Chemistry · Chemical Bonding · NEET
Find the steric number. Steric number = (number of sigma bonds made by the central atom) + (number of lone pairs on the central atom). Then match: 2 = sp, 3 = sp2, 4 = sp3, 5 = sp3d, 6 = sp3d2, 7 = sp3d3. Example: In NH4+, nitrogen has 4 sigma bonds and 0 lone pairs, so steric number = 4, hybridisation = sp3. This one step works for almost every NEET question, so learn it well.
No. Only count sigma bonds, never pi bonds. A double bond has 1 sigma + 1 pi, and a triple bond has 1 sigma + 2 pi. For hybridisation you count each double or triple bond as just ONE sigma. That is why in NO2+ nitrogen has two double bonds (O=N=O) but only 2 sigma bonds and 0 lone pairs, giving steric number 2 = sp. This trap costs many students marks in NEET.
Use this formula for lone pairs on the central atom: LP = (V - number of bonds it forms) / 2, where V is the valence electrons of the central atom, adjusted for charge. Or simply draw the Lewis structure and count the non-bonding pairs. Example: In H2O, oxygen has 6 valence electrons, uses 2 for bonds, leaving 4 electrons = 2 lone pairs. So steric number = 2 sigma + 2 lone pairs = 4 = sp3.
Adjust the central atom's electron count. For a negative charge, add that many electrons; for a positive charge, subtract. A short formula for steric number is H = (1/2)[V + M - C + A], where V = valence electrons of central atom, M = number of monovalent atoms (like H or halogen) bonded, C = positive charge, A = negative charge. For NH4+: (1/2)[5 + 4 - 1 + 0] = 4, so sp3. For NO3-: count 3 sigma bonds + 0 lone pairs = 3, so sp2.
By the total number of electron pairs (bond pairs + lone pairs) around the central atom, which is the steric number. Shape is decided AFTER, because lone pairs are invisible in the shape. Example: XeF4 has steric number 6 (sp3d2), so its electron geometry is octahedral, but because 2 positions are lone pairs, the visible shape is square planar. NEET sometimes asks hybridisation and shape in the same question, so keep them separate.
Yes. More s-character means a larger bond angle. sp (50% s) gives 180 degrees, sp2 gives 120 degrees, sp3 gives about 109.5 degrees. Lone pairs then push bond pairs closer and reduce the angle a little (like 104.5 in H2O). So first find hybridisation by steric number, then remember lone pairs squeeze the angle smaller.
The hybridizations of the atomic orbitals of nitrogen in NO2+, NO3- and NH4+ respectively are:
BF3 is planar and an electron-deficient compound. The hybridization and number of electrons around the central atom, respectively, are:
Which of the following molecules represents the order of hybridisation sp2, sp2, sp, sp from left to right atoms?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Steric number = number of sigma bonds on the central atom + number of lone pairs on it. Then 2 = sp, 3 = sp2, 4 = sp3, 5 = sp3d, 6 = sp3d2, 7 = sp3d3. For ions, use H = (1/2)[V + M - C + A].
Hybridisation forms sigma bonds and holds lone pairs. Pi bonds are made by side-on overlap of unhybridised p orbitals, so they do not use hybrid orbitals. A double or triple bond counts as only one sigma bond.
Xenon in XeF4 has 4 sigma bonds + 2 lone pairs = steric number 6 = sp3d2. The electron geometry is octahedral, but the visible molecular shape is square planar because of the two lone pairs. This was tested in NEET 2016.
Lone pairs are counted in the steric number, so they DO decide the hybridisation. But they are invisible in the drawn shape, so the shape looks different from the electron geometry. Example: NH3 is sp3 but shaped pyramidal, not tetrahedral.
Bigger. sp has 50% s-character and 180 degree angle; sp2 has about 120 degrees; sp3 has about 109.5 degrees. So sp > sp2 > sp3 in bond angle. Lone pairs then reduce the angle slightly.
