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Consider the species NO ₂^+ , NO ₂ , and NO ₂^- . Which of these is an odd-electron species, and what is its hybridization and molecular shape?

Options

  1. ANO ₂ , sp , linear
  2. BNO ₂ , sp ^2 , bent
  3. CNO ₂^- , sp ^2 , bent
  4. DNO ₂^+ , sp , linear

Correct answer

B. NO ₂ , sp ^2 , bent

Step-by-step solution

First, calculate the total number of valence electrons for each species to identify the odd-electron molecule: NO ₂^+ = 5 + 2(6) - 1 = 16 NO ₂ = 5 + 2(6) = 17 NO ₂^- = 5 + 2(6) + 1 = 18 NO ₂ has 17 valence electrons, which is an odd number, making it an odd-electron species. Next, determine the hybridization and shape of NO ₂ using VSEPR theory. The central nitrogen atom forms two -bonds with the oxygen atoms and possesses one unpaired electron. The odd electron occupies a hybrid orbital, giving a steric number of

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