p^4 ). It has 6 valence electrons and needs 2 more electrons to complete its octet, forming a stable configuration like argon (Ar).
Ionic formation:
- Aluminum loses 3 electrons to form ( Al^{3+} ).
- Sulfur gains 2 electrons to form ( S^{2-} ).
To balance the charges, we need 2 aluminum ions (each with a charge of +3) to combine with 3 sulfur ions (each with a charge of -2). The ratio of ions is 2:3.
Formula: ( \text{Al}_2\text{S}_3 )
2. Radium (Ra) and Oxygen (O):
- Radium (Ra) has an electron configuration of ( [Rn] 7s^2 ). It has 2 valence electrons and tends to lose them to form ( Ra^{2+} ).
- Oxygen (O) has an electron configuration of ( [He] 2s^2 2p^4 ). It needs 2 electrons to complete its octet, forming ( O^{2-} ).
Ionic formation:
- Radium loses 2 electrons to form ( Ra^{2+} ).
- Oxygen gains 2 electrons to form ( O^{2-} ).
Since both ions have equal but opposite charges, they combine in a 1:1 ratio.
Formula: ( \text{RaO} )
3. Calcium (Ca) and Fluorine (F):
- Calcium (Ca) has an electron configuration of ( [Ar] 4s^2 ). It has 2 valence electrons and tends to lose them to form ( Ca^{2+} ).
- Fluorine (F) has an electron configuration of ( [He] 2s^2 2p^5 ). It needs 1 electron to complete its octet, forming ( F^{-} ).
Ionic formation:
- Calcium loses 2 electrons to form ( Ca^{2+} ).
- Fluorine gains 1 electron to form ( F^{-} ).
Since each calcium ion can pair with 2 fluorine ions (to balance the charges), the ratio is 1:2.
Formula: ( \text{CaF}_2 )
4. Cesium (Cs) and Nitrogen (N):
- Cesium (Cs) has an electron configuration of ( [Xe] 6s^1 ). It has 1 valence electron and tends to lose it to form ( Cs^{+} ).
- Nitrogen (N) has an electron configuration of ( [He] 2s^2 2p^3 ). It needs 3 electrons to complete its octet, forming ( N^{3-} ).
Ionic formation:
- Cesium loses 1 electron to form ( Cs^{+} ).
- Nitrogen gains 3 electrons to form ( N^{3-} ).
To balance the charges, we need 3 cesium ions (each with a +1 charge) to combine with 1 nitrogen ion (with a -3 charge). The ratio of ions is 3:1.
Formula: ( \text{Cs}_3\text{N} )
Summary:
- Aluminum and Sulfur: ( \text{Al}_2\text{S}_3 )
- Radium and Oxygen: ( \text{RaO} )
- Calcium and Fluorine: ( \text{CaF}_2 )
- Cesium and Nitrogen: ( \text{Cs}_3\text{N} )
In each case, the formula reflects the balance of charges between the cations and anions formed by the electron transfer, ensuring electrical neutrality in the resulting ionic compounds.