Final answer:
To illustrate electron transfer, arrows are drawn to show potassium and calcium donating electrons to sulfur and chlorine, respectively, leading to the formation of potassium sulfide (K2S) and calcium chloride (CaCl2). Half-headed arrows represent the movement of single electrons, whereas double-headed arrows represent the movement of electron pairs during bond formation.
Step-by-step explanation:
The process of electron transfer to form ionic compounds involves the movement of electrons from one atom to another, resulting in the formation of cations and anions. In the formation of potassium sulfide (K2S), two potassium (K) atoms each donate one electron to a sulfur (S) atom, which needs two electrons to complete its octet. The two potassium atoms become potassium cations (K+), and the sulfur atom becomes a sulfide anion (S2-).
For the formation of calcium chloride (CaCl2), a calcium (Ca) atom donates two electrons, one to each of the two chlorine (Cl) atoms. As a result, the calcium atom becomes a calcium cation (Ca2+), and each chlorine atom becomes a chloride anion (Cl-).
Half-headed arrows are used to indicate the movement of a single electron to form free radicals, while double-headed arrows show the movement of an electron pair during heterolytic bond formation. The formal charge changes accordingly as atoms transform into ions.
- 2 K → 2 K+ + 2e- (Loss of electrons - Oxidation)
- S + 2e- → S2- (Gain of electrons - Reduction)
- Ca → Ca2+ + 2e- (Loss of electrons - Oxidation)
- 2 Cl + 2e- → 2 Cl- (Gain of electrons - Reduction)
In both reactions, electrons are balanced by ensuring the number lost by the oxidized species is equal to the number gained by the reduced species.