A) \[Si{{F}_{4}}\]
B) \[N{{a}_{2}}Si{{F}_{6}}\]
C) \[{{H}_{2}}Si{{O}_{3}}\]
D) \[Si{{(OH)}_{4}}\]
Correct Answer: B
Solution :
In etching of glass, HF reacts with \[N{{a}_{2}}Si{{O}_{3}}\] and \[CaSi{{O}_{3}}\]present in glass to give \[N{{a}_{2}}Si{{F}_{6}}\] and \[CaSi{{F}_{6}}\] \[N{{a}_{2}}Si{{O}_{3}}+6HF\xrightarrow{{}}N{{a}_{2}}Si{{F}_{6}}+3{{H}_{2}}O\] \[CaSi{{O}_{3}}+6HF\xrightarrow{{}}CaSi{{F}_{6}}+3{{H}_{2}}O\]
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