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Question
Complete the following equation:
XeF2 + H2O →
Solution
2XeF2 + 2H2O → 2Xe + 4HF + O2
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(B) XeO3 | (2) sp3d2 – square planar |
(C) XeOF4 | (3) sp3 – pyramidal |
(D) XeF4 | (4) sp3d2 – square pyramidal |
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Match List - I with List - II:
List - I | List - II | ||
(Species) | (Number of lone pairs of electrons on the central atom) |
||
(A) | XeF2 | (i) | 0 |
(B) | XeO2F2 | (ii) | 1 |
(C) | XeO3F2 | (iii) | 2 |
(D) | XeF4 | (iv) | 3 |
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\[\ce{XeF6 + H2O ->[Partial][Hydrolysis] \underline{}\underline{}\underline{}\underline{} + \underline{}\underline{}\underline{}\underline{}}\]
\[\ce{XeF4 + H2O - \underline{}\underline{}\underline{}\underline{} + \underline{}\underline{}\underline{}\underline{}}\]
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