Chemistry · Chemical Bonding · NEET
Follow 5 steps every time. 1) Pick the central atom (usually the least electronegative one, or the single atom like C in CH4, N in NH3). 2) Count bond pairs = number of atoms directly bonded to it (a double or triple bond still counts as ONE region). 3) Count lone pairs on the central atom using: lone pairs = (valence electrons of central atom - electrons used in bonds) / 2. 4) Add bond pairs + lone pairs = steric number, and arrange these regions as far apart as possible. 5) Look only at where the ATOMS are (ignore lone pairs) to name the final shape. This order never changes, so practise it until it is automatic.
Use this shortcut: L = (V - B) / 2, where V = valence electrons of the central atom and B = number of single bonds (sigma bonds) to surrounding atoms, adjusted for charge. Example NH3: N has 5 valence electrons, makes 3 N-H bonds, so L = (5 - 3)/2 = 1 lone pair. Example H2O: O has 6, makes 2 O-H bonds, so L = (6 - 2)/2 = 2 lone pairs. For an ion, add one electron for each negative charge and subtract one for each positive charge before dividing.
Electron geometry counts ALL regions (bond pairs AND lone pairs) and tells you how they spread in space. Molecular shape looks only at the ATOMS, so it ignores the lone pairs. Example: H2O has 4 electron regions (2 bonds + 2 lone pairs) so its electron geometry is tetrahedral, but because we only see the atoms, its molecular shape is bent. This is why NEET can ask you for shape and geometry separately, so always answer for atoms only when they ask for shape.
A lone pair belongs to only one atom, so it spreads out more and pushes harder than a bond pair (which is shared between two atoms). The repulsion order is lone pair-lone pair > lone pair-bond pair > bond pair-bond pair. Because lone pairs push the bonds closer together, the bond angle drops. That is why the angle goes 109.5 degrees (CH4, no lone pair) to 107 degrees (NH3, 1 lone pair) to 104.5 degrees (H2O, 2 lone pairs).
A = central atom, B = each bonded atom, x = number of bond pairs, E = lone pair, y = number of lone pairs. So CH4 is AB4 (4 bonds, 0 lone pairs = tetrahedral), NH3 is AB3E (3 bonds, 1 lone pair = trigonal pyramidal), H2O is AB2E2 (2 bonds, 2 lone pairs = bent), SF4 is AB4E (see-saw), ClF3 is AB3E2 (T-shape), XeF2 is AB2E3 (linear). Once you write the correct ABxEy label, the shape is fixed and you just recall it.
No. In VSEPR a double bond or triple bond counts as ONE electron region, just like a single bond, when you decide the shape. Example CO2: carbon has 2 double bonds, that is 2 regions and 0 lone pairs, so it is linear (180 degrees). The extra pi electrons do not add a new direction. They only make that region slightly fatter, which can shrink the other angles a little, but they never count as a separate corner.
Predict the correct order among the following (repulsion between electron pairs as per VSEPR theory):
Amongst the following, which one will have maximum 'lone pair-lone pair' electron repulsion?
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Steric number = number of bond pairs + number of lone pairs on the central atom. It tells you the basic arrangement: 2 = linear, 3 = trigonal planar, 4 = tetrahedral, 5 = trigonal bipyramidal, 6 = octahedral. You find the shape by then removing the lone pairs.
Choose the atom that is usually least electronegative and can form the most bonds, often the single atom in the formula. In CH4 it is C, in NH3 it is N, in H2O it is O, in SF4 it is S. Hydrogen is never the central atom because it forms only one bond.
Yes. First adjust the electron count for the charge: subtract one electron for each + charge and add one for each - charge, then count bond pairs and lone pairs as usual. NH4+ is AB4 (tetrahedral) and CO3^2- is AB3 with 0 lone pairs (trigonal planar).
In a trigonal bipyramid the equatorial positions have more room (only 2 close neighbours at 90 degrees instead of 3). A lone pair needs the most space, so it goes equatorial. This is why SF4 is see-saw, ClF3 is T-shaped and XeF2 is linear.
Shape questions appear almost every year, often as match-the-column with hybridisation. If you can quickly count bond pairs and lone pairs and read off the shape, you can solve these in seconds and also predict polarity and bond angle, which are common follow-up traps.
