Ionic bond: how sodium and chlorine make NaCl
Bring a sodium atom close to a chlorine atom and watch one electron jump across, the ions change size and attract each other, then see the NaCl crystal and the energy curve of the ion pair. Free for every student.
Also searched as: ionic bond · ionic bond simulation · ionic bond class 11 · formation of nacl · electrovalent bond · lattice energy of nacl
About this experiment
| Subject | Chemistry |
|---|---|
| Class | 11 |
| Unit | 4. Chemical Bonding and Molecular Structure |
| NCERT section | 4.2 |
| Level | NCERT level |
| Time | about 15 minutes |
| Access | Free, no card needed |
Aim
To see how an ionic bond is formed between sodium and chlorine by transfer of an electron, to follow the energy changes, and to understand why ionic compounds form crystals.
Theory
An ionic (electrovalent) bond is formed when one atom gives electrons to another. Sodium (2, 8, 1) loses its single outer electron to become Na⁺ (2, 8, like neon). Chlorine (2, 8, 7) gains one electron to become Cl⁻ (2, 8, 8, like argon). Both reach a noble gas arrangement, and the opposite charges attract each other.
Making the ions costs energy: it takes 5.14 eV (496 kJ/mol) to remove an electron from sodium (ionisation enthalpy) and chlorine releases 3.61 eV (349 kJ/mol) when it gains one (electron gain enthalpy). So far the ions have 1.53 eV more energy than the atoms.
The attraction between Na⁺ and Cl⁻ is the Coulomb force. The energy falls as 14.4 eV·Å / r. At r = 9.4 Å the lowering by attraction is exactly 1.53 eV, so closer than this the ion pair is lower in energy than the atoms and the electron can jump across. Real NaCl molecules in the gas have a bond length of 2.36 Å. At smaller distances the electron clouds repel each other and the energy rises again.
Formulas used
- Na → Na⁺ + e⁻ ionisation energy = 5.14 eV (496 kJ/mol)
- Cl + e⁻ → Cl⁻ electron gain energy = −3.61 eV (−349 kJ/mol)
- Coulomb energy of the ion pair = − e²/(4πε₀ r) = −14.4 eV·Å / r
- U(r) = (IE − EA) − 14.4/r + repulsion
- crossing distance r = 14.4/(IE − EA) = 9.4 Å
- 1 eV per particle = 96.5 kJ/mol
Procedure: what to do in the virtual lab
- Start with the atoms far apart (about 10.5 Å). Look at the graph: the atoms (yellow) have lower energy than the ions (blue).
- Reduce the distance slowly. At about 9.4 Å the blue curve crosses the yellow line: the electron jumps from Na to Cl.
- Notice that the sodium ion becomes smaller and the chloride ion becomes bigger.
- Keep going to the lowest point of the blue curve (2.4 Å): this is the ion pair bond length.
- Switch to the crystal picture and count the neighbours around one ion.
Observations to make
- The electron does not move until the atoms are close enough; then the transfer is suddenly favourable.
- The positive ion (cation) is smaller than its atom and the negative ion (anion) is bigger than its atom.
- Each ion in the crystal is surrounded by six ions of the opposite charge, so the bond is not a bond between a pair but a whole lattice.
What you can change in this experiment
- Picture: One Na atom and one Cl atom / The NaCl crystal
- Distance between the Na and Cl nuclei (Å)
Viva questions on Ionic bond
Questions an examiner may ask in the Class 11 chemistry practical. Run the experiment in the lab to find the answers yourself, then check them in the Learn tab.
- How is an ionic bond formed? Give an example.
- Why is the Na⁺ ion smaller than the Na atom and Cl⁻ bigger than the Cl atom?
- What is lattice energy and why does it matter?
- What is the coordination number of Na⁺ in NaCl?
- Why do ionic compounds have high melting points?
Frequently asked questions
What is the aim of the Ionic bond experiment?
To see how an ionic bond is formed between sodium and chlorine by transfer of an electron, to follow the energy changes, and to understand why ionic compounds form crystals.
Which formulas are used in the Ionic bond experiment?
Na → Na⁺ + e⁻ ionisation energy = 5.14 eV (496 kJ/mol); Cl + e⁻ → Cl⁻ electron gain energy = −3.61 eV (−349 kJ/mol); Coulomb energy of the ion pair = − e²/(4πε₀ r) = −14.4 eV·Å / r; U(r) = (IE − EA) − 14.4/r + repulsion; crossing distance r = 14.4/(IE − EA) = 9.4 Å
What viva questions are asked on Ionic bond?
How is an ionic bond formed? Give an example. Why is the Na⁺ ion smaller than the Na atom and Cl⁻ bigger than the Cl atom? What is lattice energy and why does it matter? What is the coordination number of Na⁺ in NaCl?
How do I do the ionic bond experiment for Class 11 online?
Open the Ionic bond virtual lab on PractiSci, follow the procedure above, change the values and read the live graph. It matches the Class 11 chemistry practical, with the aim, theory, observations and viva questions on this page.
Can I do the Ionic bond experiment online?
Yes. Open it as a 3D virtual lab in your browser or on your phone, change the values and read the results from live graphs. This lab is free.
Syllabus: Class XI Chemistry, Chapter 4 — Chemical Bonding and Molecular Structure, §4.2 (ionic or electrovalent bond), lattice enthalpy
What you do in the lab
Change the controls and watch the chemistry behave in real time in 3D. Live readouts and graphs show the numbers behind it, and the Learn tab has the theory, procedure, observations and viva questions. Works on a phone or a laptop, in English and Hindi.
Topics: ionic bond · electrovalent bond · NaCl · sodium chloride · ions · lattice energy · ionisation enthalpy · electron gain enthalpy · chemical bonding
More from this unit
- Covalent bond: sharing electrons, bond length and bond energy
- Shapes of molecules (VSEPR): CO₂, BF₃, CH₄, NH₃, H₂O, PCl₅, SF₆, XeF₄