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Chemistry LibreTexts

Ionic Compounds

  • Page ID
    33778
  • Skills to Develop

    • Explain stability of ions in terms of electronic configuration.
    • Define lattice energy.
    • Point out the trend of lattice energies in a series of compounds.
    • Draw the Born Haber Cycle and calculate the energy that can not be measured for a process.

    Ionic Compounds

    The common table salt, \(\ce{NaCl}\), is a representative of an important class of compounds called salts. A salt consists of positive and negative ions, \(\ce{Na+}\) and \(\ce{Cl-}\) in \(\ce{NaCl}\) . Salts such as \(\ce{KCl}\) , \(\ce{CsCl}\) , \(\ce{CaCl2}\), \(\ce{CsF}\), \(\ce{KClO4}\), \(\ce{NaNO3}\), and \(\ce{CaSO4}\) are generally considered ionic compounds, which are composed of (positive) cations and (negative) anions. When they dissolve in water, the solutions contain hydrated positive and negative ions.

    A salt is produced in a neutralization reaction or in a direct reaction. For example, \(\ce{NaCl}\) solid can be obtained by

    \(\ce{NaOH + HCl \rightarrow H2O + NaCl}\) (solid after evaporation of solvent)
    \(\ce{2 Na + Cl_{2\large{(g)}} \rightarrow NaCl}\) (solid) (burning metallic sodium in a chlorine gas.)

    Monoatomic Ions

    Halide ions are formed when a group 7A element acquires an electron to attain the electronic configuration of a noble gas. Group 6A elements need to acquire two, instead of one, electron to achieve the same, whereas a nitride needs to gain three electrons. Similarly, alkali metals, alkali earth metals, and group 3A elements lose 1, 2, and 3 electrons respectively to attain the same stablilty. The following groups of ions have the same electronic configurations.

    \(\mathrm{[He] = 1s^2}\) \(\ce{H-}\) \(\ce{He}\) \(\ce{Li+}\) \(\ce{Be^2+}\) \(\ce{B^3+}\)
    \(\mathrm{[Ne] = 2s^22p^6}\) \(\ce{N^3-}\) \(\ce{O^2-}\) \(\ce{F-}\) \(\ce{Ne}\) \(\ce{Na+}\) \(\ce{Mg^2+}\) \(\ce{Al^3+}\)
    \(\mathrm{[Ar]}\) \(\ce{S^2-}\) \(\ce{Cl-}\) \(\ce{Ar}\) \(\ce{K+}\) \(\ce{Ca^2+}\)
    \(\mathrm{[Kr]}\) \(\ce{Se^2-}\) \(\ce{Br-}\) \(\ce{Kr}\) \(\ce{Rb+}\) \(\ce{Sr^2+}\)
    \(\ce{Cu+}\) \(\ce{Zn^3+}\) \(\ce{Ga^3+}\) \(\ce{Ge^4+}\)
    \(\ce{Ag+}\) \(\ce{Cd^2+}\) \(\ce{In^3+}\) \(\ce{Sn^4+}\)
    \(\mathrm{[Xe]}\) \(\ce{I-}\) \(\ce{Xe}\) \(\ce{Cs+}\) \(\ce{Ba^2+}\)

    This list gives you positive and negative ions. Some of them have electronic configurations of some noble gases. Thus, the formation of these ions can be attributed to the tendency of an atom to acquire the electronic configuration of its nearest noble gas on the periodic table of elements.

    Ions of Transition Metals

    Transition metal ions also appear frequently; some of the common mono-atomic ions of transition metals are given below with their electronic configurations indicated. The stable cores of noble gases are indicated by [], whereas the number of d electrons are indicated by a superscript number after d.

    • Transition-metals lose n s electrons first, and then they may lose one or more (n-1) d electreons to form ions.
    • Transition metal ions are usually colored, due to the d-d transitions. The d-sub shell is split for ions in solutions.
    • Electronic configurations for transition metal ions.
    Ions Electronic
    Configuration
    Remark
    \(\ce{Sc}\) \(\ce{[Ar]}4s^2 3d^1\) Metal
    \(\ce{Sc^2+}\) \(\ce{[Ar]} 3d^1\) Color ion
    \(\ce{Sc^3+}\) \(\ce{[Ar]}\) Colorless
    \(\ce{Ti}\) \(\ce{[Ar]}4s^2 3d^2\)
    \(\ce{Ti^4+}\) \(\ce{[Ar]}\) Colorless
    \(\ce{V}\) \(\ce{[Ar]}4s^2 3d^3\)
    \(\ce{V^2+}\) \(\ce{[Ar]} 3d^3\) \(\ce{V^2+_{\large{(aq)}}}\)
    \(\ce{V^3+}\) \(\ce{[Ar]} 3d^2\) \(\ce{V^3+_{\large{(aq)}}}\)
    \(\ce{Cr}\) \(\ce{[Ar]}4s^1 3d^5\)
    \(\ce{Cr^3+}\) \(\ce{[Ar]} 3d^3\) \(\ce{Cr^3+_{\large{(aq)}}}\)
    \(\ce{Mn}\) \(\ce{[Ar]}4s^2 3d^5\)
    \(\ce{Mn^2+}\) \(\ce{[Ar]} 3d^5\) \(\ce{Mn^2+_{\large{(aq)}}}\)
    \(\ce{Fe}\) \(\ce{[Ar]}4s^2 3d^6\)
    \(\ce{Fe^2+}\) \(\ce{[Ar]} 3d^6\) \(\ce{Fe^2+_{\large{(aq)}}}\)
    \(\ce{Fe^3+}\) \(\ce{[Ar]} 3d^5\) \(\ce{Fe^3+_{\large{(aq)}}}\)
    \(\ce{Co}\) \(\ce{[Ar]}4s^2 3d^7\)
    \(\ce{Co^2+}\) \(\ce{[Ar]} 3d^7\) \(\ce{Co^2+_{\large{(aq)}}}\)
    \(\ce{Ni}\) \(\ce{[Ar]}4s^2 3d^8\)
    \(\ce{Ni^2+}\) \(\ce{[Ar]} 3d^8\) \(\ce{Ni^2+_{\large{(aq)}}}\)
    \(\ce{Cu}\) \(\ce{[Ar]}4s^1 3d^{10}\)
    \(\ce{Cu^1+}\) \(\ce{[Ar]} 3d^{10}\) Colorless
    \(\ce{Cu^2+}\) \(\ce{[Ar]} 3d^9\) \(\ce{Cu^2+_{\large{(aq)}}}\)
    \(\ce{Zn}\) \(\ce{[Ar]}4s^2 3d^{10}\)
    \(\ce{Zn^2+}\) \(\ce{[Ar]} 3d^{10}\) Colorless

    Ionic Radii

    • Ionic radii of the same charge increase for a group as the atomic numbers increase.
    • Isoelectronic ions are ions with the same electronic configuration.
    • Across a period, the cations decrease in radius, even if they have the same charge. The higher the charge, the smaller the ion.
    • Sizes of anions are much larger than their isoelectronic cations.

    Polyatomic Ions

    The ammonium ion and some anions from oxy-acids are typical poly-atomic ions. Some of these are given so that you will be familar with them.

    \(\ce{NH4+}\) ammonium ion
    \(\ce{N(CH3)H3+}\) Methyl ammonium ion,
    any number of \(\ce{H}\) can be replaced by a methy group
    \(\ce{N(CH3)4+}\) Tetramethyl ammonium
    \(\ce{SO4^2-}\) Sulfate
    \(\ce{SO3^2-}\) Sulfite
    \(\ce{PO4^3-}\) Phosphate
    \(\ce{PO3^3-}\) Phosphite
    \(\ce{ClO4-}\) Perchlorate
    \(\ce{ClO3-}\) Chlorate
    \(\ce{ClO2-}\) Chlorite
    \(\ce{NO3-}\) Nitrate
    \(\ce{NO2-}\) Nitrite

    A salt consists of positive and negative ions, and the stoichiometry is determined by balancing the charges so that the salt as a whole is neutral.

    Lattice Energy and Energy of Crystallization

    The Lattice energy, U, is the amount of energy required to separate a mole of the solid (s) into a gas (g) of its ions.

    \(\mathrm{NaCl_{\large{(s)}} \rightarrow Na^+_{\large{(g)}} + Cl^-_{\large{(g)}}}\) U kJ mol-1

    Wishful thinking would be to separate the ions in a salt so that there is no interaction between the positive and negative ions. After the separation, the ions must be in a gaseous sate. However, the process requires energy, and the lattice energies are positive.

    The same amount of energy will be released when the ions are condensed from a gaseous state to a solid state. In this view, the released energy is called energy of crystallization, Ecryst. Since energy is released, the sign is negative for such a quantity. Thus, we have

    \(\mathrm{a M^{b+}_{\large{(g)}} + b X^{a-}_{\large{(g)}} \rightarrow M_aX_{b\large{(s)}}}\) Ecryst kJ mol-1

    Therefore, we have this relationship:

    \(E_{\large\textrm{cryst}} = - U\)

    To carry out either process experimentally is impossible, let alone measuring the energy involved. One way to evaluate lattice energy is called the Born-Haber cycle.

    Born-Haber Cycle for the Calculation of Lattice Energy

    Since the concepts of lattice energy and energy of crystallization are sound, and lattice energy is a useful quantity to judge the chemical properties of the salt, we can and usually calculate it by applying the law of conservation of energy, in a manner similar to the Hess's law. For the calculation of lattice energy, this process is called the Born-Haber cycle.

    In 1919, M. Born and F. Haber separately devised a method to calculate the lattice energy from known thermodynamic data such as,

    Hsub Enthalpy of sublimation,
    Hmelt Enthalpy of melting,
    Hvap Enthalpy of vaporization,
    Hf Enthalpy of formation,
    U Lattice energy,
    Ecryst Energy of crystallization,
    H Enthalpy of reaction,
    D Bond (dissociation) energy
    IP Ionization potential, or ionization energy, IE
    EA Electron affinity,

    For example, to calculate the lattice energy of \(\ce{NaBr}\), we start with \(\ce{Na}\) metal and \(\ce{Br2}\) liquid. To make \(\ce{NaBr}\), each of the reactants need to go through a series of processes in order to calculate the lattice energy:

    \(\ce{2 Na_{\large{(s)}} --2 \mathit{H}_{\large{sub}} \rightarrow 2 Na_{\large{(g)}} --2 \mathit{IP} \rightarrow 2 Na+_{\large{(g)}}}\)

    \(\ce{Br_{2\large{(l)}} --\mathit{H}_{\large{vap}} \rightarrow Br_{2\large{(g)}} --\mathit{D} \rightarrow 2 Br_{\large{(g)}} --2\mathit{EA} \rightarrow} \mathrm{2 Br^-_{\large{(g)}}}\)

    One other reaction is required

    \(\ce{2 Na + Br2 \rightarrow 2NaBr}\), \(2H_{\textrm f}\)

    The lattice energy corresponds to the reaction

    \(\mathrm{2 NaBr_{\large{(s)}} \rightarrow 2Na^+_{\large{(g)}} + 2 Br^-_{\large{(g)}}}\)

    To put all the information together, we may now draw the cycle:

              THE BORN-HABER CYCLE
    
         -----------2Na+ + 2 Br(g)--------
             ­                       |
             |2IP + D                |2EA
             |                       ¯
      2Na(g) + Br2(g)            2Na+(g) + 2Br-(g)
             ­                       |
             |2Hsub+Hvap              |
             |                       |
       2Na(s) + Br2(l)               |2Ecryst 
             |                       |
             |2Hf                    |
             ¯                       ¯
         ----------2 NaBr(s)---------------
    

    Thus,

    \(2H_{\large\textrm f} - (2H_{\large\textrm{sub}} + H_{\large\textrm{vap}} + 2IP + D) = 2EA + 2E_{\large\textrm{cryst}}\)

    Note that we have used two moles of \(\ce{NaBr}\) in the above diagram.

    This scheme shows that we can calculate the lattice energy of \(\ce{NaBr}\) from some known thermodynamic data. The same can be calculated from reaction equations and their associated energies. This is illustrated below

    \(\ce{2 Na_{\large{(s)}} + Br_{2\large{(l)}} \rightarrow 2 NaBr_{\large{(s)}}}\) \(2 H_{\textrm f}\)
    \(\ce{2 Na_{\large{(g)}} \rightarrow 2 Na_{\large{(s)}}}\)
    or \(\ce{2 Na_{\large{(s)}} \rightarrow 2 Na_{\large{(g)}}}\)
    \(- 2 H_{\large\textrm{sub}}\)
    \(\ce{2 Na+_{\large{(g)}} + 2e- \rightarrow 2 Na_{\large{(g)}}}\)
    or \(\ce{2 Na_{\large{(g)}} \rightarrow 2 Na+ + 2e-}\)
    \(- 2\, IP\)
    \(\ce{Br_{2\large{(g)}}\rightarrow Br_{2\large{(l)}}}\)
    or \(\ce{Br_{2\large{(l)}} \rightarrow Br_{2\large{(g)}}}\)
    \(- H_{\large\textrm{vap}}\)
    \(\ce{2 Br_{\large{(g)}} \rightarrow Br_{2\large{(g)}}}\)
    or \(\ce{Br_{2\large{(g)}} \rightarrow 2 Br_{\large{(g)}}}\)
    \(- D\)
    \(\mathrm{2 Br^-_{\large{(g)}} \rightarrow 2 Br_{\large{(g)}} + 2 e^-}\)
    or \(\ce{2 Br_{\large{(g)}} + 2 e- \rightarrow} \mathrm{2 Br^-_{\large{(g)}}}\)
    \(- 2\, EA\)
    Add all the above equations leading to
    \(\mathrm{2 Na^+_{\large{(g)}} + 2 Br^-_{\large{(g)}} \rightarrow 2NaBr_{\large{(s)}}}\) \(2 E_{\large\textrm{cryst}}\)

    Thus,

    \(2 H_{\textrm f} - 2H_{\textrm{sub}} - 2 IP - H_{\textrm{vap}} - D - 2 EA = 2 E_{\textrm{cryst}}\)
    \(E_{\textrm{cryst}} = H_{\textrm f} - H_{\textrm{sub}} - IP - \dfrac{(H_{\textrm{vap}} + D)}{2} - EA\)

    This is the same result as shown in the diagram.

    Example 1

    Calculate the lattice energy of \(\ce{NaCl}\).

    Solution
    To solve a problem like this one, you have to know what data are required, and where to find them. A handbook is a good source, but you will have to look them up in several tables. To solve this problem requires the following data:

    \(\mathrm{\mathit{H}_{\large{sub}}\: of\: Na = 108\: kJ/mol}\)
    \(D\mathrm{\: of\: Cl_2 = 244}\)
    \(IP \mathrm{\:of\: Na_{\large{(g)}} = 496}\)
    \(EA \mathrm{\:of\: Cl_{\large{(g)}} = -349}\)
    \(\mathrm{\mathit{H}_f\: of\: NaCl = -411}\)

    The Born-Haber cycle to evaluate Elattice is shown below:

         -----------Na+ + Cl(g)--------
             ­                       |
             |                       |-349
             |496+244/2              ¯
             |                 Na+(g) + Cl-(g)
             |                       |
       Na(g) + 0.5Cl2(g)             |
             ­                       |
             |108                    |
             |                       |Ecryst= -788 
       Na(s) + 0.5Cl2(g)             |
             |                       |
             |-411                   |
             ¯                       ¯
         -------------- NaCl(s) --------------
    

    \(\begin{align}
    &E_{\large\textrm{cryst}}
    = \mathrm{-411-\left(108+496+\dfrac{244}{2}\right)-(-349)\: kJ/mol}\\
    &\hspace{44px} = \mathrm{-788\: kJ/mol}\\
    &U = - E_{\large\textrm{cryst}}\\
    &\hspace{12px} = \textrm{788 kJ/mol (lattice energy)}
    \end{align}\)

    Discussion
    The value calculated for U depends on the data used. Data from various sources differ slightly, and so does the result. The lattice energies for \(\ce{NaCl}\) most often quoted in other texts is about 765 kJ/mol.

    Compare with the method shown below

    \(\mathrm{Na_{\large{(s)}} + 0.5 Cl_{2\large{(l)}} \rightarrow NaCl_{\large{(s)}}}\) - 411 Hf
    \(\mathrm{Na_{\large{(g)}} \rightarrow Na_{\large{(s)}}}\) - 108 -Hsub
    \(\mathrm{Na^+_{\large{(g)}} + e^- \rightarrow Na_{\large{(g)}}}\) - 496 -IP
    \(\mathrm{Cl_{\large{(g)}} \rightarrow 0.5\, Cl_{2\large{(g)}}}\) - 0.5 * 244 -0.5*D
    \(\mathrm{Cl^-_{\large{(g)}} \rightarrow Cl_{\large{(g)}} + 2 e^-}\) 349 -EA
    Add all the above equations leading to
    \(\mathrm{Na^+_{\large{(g)}} + Cl^-_{\large{(g)}} \rightarrow NaCl_{\large{(s)}}}\) -788 kJ/mol = Ecryst

    There is a another method based on principles of physics to evaluate the lattice energy, and some examples are given in the discussion enthalpy of hydration and lattice energy.

    The Born-Haber cycle enables us to calculate lattice energies of various compounds. For salts containing polyatomic ions, the Born-Haber cycle is not as useful. Some other means have to be used to evaluate the lattice energy or energy of crystallization.

    Comparison of Lattice Energies (U in kJ/mol) of Some Salts
    Solid U Solid U Solid U Solid U
    \(\ce{LiF}\) 1036 \(\ce{LiCl}\) 853 \(\ce{LiBr}\) 807 \(\ce{LiI}\) 757
    \(\ce{NaF}\) 923 \(\ce{NaCl}\) 786 \(\ce{NaBr}\) 747 \(\ce{NaI}\) 704
    \(\ce{KF}\) 821 \(\ce{KCl}\) 715 \(\ce{KBr}\) 682 \(\ce{KI}\) 649
    \(\ce{MgF2}\) 2957 \(\ce{MgCl2}\) 2526 \(\ce{MgBr2}\) 2440 \(\ce{MgI2}\) 2327

    The lattice energies of some salts are given in the table on the right here.

    Among the mono-valent salts, the lattice energies decrease when the sizes of the ions increase.

    Comparing the lattice energy for salts with one divalent ion leads to the same conclusion. The lattice energies decrease when the sizes of the ions increase.

    The lattice energy of salts involving a divalent ion are much higher than those of monovalent salts, because much more energy is required to separate these ions.

    Confidence Building Questions

    1. Which of the following is not an ionic compound?
      \(\ce{NaCN,\ CaSO4,\ LiCl,\ MgO,\ HCl,\ CO,\ NH4NO3}\)
      Hint: \(\ce{CO}\)

      Discussion: Carbon monoxide is a gas, and \(\ce{C-O}\) bond is predominantly covalent. Some covalent character remains in the ionic compound.

    2. Give the symbol of the noble gas that has the same electronic configuration as the following ions.
      \(\ce{S^2- ,\ Cl- ,\ K+,\ Ca^2+,\ Ga^3+}\)
      Hint: \(\ce{Ar}\)

      Skill: Identify the electronic configuration of ions.
      The noble gas beside \(\ce{Cl}\) is argon.
      What ions have the same electronic configuration as \(\ce{Ne}\)?

    3. Choose the type of energy for the process \(\ce{Mg_{\large{(s)}} \rightarrow Mg_{\large{(l)}}}\).
      1. Enthalpy of sublimation
      2. Enthalpy of melting
      3. Enthalpy of vaporization
      4. Enthalpy of formation
      5. Enthalpy of reaction
      Hint: b. Enthalpy of melting

      Discussion: Is the process endothermic or exothermic?

    4. Choose the type of energy for the process
      \(\ce{Mg_{\large{(g)}} \rightarrow Mg+_{\large{(g)}} + e- }\).
      1. Enthalpy of formation
      2. Enthalpy of reaction
      3. Bond (dissociation) energy
      4. Ionization potential
      5. Electron affinity
      Hint: d. Ionization potential

      Skill: Give appropriate names for various chemical processes and know if the process endothermic or exothermic.

    5. The energy called electron affinity is the energy released for which one of the following steps?
      1. \(\mathrm{F_{2\large{(g)}} \rightarrow 2 F_{\large{(g)}}}\)
      2. \(\mathrm{Cl_2 \rightarrow 2 Cl}\)
      3. \(\mathrm{Na_{\large{(s)}} \rightarrow Na^+_{\large{(s)}} + e^-}\)
      4. \(\mathrm{F_{2\large{(g)}} + e^- \rightarrow F_2^-}\)
      5. \(\mathrm{Br_{\large{(g)}} + e^- \rightarrow Br^-_{\large{(g)}}}\)
      Hint: e.

      Discussion: EA is the amount of energy released when an atom acquires an electron. Is the process exothermic?

    6. Which one of the following is more than a single step for making up the Born-Haber cycle?
      1. \(\mathrm{Mg_{\large{(s)}} \rightarrow Mg_{\large{(l)}}}\)
      2. \(\mathrm{Mg_{\large{(g)}} \rightarrow Mg^+_{\large{(g)}} + e^-}\)
      3. \(\mathrm{Br_{2\large{(l)}} \rightarrow 2 Br_{\large{(g)}}}\)
      4. \(\mathrm{Ca_{\large{(s)}} + Cl_{\large{(g)}} \rightarrow Ca_{\large{(g)}} + Cl_{\large{(g)}}}\)
      5. \(\mathrm{Ca_{\large{(s)}} + Cl_{2\large{(g)}} \rightarrow CaCl_2}\)
      Hint: c.

      Skill: Identify the chemical reactions involving these types of energy:
      \(H_{\large\textrm{sub}}\), \(H_{\large\textrm{melt}}\), \(H_{\large\textrm{vap}}\), \(H_{\large\textrm f}\), \(H\) (enthalpy of reaction), \(D\), \(IP\), and \(EA\)
      Note: \(\mathrm{Br_{2\large{(l)}} \rightarrow Br_{2\large{(g)}} \rightarrow 2Br_{\large{(g)}}}\)

    Contributors