Chemistry · Chemical Bonding · NEET
Very easy. Just count how many atoms are directly bonded to the central atom. That number decides the shape. 2 atoms = linear (BeCl2), 3 atoms = trigonal planar (BF3), 4 atoms = tetrahedral (CH4), 5 atoms = trigonal bipyramidal (PCl5), 6 atoms = octahedral (SF6). No lone pair means nothing bends the shape, so the bonded atoms simply move as far apart as possible.
AB2 linear = 180 degrees. AB3 trigonal planar = 120 degrees. AB4 tetrahedral = 109.5 degrees. AB5 trigonal bipyramidal = 120 degrees (equatorial, in the middle plane) and 90 degrees (axial, top and bottom). AB6 octahedral = 90 degrees. Remember: TBP has TWO angles because its positions are not all the same.
The 4 bond pairs repel each other and want to be as far apart as possible in 3D space. A flat square would give 90 degree angles, which is too close. Spreading into a 3D tetrahedron gives 109.5 degrees, which is the largest angle possible for 4 pairs. VSEPR always picks the shape with least repulsion.
In PCl5 (TBP), the 5 atoms are not equal. Three atoms lie in a middle flat triangle (equatorial, 120 degrees apart). Two atoms point straight up and down (axial, 90 degrees to the plane). Axial bonds face 3 nearby equatorial bonds, so they feel more repulsion and are slightly longer/weaker. This is the only common no-lone-pair shape where all bonds are NOT identical.
They match one-to-one when there is no lone pair. AB2 = sp = linear. AB3 = sp2 = trigonal planar. AB4 = sp3 = tetrahedral. AB5 = sp3d = trigonal bipyramidal. AB6 = sp3d2 = octahedral. So if a question gives you the hybridisation, you can read off the shape directly, and vice versa. This shortcut works ONLY when lone pairs are zero.
Do a quick electron count. For BF3: B has 3 valence electrons, all used to bond 3 F atoms, so 0 lone pairs left. For SF6: S has 6 valence electrons, all 6 go to bond 6 F atoms, so 0 lone pairs. If every valence electron of the central atom is used in bonding, the lone pair count is zero and you use these 5 clean shapes.
Match List-I with List-II. List-I: (a) PCl5, (b) SF6, (c) BrF5, (d) BF3. List-II: (i) Square pyramidal, (ii) Trigonal planar, (iii) Octahedral, (iv) Trigonal bipyramidal.
BF3 is planar and an electron-deficient compound. The hybridization and number of electrons around the central atom, respectively, are:
Match the species in List-I with their geometry in List-II. List-I: A. PCl5, B. BrF5, C. BF4-, D. [Ni(CN)4]2-. List-II: I. Tetrahedral, II. Square planar, III. Trigonal bipyramidal, IV. Square pyramidal.
Try the real previous-year questions from this chapter — each with the answer and a full solution.
Yes. With zero lone pairs the bonded atoms spread out evenly, giving perfectly symmetric shapes: linear, trigonal planar, tetrahedral, trigonal bipyramidal and octahedral. This also makes many such molecules (like BF3, CO2, CH4, SF6) non-polar even if the bonds are polar, because the bond dipoles cancel.
AB2 linear: BeCl2, CO2. AB3 trigonal planar: BF3, BCl3. AB4 tetrahedral: CH4, CCl4, BF4-, NH4+, SiCl4. AB5 trigonal bipyramidal: PCl5, PF5. AB6 octahedral: SF6, [Al(H2O)6]3+. Memorise one clean example per shape.
Because its 5 positions are of two kinds. The 3 equatorial positions sit 120 degrees apart in a flat triangle, and the 2 axial positions are 90 degrees above and below that plane. So a TBP molecule has both 120 and 90 degree angles, unlike the other four shapes which have a single angle.
Almost always as a match-the-column question: molecule to shape, or molecule to hybridisation to shape. PCl5, SF6, BF3 and BF4- appear repeatedly. If you know the 5 no-lone-pair shapes cold, you can lock those matches instantly and only think hard about the lone-pair ones.
