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Question
Give the formula and describe the structure of a noble gas species which is isostructural with:
`IBr_2^(-)`
Solution 1
XeF2 is isoelectronic to `IBr_2^(-)` and has a linear structure.
Solution 2
IBr2–: In IBr2–, central atom I has eight electrons. Two of these are utilized in forming two single bonds with two Br atom. Six remaining electrons constitutes three lone pairs. It is arranged in linear structure.
IBr2– has 22 valence electrons. A noble gas species having 22 valence electrons is XeF2 (8+2 x 7=22).
XeF2 is also linear.
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| Column I | Column II |
| (A) XeF6 | (1) sp3d3 – distorted octahedral |
| (B) XeO3 | (2) sp3d2 – square planar |
| (C) XeOF4 | (3) sp3 – pyramidal |
| (D) XeF4 | (4) sp3d2 – square pyramidal |
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| 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 |
Choose the most appropriate answer from the options given below:
\[\ce{XeF6 + H2O ->[Partial][Hydrolysis] \underline{}\underline{}\underline{}\underline{} + \underline{}\underline{}\underline{}\underline{}}\]
\[\ce{XeF4 + H2O - \underline{}\underline{}\underline{}\underline{} + \underline{}\underline{}\underline{}\underline{}}\]
