**Chapter 2:- Periodic properties of the elements**

**B.sc 1st year Book**

**(Page 3)**

**Ionic radii by Lande’s Method and Pauling’s Method**

**Ionic radii:**

- Though it is possible to measure inter-nuclear distance in an ionic crystal very accurately by
**X-ray**diffraction there is no universally accepted formula for measuring the radius of anions or cations. There are several different sets of ionic radii which have been estimated. - Correction should be made if the charge on an ion is changed e.g.
**Fe**and^{3+ }**Fe**.^{2+} - Correction should be made for the coordination number and geometry of an ion.
- The loans should be spherical which is true for ions of s and p-block with noble gas configuration but not true for transitional elements.
- It is also not accurate in the case of delocalization of d-electrons where they give rise to 10 metallic conduction or form cluster compounds.

Thus, the size of an ion can be determined only when the size of its opposite ion is known. The following methods are used to measure the ionic radii.

**(a) Lande’s Method (1920):**

It was its first attempt to determine ionic radii by Lande in 1920. He determined intemuclear. distance of an ionic crystal in which the anion is very large as compared to the cation because there Is anion-anion contact in the crystal lattice For example: in lithium iodide ionic crystal, Li* ion is very small {Li+=0.63 A) and F ion is large (1=2.16 A) due to which there is anion-anion contact in the crystal lattice as shown in fig 2.03a.^{ }

^{ $$ r_{I^-}\:=\frac{d_{I^-}-_{I^-}}{2}=2.16A^{\circ }$$ }

^{$$ d_{Li^+}_{-\:I^-}\:=r_{I^+}+_{Li^+}=0.60A^{\circ } $$}

**By X-ray**, crystallography it is found that the edge length AB of **L+1** – unit cell is equal to 6.00 A, and △ABC is an equilateral triangle. The simple geometry allows us to fix the diagonal of the square as **2η−+2r+11**. It is possible when the angle formed by the diagonal in the corner is equal to **45∘**.

Therefore

AC^{2} = AB^{2} + BC^{2}

(AB = BC = CD = AD = d_{1} – 1 – m6.00A)

AC^{2} = 6^{2} + 6^{2} = 72A

$$ AC\:=\sqrt{72}=2r_1+2r_1\:=4r_1 $$

$$ r_{1=}\frac{1}{4}\sqrt{72}=2.12A $$

^{+}1

^{–}, K

^{+}I

^{–}, Cs

^{+}1

^{–}, etc) which contain larger cations as compared to Li cation: halide ions do not touch each other but cation-anion contact will occur in their crystal lattice as shown in figure 2.04. Hence, the sum of the radii of the cation and the anion would be equal to the inter-ionic distance between the cation and anion thus in the case of K

^{+}1

^{– }the ionic crystal dK

^{+}T

^{– }is given as:

`d`_{k} _{+} _{1-} = r_{k} + r_{1-}

_{K+ – 1–}= 3.50 Å and r

_{1–}= 2.12 Å, the ionic radius of K

^{+}, f

_{K}+ thus calculated as –

_{K}+ = d

_{K+ -I }– r

_{1–}

**(b) Pauling’s Method :**

Pauling’s method has been utilized for calculating ionic radii of such logic crystals which have monovalent isoelectronic ions e.g. **Na+F−, K+Cr, Rb+Br, **and **Cs++– e.** the iconic pairs having the same number of electrons as shown below :

**(i)**The cation,

**c**and anion,

^{+},**a**of the ionic crystal, c+a−are considered to be charged spheres and very closely packed to each other.

^{–}^{+}and a

^{–}in the ionic crystal is c

^{+}a

^{−}is d

_{c+a−}. which is equal to the sum of the ionic radii of a cation and anion.

_{c+ – H–}= fc

^{+}+ r

_{a– }

^{ ……… (i)}

**r**is the radius of the cation and

_{c+ }**r**is the radius of the anion.

_{a–}**(ii).**If an ionic crystal,

**c+a-**contains isoelectronic Ionic pair which have electronic configuration

**ns**, the radius of the cation or anion is inversely proportional to their effective nuclear charge

^{2}or ns^{2}np^{6}**Zeff (or Z )**.

**(ii)\**

**(ii)**to cation and anion, we get:

**Z**are effective nuclear charges of anion and cation respectively and σ

^{∗}_{a− }and Z^{∗}_{c+}_{a−}and σ

_{c+}are screening constants for anion and cation respectively.

**(i)**and

**(iii)**we can calculate the value of

**r**and

_{c+}**r**by knowing the isoelectronic ions and their radii are given in table 2.5.

_{a−}

**Illustration:** Let us apply Pauling’s method to calculate the radii of the isoelectronic **Na ^{+ }**and

**F**ions in

^{− }**Na**crystal. In this ionic crystal, the internuclear distance between

^{+}F^{−}**Na+**and

**F−**ions;

**dNa**and nuclear charges are 11 and 9 respectively. Since

^{+}-F = 2.13A,**Na+ F–**ions both have a stable electronic configuration of neon

**(1s**for which using

^{2}2s^{2}2p^{6})**Slater’s rule**. screening constant

**(σ)**is found to be :

σ = 2(0.85) + 8(0.35) = 1.70 + 2.80

= 4.50

Hence, Z*_{Na+} = (Z_{Na+} – σ_{Na+})

= 11.0 – 4.5 = 6.5

and Z*_{F–} = (Z_{F–} – σ_{F–} )

= 9.0 – 4.5 = 4.5

On putting the above values in Pauling’s Equation, we get

**(i)**and

**(ii)**we can get the values of

**r**

_{Na+}and

**r**equal to

_{f-}**0.95**and

**1.36 A**respectively.