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Question

What is the hybridization of atomic orbitals of nitrogen in NO2+, NO3-and NH4+?


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Solution

Step1: Finding Hybridization

  • Hybridization is the concept of mixing several atomic orbitals to form a new hybrid orbital with a new hybridization contributed by all atomic orbitals.
  • In order to find the hybridization, we need to calculate the steric number of the given compound.
  • The steric number is the total number of sigma bonds and lone pairs in that atom we are considering.
  • Stericnumber=numberofsigmabonds+numberoflonepairs
  • The relation between steric number and hybridization is given below.
SeF6 Lewis Structure, Geometry, Hybridization, and Polarity -  Techiescientist

Step 2: Hybridization of NO2+,NO3- and NH4+

NO2+

  • Here there is one positive charge on nitrogen(one electron is lost). Thus the lone pair is absent on nitrogen.
  • There will be (5-1)=4 valence electron on nitrogen. So there are two sigma bond and two pi bonds in structure as follows,
MakeTheBrainHappy: The Lewis Dot Structure for NO2+
  • Thus the steric number is (2+0)=2
  • Therefore the hybridization is sp.

NO3-

  • Number of valence electron in nitrogen is (5-1)=6.
  • Thus there will be three sigma bonds and one pi bonds present. And no lone pairs on nitrogen.
How to draw the structure for NO3 - Quora
  • So the steric number is (3+0)=3
  • Therefore hybridization of nitrogen in NO3- is sp2

NH4+

  • Here there is a total of four sigma bonds and no lone pairs present as follows,
Ammonium - Wikipedia
  • The steric number is (4+0)=4
  • Thus the hybridization is sp3.

Therefore the hybridization of nitrogen in NO2+,NO3-and NH4+ are sp,sp2, and sp3respectively.


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