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प्रश्न
What inspired N. Bartlett for carrying out reaction between Xe and PtF6?
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उत्तर
PtF6 interacts with oxygen to generate \[\ce{O^+_2[PtF6]-}\], according to N. Bartlett's 1962 discovery. Because the first ionisation enthalpy of Xe (1170 kJ mol−1) is so near to that of the O2 molecule (1175 kJ), Bartlett was motivated to conduct a reaction with Xe and PtF6. He was successful in isolating the orange-yellow complex XePtF6.
\[\ce{Xe + PtF6 ->[278 K] XePtF6}\]
संबंधित प्रश्न
Draw the structure of the following molecule:
XeF4
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\[\ce{XeF2 + H2O ->}\]
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test q 1234
Balance the following equation:
\[\ce{XeF6 + H2O -> XeO2F2 + HF}\]
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\[\ce{ICI^-_4}\]
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\[\ce{BrO^-_3}\]
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On partial hydrolysis, XeF6 gives ______.
The atomic radii of Zr and Hf are almost identical. This is because of
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 |
Choose the most appropriate answer from the options given below:
The element expected to form the largest ion to achieve the nearest noble gas configuration is:
