orbital hybridization?

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Which one of the following statements about orbital hybridization is incorrect?

a. The carbon atom in CH4 is sp3 hybridized.

b. The carbon atom in CO2 is sp hybridized.

c. The nitrogen atom in NH3 is sp2 hybridized.

d. sp hybrid orbitals lie at 180° to each other.

e. sp2 hybrid orbitals are coplanar, and at 120° to each other.

4 Answers

  • C. In NH3, if we consider the lone pair of electrons, it would be similar to tetrahedral geometry (Differs from exact tetrahedral geometry due to VSEPR theory).

    – In tetrahedral geometry, there are 4 vertices (3 H and 1 lone pair), hence N is sp3 hybridised. [Hint: always include 1 hybrid orbitals for each lone pair]

    CH4 is tetrahedral => sp3 hybridised

    CO2 is linear => no lone pair. [Note: Electrons that are unpaired remain in the unhybridised (px, py or pz) orbitals to form pi-bonds]

    CO2 is an example of a linear molecule (bond angle = 180 degrees) and the C atom is sp hybridised.

    In BF3 (bond angle = 120 degrees), the B is sp2 hybridised because it has only 3 unpaired electrons.

  • Statement (c) is incorrect. The N atom in NH3 is sp3 hybridized

  • as a manner to determine hybridization, count selection up the electron domain names around the atom of interest (many times the necessary atom). A single bond = a million area double bond = a million area triple bond = a million area lone pair of electrons = a million area count form of the domain names and fill in with letters. you ought to use right here: maximum a million s maximum 3 p maximum 5 d maximum 7 f those numbers come from the reality that for a given point or quantum selection, there is completely a million conceivable s orbital, there are 3 conceivable p orbitals (px, py, pz) 5 conceivable d orbitals (dxy, dyz, dxz, dx2-y2, dz2) and 7 conceivable f orbitals (no longer seen in hybridization) available to be “smooshed” mutually to variety a hybrid orbital So, H2-C=O – there are 3 electron domain names around the C. Fill in with letters, spp so sp2 H2C=CH2 – there are 3 domain names around the 1st carbon and additionally around the 2nd carbon, so we want 3 letters, spp returned so sp2 CH4 there are 4 domain names around the C, fill in with letters, sppp, so sp3 HCN there are 2 domain names around the C – fill in with letters sp O=C=O there are 2 domain names around the C, fill in with letters so sp undergo in thoughts that all and sundry domain names purely count selection as ONE. A single bond = a million area, a double bond = a million area, a triple bond = a million area, a lone pair = a million area. Fill in with letters: spppddddd consequently 2 domain names: sp 3 domain names: spp = sp2 4 domain names: sppp = sp3 5 domain names: spppd = sp3d 6 domain names: spppdd = sp3d2

  • c

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