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Chemical Bonding and Molecular Structure question

2011 · Shift 0 · Q25
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Chemical Bonding and Molecular Structure question

2011 · Shift 0 · Q25

JEE MainChemistryChemical Bonding and Molecular StructureMCQ+4 / −1
The hybridization of orbitals of N atom in NO3−NO_3^-NO3−​, NO2+NO_2^+NO2+​ and NH4+NH_4^+NH4+​ are respectively :
  1. A
    sp , sp2, sp3
  2. B
    sp2, sp , sp3
  3. C
    sp , sp3, sp2
  4. D
    sp2, sp3, sp
View written solutionFree

Correct answer: B

  1. Find the hybridization of N in NO3−NO_3^-NO3−​

    • In nitrate ion, nitrogen is the central atom bonded to three oxygen atoms.
    • Due to resonance, all three N−ON-ON−O bonds are equivalent.
    • Around nitrogen, there are 3 regions of electron density and no lone pair on N.
    • Geometry is trigonal planar.

    Therefore, hybridization of N is: sp2sp^2sp2

  2. Find the hybridization of N in NO2+NO_2^+NO2+​

    • In nitronium ion, structure is: O=N=O+O=N=O^+O=N=O+
    • Nitrogen is bonded to two oxygen atoms with double bonds.
    • So nitrogen has 2 regions of electron density and no lone pair.
    • Geometry is linear.

    Therefore, hybridization of N is: spspsp

  3. Find the hybridization of N in NH4+NH_4^+NH4+​

    • In ammonium ion, nitrogen forms four N−HN-HN−H sigma bonds.
    • There are 4 regions of electron density around nitrogen and no lone pair.
    • Geometry is tetrahedral.

    Therefore, hybridization of N is: sp3sp^3sp3

  4. Collect the results

    NO3−:sp2NO_3^- : sp^2NO3−​:sp2 NO2+:spNO_2^+ : spNO2+​:sp NH4+:sp3NH_4^+ : sp^3NH4+​:sp3

    So the correct sequence is: sp^2, sp, sp^3

  5. Match with the options

    This corresponds to Option B.

  6. Comparison with stored answer

    Stored correct answer: B

    Our derived answer: B

    Hence, the derived answer agrees with the stored correct answer.

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