SATHEE: Chemical Bonding And Molecular Structure Question 471

Chemical Bonding And Molecular Structure Question 471

Question: The hybridization of atomic orbital’s of nitrogen in $N O_{2}^{+}, N O_{2}^{-}$ and $N H_{4}^{+}$ are

Options:

A) $s p^{2}, s p^{3}$ and $s p^{2}$ respectively

B) $s p, s p^{2}$ and $s p^{3}$ respectively

C) $s p^{2}, s p$ and $s p^{3}$ respectively

D) $s p^{2}, s p^{3}$ and sp respectively

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Answer:

Correct Answer: B

Solution:

$N O_{2}^{+} = \frac{1}{2} [5 + 0 + 0 - 1] = 2 s p;$

$N O_{2}^{-} = \frac{1}{2} [5 + 0 + 1 - 0] = 3 s p;$

$N H_{4}^{+} = \frac{1}{2} [5 + 4 + 0 - 1] = 4 s p^{3}$

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Chemical Bonding And Molecular Structure Question 471

Question: The hybridization of atomic orbital’s of nitrogen in NO2+,NO2− and NH4+ are

Options:

Answer:

Solution:

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Question: The hybridization of atomic orbital’s of nitrogen in $N O_{2}^{+}, N O_{2}^{-}$ and $N H_{4}^{+}$ are