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Define ligancy and critical radius ratio in case of ionic solid. Write the conditions for stability of ionic crystal in 3-D ? Determine critical radius ratio for ligancy 6.
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Ligancy –

Every atom in crystalline structure is surrounded by other atoms. Coordinate number of equidistant nearest neighbour atoms which are simultaneously surrounding the given atom.

When CN is large, the structure is more closely packed. In Bravais Lattice, all atom have the same CN.

Critical Radius Ratio (CRR) –

It is defined as the cation-anion radius ratio. It is the ratio of ionic radius of cation $(r_c)$ to ionic radius of anion $(r_a)$ in cation-anion compound.

Thus, cation-anion radius ratio $= \dfrac {r_{c}}{r_{a}}$

Consider, following figures –

There are three cases –

1)   Case 1 – (\left (\dfrac {r_{c}}{r_{a}} < 0.155 \right ))

If the cation is too small then it will attract the anion into each other & they will collide causing unstable compound, fig (a).

2)  Case 2 – $\left ( \dfrac {r_{c}}{r_{a}}=0.155 \right )$

At the stability limit, the cation is touching all anions which are just touching at their edges. In this case, the radius ratio = 0.155 and is called as C.R.R. because compound is just stable, fig (b).

3)  Case 3 – (\left ( \dfrac {r_{c}}{r_{a}}>0.155 \right ))

Beyond the stability limit i.e. radius ratio > 0.155, the compound will be stable, fig(c).

Ligancy=6

Consider the plane passing through the central sphere occupying void and the four spheres co-ordinating to it. In isosceles right-angled triangle LMN

$\dfrac {LM}{LN}= \dfrac {\sqrt{2}}{1}=\dfrac {2(R+r)}{2R}$

$\therefore \dfrac {r}{R}=\sqrt{2}-1=0.414$

 

∴ The critical radius ratio for ligancy 6 is 0.414.

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