Radius Ratio Rules
The three physical states of matter are solid, liquid, and gas. Any state of matter may be transformed into another by varying the temperature and pressure. At lower temperatures, the most prevalent state of matter is solid. By elevating the temperature of a solid to its melting point, enough energy is injected into the solid to overcome intermolecular forces of attraction and melt the solid and change it into a liquid. When liquids are heated to their boiling temperatures, they evaporate and become gaseous. On the other hand, when gases are cooled and subjected to high pressure, they can be changed into liquids, which can then be further cooled to produce solids.
Because the locations of their component particles, atoms, or molecules cannot be modified, solids cannot be squeezed like gases or poured like liquids. The physical state of matter results from the interaction of intermolecular forces of attraction such as dipole-dipole interactions, dipole-induced dipole interactions, London forces, hydrogen bonding, and so on.
Properties of solid
- The majority of pharmacological medications are solid.
- A fixed-composition solid has a set mass, volume, form, and density. [In general, the solid-state of a material is less dense than the liquid and gaseous states.]
- The majority of solids are hard, stiff, and incompressible.
- Intermolecular forces of attraction hold the component particles of a solid together securely.
- All pure solids have a distinct melting point that is determined by the strength of intermolecular forces present in the solid-state.
- In the solid-state, the intermolecular force of attraction between component particles is greater than in the liquid and gaseous phases.
Radius Ratio rule
The structure of each ionic compound is determined by stoichiometry and ion sizes. Larger cations can fit into cubic or octahedral holes. In tetrahedral holes, smaller cations can be accommodated. If we examine an array of anions in the form of cubic tight packing, the diameters of the tetrahedral and octahedral holes will vary. As a result, the cations will only occupy the voids if there is adequate room for them.
The Radius Ratio can be used to determine whether the ions will be able to retain the cations. The shape of the unit cell is also determined by the ion’s coordination number in the crystal structure. For a particular coordination number, there is a limiting value of the ratio of cation radius to anion radius, i.e. r^{+ }/ r^{–}. The ionic structure becomes unstable if the value of the ratio r^{+ }/ r^{–} is smaller than the predicted value.
Hence, the radius ratio is defined as the ratio of a smaller ionic radius (cation) to a larger ionic radius (anion), and is given by,
ρ = r^{+ }/ r^{–}
where,
- ρ is the radius ratio,
- r^{+} is the radius of cation, and
- r^{–} is the radius of the anion.
Given below are the limiting values of r^{+ }/ r^{–} and their coordination numbers.
Coordination Number of cation | Limiting value of r^{+} / r^{–} | Type of hole occupied (void) |
2 | <0.155 | Linear |
3 | 0.155 to 0.225 | Planar Triangular |
4 | 0.225 to 0.141 | Tetrahedral |
6 | 0.414 to 0.732 | Octahedral |
8 | 0.732 to 1.000 | Cubic |
12 | >1 | Close Packing |
Examples are B_{2}O_{3}, ZnS, NaCl, CsCl, MgO, CuCl
Some properties are:
- The radius ratio rule only applies to ionic substances.
- If the bonds are covalent, the rule is broken.
- The rule may be used to predict the structures of various ionic solids.
Sample Problems
Problem 1: If a solid “X^{+}Y^{–}” has a structure similar to NaCl and the radius of anion is 250 pm. Then, find the ideal radius of the cation in the structure. Also, state your reason, is it possible to fit a cation Z^{+} of radius 180 pm in the tetrahedral site of the structure(X^{+}Y^{–})?
Solution:
Given that,
Structure X^{+}Y^{–}
Radius of anion = 250 pm = 2.5 A° [1 picometer = 0.01A°]
Limiting ratio = 0.414 [refer to the above table]
If the X^{+}Y^{–} structure is comparable to that of the Na^{+}Cl^{–} ion, then six Cl^{–} ions will surround the Na^{+} and vice versa.
Thus, octahedral void is occupied.
Radius ration ρ = r^{+} / r^{–}
r^{+} = 0.414 × 2.5
= 1.035 A°
Now, the limiting ratio for tetrahedral site is 0.225 [refer to table]
As a result, r^{+} / r^{–} = 0.225
r^{+} = 0.225 × 2.5
= 0.5625 A° or 56.25 pm
Thus, for a tetrahedral site, the optimum radius for the cation in the given structure is 56.25 pm.
We know that the radius of Z^{+} is 180 pm. This implies that Z^{+}‘s radius is substantially bigger than 56.25 pm.
Thus, cation Z^{+} cannot be accommodated in a tetrahedral location.
Problem 2: Predict the coordination number of Cs^{+} ion and the structure of CsCl if r_{cs+ }= 1.69 A° and r_{Cl–} = 1.81A°
Solution:
Given that:
r_{cs+} = 1.69 A°
r_{Cl–} = 1.81A°
Calculation:
Radius ration ρ = r^{+} / r^{–}
= 1.69 / 1.81
= 0.9337 A°
As 0.9337 > 0.732, hence the coordination number is 8 and geometry of CsCl is cubic.
Problem 3: Predict the coordination number of Na^{+} ion and the structure of NaCl crystal if r_{Na+} = 0.95 A° and r_{Cl–} = 1.81A°
Solution:
Given that:
r_{Na+} =0.95A°
r_{Cl–} = 1.81A°
Radius ration ρ = r^{+} / r^{–}
= 0.95 / 1.81
= 0.5248 A°
As 0.5248 lies between (0.414 to 0.732), hence the coordination number is 6 and geometry of NaCl is Octahedral.
Problem 4: In silicates, the oxygen atom forms a tetrahedral void. The limiting radius ratio for tetrahedral void is 0.22. The radius of oxide is 1.4 Å. Find out the radius of the cation.
Solution:
Given that:
Radius of oxide (r^{−}) = 1.4 ÅRadius ratio = 0.22
Radius ratio = r^{+} / r^{–}
0.22 = r^{+} / 1.4
r^{+} = 0.22 × 1.4
r^{+} = 0.308 Å
Problem 5: If the radius of cation is 96 pm and that of anion is 618 pm. Determine the coordination number and structure of the crystal lattice.
Solution:
Given that:
Radius of cation (r^{+}) = 96 pm
Radius of anion (r^{−}) = 618 pm
Radius ratio = r^{+} / r^{–}
= 96 / 618
= 0.1553
Since the radius ratio lies in between the range 0.155 to 0.225.
The coordination number of crystal is 3 And the structure of crystal lattice is Trigonal planar.
Problem 6: Br^{−} ion forms a close-packed structure. If the radius of Br^{−} ions is 195 pm. Calculate the radius of the cation that just fits into the tetrahedral hole. Can a cation A^{+} having a radius of 82 pm be slipped into the octahedral hole of the crystal A^{+}Br^{–}?
Solution:
Given that:
Radius of anion (Br^{–})(r^{–}) = 195 pm
Radius of cation (A) (r^{+}) = 82 pm
Here we have to find the radius of the cation that just fits into the tetrahedral hole and determine whether the cation A^{+} having a radius of 82 pm can be slipped into the octahedral hole of the crystal
Limiting value for r^{+} /r^{– }for tetrahedral hole is 0.225 – 0.414
So, Radius of the tetrahedral hole = Radius ratio × r^{–}
= 0.225 × 195
= 43.875 pm
For cation A^{+} with radius = 82 pm
Radius ratio = r^{+}/r^{–}
= 82 / 195
= 0.4205
As it lies in the range 0.414 – 0.732, hence the cation A^{+} can be slipped into the octahedral hole of the crystal A^{+}Br^{−}.
Problem 7: Determine the structure and coordination number of MgS on the basis of radius ratio in which radius of Mg2+ and S2– is 65 pm and 184 pm respectively.
Solution:
Given that:
Radius of cation Mg_{2}^{+} (r^{+}) = 65 pm
Radius of anion S_{2}^{−} (r^{–}) = 184 pm
Radius ratio = r^{+} /r^{–}
= 65 / 184
= 0.3533
Since the radius ratio lies in between 0.225 – 0.414 the coordination number of MgS is 4 and the structure of MgS is Tetrahedral.
Problem 8: A solid AB has ZnS type structure. If the radius of cation is 50 pm, calculate the maximum possible value of the radius of anion B^{–}.
Solution:
Given that:
Radius of cation (r+) = 50 pm
Radius ratio = r^{+} /r^{– }
ZnS has tetrahedral arrangement.
The range of r^{+}/r^{–} for stable four fold coordination is 0.225 to 0.414
Hence the radius of anion can be calculated by
taking r^{+} / r^{–} = 0.225
∴ r^{–} = r^{+} / 0.225
= 50 / 0.225
= 222.22 pm