| I. \[H{{g}_{2}}C{{l}_{2}}\]in aqueous solution. |
| II. Boron in non-aqueous solution |
A) \[2.2,\,\,0.45\]
B) \[2,\,\,0.54\]
C) \[2.8,\,\,0.55\]
D) \[3,0.5\]
Correct Answer: A
Solution :
[a] (I) \[HgC{{l}_{2}}H{{g}_{2}}^{2+}+2C{{l}^{\bigcirc -}}(3\,ions)\] \[2=(Number\,\,of\,\,ions\,\,\times \alpha )+(1-\alpha )\] \[=(3\times 0.6)+0.4=2.2\] (II). \[12B{{B}_{12}}\] (Boron exists as icosahedron structure) \[i=(Number\,\,of\,\,ions\,\,\times \alpha )+(1-\alpha )\] \[=\left( \frac{1}{12}\times 0.6 \right)+(1-0.6)\] \[=\frac{1}{20}+0.4=0.05+0.4=0.45\]
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