(O Level Chem) Formulas and dot-and-cross diagrams of ionic compounds

Ionic compounds contain cations (usually metal ions) and anions (non-metal ions).

When atoms form ions, they tend to achieve the electronic configuration of noble gases.

Examples:

• Beryllium (electronic configuration: 2,2) loses 2 electrons to form the duplet structure of helium (electronic configuration: 2). 

• Oxygen (electronic configuration: 2,6) gains 8 - 6 = 2 electrons to form the octet structure of neon (electronic configuration: 2,8).
  

Be --> Be²⁺ + 2e (loss of electrons: oxidation)

O + 2e --> O^2- (gain of electrons: reduction) 

What if you're asked to write out the formulas of unfamiliar ionic compounds like strontium chloride, silver astatide or gallium(III) sulfide? First, refer to the Periodic Table to find out which Group each element belongs to. Then decide the number of each type of ions needed to balance each others' charges.

• Strontium chloride: Sr (Group II) forms Sr²⁺ and chlorine (Group VII) forms Cl^-. One Sr²⁺ balances the 2 x -1 = -2 charge of two Cl^-. Formula: SrCl₂.
  
• Silver astatide: Silver is known to form Ag⁺ as in AgCl and AgNO₃. Astatide (Group VII) forms At^-. One Ag⁺ balances the -1 charge of one At^-. Formula: AgAt.
  
• Gallium(III) sulfide: As indicated by “(III)”, we write Ga³⁺. Sulfur (Group VI) forms S^2-. Two Ga³⁺ balances the 3 x -2 = -6 charge of three S^2-. Formula: Ga₂S₃.
  

What if you're asked to draw the dot-and-cross structures of unfamiliar ionic compounds and the question asks for only the valence (outer-most) electrons? We shall use dots for electrons from the metal atom and crosses for electrons from the non-metallic atom. The valence shell of the resultant cation should be empty, while the valence shell of the resultant anion should be full (have both dots and crosses). Remember to include charges as superscripts and number of ions as subscripts.

In the formation of ionic compounds, the outer electrons of the metal atom is transferred to the non-metallic atom.

Now, complete this table. Then form dot-and-cross ionic structures of your own using only the Groups I-VII elements.

Group number	Number of electrons involved to form an ion	Formation of ions
I (alkali metals)	1 valence electron ==> Atom will lose 1 electron	M --> M+ + e e.g. K --> K+ + e (oxidation)
II (alkaline metals)
III
IV
V
VI
VII (halogens)
0	2 (duplet) or 8 (octet) ==> Atom will neither lose nor gain electrons

Answer:

Group number	Number of electrons involved to form an ion	Formation of ions
I (alkali metals)	1 valence electron ==> Atom will lose 1 electron	M --> M+ + e e.g. K --> K+ + e (oxidation)
II (alkaline metals)	2 valence electrons ==> Atom will lose 2 electrons	M --> M2+ + 2e e.g. Ca --> Ca2+ + 2e (oxidation)
III	3 valence electrons ==> Atom will lose 3 electrons	M --> M3+ + 3e e.g. Al --> Al3+ + 3e (oxidation)
IV	4 valence electrons ==> Atom will gain 8 – 4 = 4 electrons	X + 4e --> X4- (reduction)
V	5 valence electrons ==> Atom will gain 8 – 5 = 3 electrons	X + 3e --> X3- e.g. N + 3e --> N3- (reduction)
VI	6 valence electrons ==> Atom will gain 8 – 6 = 2 electrons	X + 2e --> X2- e.g. O + 2e --> O2- (reduction)
VII (halogens)	7 valence electrons ==> Atom will gain 8 – 7 = 2 electrons	X + e --> X- e.g. Cl + e --> Cl- (reduction)
0	2 (duplet) or 8 (octet) ==> Atom will neither lose nor gain electrons