Humboldt State University ® Department of Chemistry

Richard A. Paselk

	General Chemistry	Fall 2008
Exercise: Nomenclature	© R. Paselk 2008
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Chemical Nomenclature - Binary Ionic Compounds

The Big Picture

• You can guess the charge on most elemental ions based on their positions on the Periodic table.
  • For positive ions (those in blue in the table above) the charge is the number of the group for the "A" group (Representative) elements.
    • For representative elements with more than one ion the higher charge will be the group number, the lower will be two lower, e.g. Bi(V) and Bi(III). Note that these ions occur near the bottom of the Periodic table (Periods 5 & 6).
  • For negative ions (those in purple-pink) the charge is the group number - 8. Thus for Sulfur in group VI the charge is 6 - 8 = -2.

    1

    2

    13

    14

    15

    16

    17

    18

    IA

    IIA

    IIIA

    IVA

    VA

    VIA

    VIIA

    VIIIA

    +1

    +2

    +3

    (+4)

    5-8 =-3

    6-8 =-2

    7-8 =-1

    Li

    Be

    C

    N

    O

    F

    Ne

    Na

    Mg

    IIIB

    IVB

    VB

    VI

    VIIB

    VIIIB

    IB

    IIB

    Al

    Si

    P

    S

    Cl

    Ar

    K

    Ca

    Fe

    Ni

    Cu

    Zn

    Ga

    As

    Se

    Br

    Kr

    Rb

    Sr

    Ag

    Cd

    In

    Sn

    Sb

    I

    Xe

    Cs

    Ba

    Hg

    Pb

    Bi

• Elemental ions are independent entities-subscripts show the number of independent ions.
  • FeI₃ exists as four ions = one iron and three iodides NOT one iron and a triiodide!
  • EXCEPTION: mercury(I) exists as the dimercurous ion = Hg₂²⁺
• Compounds have a charge of Zero, so the charges of the ions must add up to Zero.

Write the Formulae the Binary Ionic Compounds:

1. Sodium chloride

• First write out the formulae of the two ions.

  IA

  IIA

  IIIA

  IVA

  VA

  VIA

  VIIA

  VIIIA

  Na

  IIIB

  IVB

  VB

  VI

  VIIB

  VIIIB

  IB

  IIB

  Cl

  • Sodium (Na) is in Group IA, so it has a charge of +1 = Na⁺
  • Chloride has the -ide ending, so it is the elemental ion of Chlorine (Cl^-). Since it is in Group VIIA it has a charge of 7-8=-1 as seen on the Periodic chart above. Thus chloride = Cl^-
• Recall that all coumpound MUST have a charge of zero. Looking at the respective charges of the two ions, they will cancel each other out with a 1:1 ratio, (1+) + (1-) = 0. Thus the answer is NaCl

2. Calcium chloride

• Write out the formulae of the two ions.
  

  IA

  IIA

  IIIA

  IVA

  VA

  VIA

  VIIA

  VIIIA

  IIIB

  IVB

  VB

  VI

  VIIB

  VIIIB

  IB

  IIB

  Cl

  Ca

  • Calcium (Ca) is in Group IIA, so it has a charge of +2 = Ca²⁺
• Chloride has the -ide ending, so it is the elemental ion of Chlorine (Cl^-, note that it is NOT Cl_2^2-, rather there are TWO Cl^- ions) . As a Group VIIA it has a charge of 7-8 = -1 as seen on the Periodic chart above = Cl^-
• For a zero net charge there must be two Cl^- for each Ca²⁺ to cancel the charges. Thus the answer is CaCl₂

3. Aluminum sulfide

• Write out the formulae of the two ions.

  IA

  IIA

  IIIA

  IVA

  VA

  VIA

  VIIA

  VIIIA

  IIIB

  IVB

  VB

  VI

  VIIB

  VIIIB

  IB

  IIB

  Al

  S

  • Aluminum (Al) is in Group IIIA, so it has a charge of +3 = Al³⁺
  • Sulfide has the -ide ending, so it is the elemental ion of sulfur (S^2-) . As a Group VIA it has a charge of 6-8 = -2 as seen on the Periodic chart above = S^2-
• For a zero net charge we can cross multiply to find the coefficients to cancel the charges: 2( Al ³⁺) = 6+ and 3(S^2-) = 6-, (6+) + (6-) = 0. Thus the answer is Al₂S₃

4. Magnesium iodide

IA

IIA

IIIA

IVA

VA

VIA

VIIA

VIIIA

Mg

IIIB

IVB

VB

VI

VIIB

VIIIB

IB

IIB

I

• Ions: magnesium = Mg²⁺ (Group IIA); iodide = I^- (Group VIIA, 7-8 = -1)
• For a zero net charge need two iodides for each magnesium, (2+) + 2(1-) = 0; MgI₂

5. Potassium phosphide

IA

IIA

IIIA

IVA

VA

VIA

VIIA

VIIIA

IIIB

IVB

VB

VI

VIIB

VIIIB

IB

IIB

P

K

• Ions: Potassium = K⁺ (Group IA); phosphide = P^3- (Group VA, 5-8 = -3)
• For a zero net charge need three potassiums for each phosphide, 3(+1) + (-3) = 0; K₃P

6. Zinc oxide

IA

IIA

IIIA

IVA

VA

VIA

VIIA

VIIIA

Li

Be

C

N

O

F

Ne

Na

Mg

IIIB

IVB

VB

VI

VIIB

VIIIB

IB

IIB

Al

Si

P

S

Cl

Ar

K

Ca

Fe

Ni

Cu

Zn

Ga

As

Se

Br

Kr

Rb

Sr

Ag

Cd

In

Sn

Sb

I

Xe

Cs

Ba

Hg

Pb

Bi

• Ions: Zinc = Zn²⁺ (Group IIB-Memorize Zn charge); oxide = O^2- (Group VIA, 6-8 = -2)
• For a zero net charge need 1:1, (2+) + (2-) =0; ZnO

7. Galium selenide

• Ions: Galium = Ga³⁺ (Group IIIB); sulfide = Se^2- (Group VIA, 6-8 = -2)
• For a zero net charge we can cross multiply to find the coefficients to cancel the charges: 2( Ga³⁺) = 6+ and 3(S^2-) = 6-, (6+) + (6-) = 0; Ga₂Se₃

8. Iron(II) nitride

• Iron(II) = Fe²⁺ (II = +2); nitride = N^3- (Group VA, 5-8 = -3)
• 3(Fe²⁺) = 6+ and 2(N^3-) = 6-, (6+) + (6-) = 0; Fe₃N₂

9. Lead(IV) phosphide

• Pb(IV) = Pb ⁴⁺ (IV = +4); phosphide = P ^3- (Group VA, 5-8 = -3)
• 3(Pb⁴⁺) = 12+ and 4(P^3-) = 12-, (12+) + (12-) = 0; Pb₃P₄

10. Mercury(II) selenide

• Hg(II) = Hg²⁺ (II = +2); selenide = Se^2- (Group VIA, 6-8 = -2)
• Hg²⁺ = 2+ and Se^2- = 2-, (2+) + (2-) =0; HgSe

11. Mercury(I) sulfide

• Hg(I) = Hg₂²⁺ (I = +1; mercury(I) is exception as a di-ion); sulfide = S^2- (Group VIA, 6-8 = -2)
• Hg₂²⁺ = 2(+1) = 2+ and Se^2- = 2-, (2+) + (2-) =0; Hg₂S

12. Tin(II) fluoride

• Sn(II) = Sn ²⁺ (II = +2); flouride= F^- (Group VIIA, 7-8 = -1)
• Sn²⁺ = 2+ and 2 F^- = 2-, (2+) + 2(1-) =0; SnF₂

 

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