In tetramethyl silicane, i.e. $$Si{\left( {C{H_3}} \right)_4},\,Si$$    is $$s{p^3}$$ hybridised. Hence it has tetrahedral structure.

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White phosphorus is soluble in $$C{S_2}$$  whereas red phosphorus is insoluble in $$C{S_2}.$$

$$B{r^ - }$$  has been changed to $$B{r_2} \cdot {H_2}S{O_4}$$   is acting as an oxidising agent.

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$${H_3}P{O_4}$$  is a tribasic acid as it has $$3P - OH$$  bonds. i.e., 3 ionisable $$H$$ atoms thus, can form three series of salts.

Hybridisation is $$s{p^3}{d^2},$$  hence octahedral

\[Xe{{F}_{4}}+Sb{{F}_{5}}\to {{\left[ Xe{{F}_{3}} \right]}^{+}}{{\left[ Sb{{F}_{6}} \right]}^{-}}\]       \[\to \underset{\begin{smallmatrix} s{{p}^{3}}d \\ \text{bent }T-\text{shape} \end{smallmatrix}}{\mathop{{{\left[ Xe{{F}_{3}} \right]}^{+}}}}\,+\underset{\begin{smallmatrix} s{{p}^{3}}{{d}^{2}} \\ \text{octahedral} \end{smallmatrix}}{\mathop{{{\left[ Sb{{F}_{6}} \right]}^{-}}}}\,\]

Among the given compounds, the $$\ddot N{H_3}$$  is most basic. Hence has highest proton affinity

$$SnC{l_4}$$  is colourless volatile liquid and $$SnC{l_2}$$  is colourless solid $$Sn$$  conducts electricity and it belongs to 14 group.