ABC26IN0070 · Chemical Bonding
Subject: Inorganic Chemistry · Chapter: Chemical Bonding · Topic: Chemical Bonding – General · Exam: CSIR-NET DEC 2017 · Marks: 2 · Difficulty: Easy
Geometries of $\mathrm{SNF}_{3}$ and $\mathrm{XeF}_{2} \mathrm{O}_{2}$ respectively, are
(a)square planar and square planar
(c)square planar and trigonal bipyramidal
(d)tetrahedral and trigonal bipyramidal
(b)tetrahedral and tetrahedral
Answer
Explanation
",4,)×6 6% SNF3 = = = 4 = sp3 = Td 7 7 7,"×%,)×% 6$ XeO2F2 = = == =4+1= 5=sp3d→TBP 7 7
Explanation as extracted from the printed page; notation may be imperfect.
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