MPSC 2023 Prelims Paper-I · Question 63 of 100

Correct — option (4). Both molecules have a central atom bonded to exactly three others, so their shapes must be decided by something other than the number of bonds, and that something is the lone pair. The tool is VSEPR theory, which holds that the electron pairs in the valence shell of the central atom — bonding pairs and lone pairs alike — arrange themselves as far apart as possible, and that the shape we name is the arrangement of the atoms alone, with the lone pairs invisible but still pushing. Take ammonia first. Nitrogen has five valence electrons; three of them are used in bonds to hydrogen and the remaining two stay on nitrogen as a lone pair. That gives four electron domains around the central atom, which arrange themselves tetrahedrally, and nitrogen is accordingly sp³ hybridised. Because one of those four positions is occupied by a lone pair rather than by an atom, the four nuclei do not form a tetrahedron: the nitrogen sits at an apex above a triangular base of three hydrogens, and the shape is described as trigonal pyramidal. The lone pair also repels the bonding pairs more strongly than they repel one another, which squeezes the H–N–H angle down from the regular tetrahedral 109.5° to about 107°. Now boron trifluoride. Boron has only three valence electrons and uses all three in bonds to fluorine, so after bonding it carries six electrons in its valence shell rather than eight — it is an electron-deficient molecule and has no lone pair at all. Three electron domains and nothing else means the three fluorines spread out in a single plane at 120° to one another, boron is sp² hybridised, and the molecule is trigonal planar, with all four atoms lying flat in the same plane. So ammonia is pyramidal and boron trifluoride is planar, which is precisely what option (4) states. The consequence worth carrying away is that the difference in shape produces a difference in polarity: the individual N–H and B–F bonds are all polar, but in the symmetric planar arrangement of BF₃ the three bond dipoles cancel exactly and the molecule is non-polar, whereas in the pyramidal ammonia molecule they do not cancel and add to the lone pair's own contribution, making ammonia distinctly polar.

• (1)NH₃ – Pyramidal and BF₃ – Pyramidal — The first half of this option is right and the second half is wrong, which is what makes it the most attractive of the three. Ammonia is indeed pyramidal, but boron trifluoride is not: describing it as pyramidal would require a fourth electron domain on boron to push the three fluorines out of the plane, and boron simply has no lone pair to supply one. Having used all three of its valence electrons in bonding, boron is left with six electrons in its valence shell, and three bonding domains with nothing else present spread out flat at 120°. This option is the trap for a candidate who has learnt that molecules with three bonds tend to be pyramidal — a generalisation drawn from ammonia, phosphine and the trihalides of nitrogen and phosphorus, all of which carry a lone pair, and which fails exactly where the central atom is electron-deficient.
• (2)NH₃ – Planar and BF₃ – Planar — Here the boron trifluoride half is correct and the ammonia half is not. Ammonia cannot be planar, because a planar arrangement of three bonds around nitrogen would leave the lone pair with nowhere to go except perpendicular to that plane, and VSEPR requires the lone pair to be counted as an electron domain competing for space with the bonds. Counting it gives four domains, a tetrahedral electron arrangement and a pyramidal molecule. It is worth knowing that the pyramid is not rigid — ammonia inverts rapidly through a planar transition state, the so-called umbrella inversion — but that planar form is a fleeting configuration passed through during inversion and not the shape of the molecule, which is trigonal pyramidal with a bond angle of about 107°.
• (3)NH₃ – Planar and BF₃ – Pyramidal — This option has both halves exactly the wrong way round, assigning planar to ammonia and pyramidal to boron trifluoride, and it is what a candidate produces by knowing that one of the two molecules is pyramidal and the other planar but attaching each label to the wrong formula. The way to fix the assignment rather than guess at it is to count valence electrons on the central atom: nitrogen with five electrons and three bonds keeps a lone pair and is therefore pyramidal, while boron with three electrons and three bonds keeps nothing and is therefore planar. That count also explains the chemistry of the pair. Ammonia, with a lone pair to donate, is a Lewis base; boron trifluoride, with a vacant orbital and an incomplete octet, is a strong Lewis acid; and when the two combine to form the adduct H₃N→BF₃ the boron acquires a fourth electron domain, becomes sp³ hybridised, and the BF₃ unit itself pyramidalises.

VSEPR theory predicts molecular shape from a single idea: electron pairs in the valence shell of the central atom repel one another and settle into the arrangement that keeps them furthest apart. The procedure is mechanical. Count the electron domains around the central atom — each single, double or triple bond counts once, and each lone pair counts once — then read off the electron-pair geometry: two domains give a linear arrangement, three give trigonal planar at 120°, four give tetrahedral at 109.5°, five trigonal bipyramidal and six octahedral. The molecular shape is then the geometry of the atoms alone, obtained by deleting the lone pairs from that arrangement while leaving the positions of the bonded atoms as the lone pairs have set them. Four domains with no lone pair give a tetrahedron, as in methane; four with one lone pair give a trigonal pyramid, as in ammonia; four with two lone pairs give a bent molecule, as in water. Repulsion is not equal in all directions, and the ordering lone pair–lone pair greater than lone pair–bond pair greater than bond pair–bond pair explains the observed contraction of bond angles down the series methane 109.5°, ammonia about 107°, water about 104.5° as lone pairs are added. The same domain count fixes the hybridisation of the central atom — two domains sp, three sp², four sp³ — so shape, hybridisation and bond angle are three readings of one underlying count. Shape in turn governs polarity: a molecule whose bond dipoles are arranged symmetrically about the central atom has no net dipole even when every bond in it is polar, which is why trigonal planar BF₃ and linear CO₂ are non-polar while pyramidal NH₃ and bent H₂O are polar.

Molecular geometry is among the most reliably examined chemistry topics in MPSC's science section, because it can be tested in a single short line without any figure and rewards understanding rather than recall. This item is built on the sharpest contrast the topic offers — two molecules with the same number of bonds and different shapes — so a candidate cannot answer it by counting bonds and must actually count electrons on the central atom. The Commission's habitual construction is visible in the option set: rather than offering four unrelated shapes, it offers all four combinations of two shapes across two molecules, so that knowing one half of the answer eliminates only two options and the mark is awarded only for knowing both. Two of the wrong options are each half correct, which is why partial knowledge is worth nothing here. The efficient preparation is not a list of molecules and shapes but the counting procedure itself, applied to a dozen standard examples — methane, ammonia, water, boron trifluoride, carbon dioxide, phosphorus pentachloride, sulphur hexafluoride, sulphur dioxide — since any question the Commission can set on this topic can be worked out in seconds from the count. Note that the formulae here are printed in Latin script inside the Marathi column too, only the shape names being transliterated, so the two columns pose an identical question.

• Ammonia has four electron domains around nitrogen — three bonding pairs and one lone pair — so nitrogen is sp³ hybridised, the electron arrangement is tetrahedral and the molecular shape is trigonal pyramidal with a bond angle of about 107°.
• Boron trifluoride has three bonding domains and no lone pair on boron, so boron is sp² hybridised and the molecule is trigonal planar with F–B–F angles of 120°, all four atoms lying in one plane.
• Boron in BF₃ carries only six electrons in its valence shell after bonding, making the molecule electron-deficient and a strong Lewis acid, while ammonia's lone pair makes it a Lewis base; the two combine to form the adduct H₃N→BF₃, in which the boron becomes sp³ and the BF₃ unit pyramidalises.
• In VSEPR the repulsion order is lone pair–lone pair greater than lone pair–bond pair greater than bond pair–bond pair, which is why bond angles contract along the series methane 109.5°, ammonia about 107°, water about 104.5°.
• BF₃ is non-polar despite having three polar B–F bonds, because its symmetric trigonal planar geometry causes the bond dipoles to cancel exactly; ammonia's pyramidal geometry prevents such cancellation and the molecule has a substantial dipole moment.

• Deciding the shape from the number of bonded atoms alone, which makes ammonia and boron trifluoride look identical when the presence or absence of a lone pair is what separates them
• Confusing the electron-pair geometry with the molecular shape, and so calling ammonia tetrahedral because its four electron domains are tetrahedrally arranged
• Assuming a molecule with polar bonds must itself be polar, when a symmetric arrangement such as that of BF₃ cancels the bond dipoles exactly
• Expecting every central atom to complete an octet, when boron and beryllium form stable electron-deficient molecules with six and four valence electrons respectively
• Knowing one half of a paired-comparison option and guessing the other, when the Commission has offered all four combinations precisely so that partial knowledge earns nothing

Molecular shape reaches MPSC papers in three recurring forms: name the shape of a given molecule, pair several molecules with their shapes or hybridisations in a matching item, or compare two molecules as this question does. The molecules used are drawn from a short and stable list — methane, ammonia, water, boron trifluoride, carbon dioxide, sulphur dioxide, phosphorus pentachloride and sulphur hexafluoride — and the contrasts the Commission favours are those where the naive count misleads, above all the pyramidal-against-planar pair used here and the bent-against-linear pair of sulphur dioxide against carbon dioxide. Related questions ask which molecule is non-polar despite polar bonds, or place the bond angles of methane, ammonia and water in order, both of which follow from the same domain count. Preparing the count rather than the list covers all of them.

No directly related past PYQ was found.

• practice — not a real PYQ

  Which of the following molecules is non-polar even though each of the bonds within it is polar ?

  • (a)NH₃
  • (b)H₂O
  • (c)BF₃
  • (d)SO₂

  Answer(c) BF₃ — each B–F bond is strongly polar because fluorine is far more electronegative than boron, but the three bonds point outwards at 120° in a single plane and their dipoles cancel exactly, leaving the molecule with no net dipole moment. Ammonia and water are polar because their lone pairs make the arrangement of bonds unsymmetrical, and sulphur dioxide is bent for the same reason, so in all three the bond dipoles fail to cancel.

• practice — not a real PYQ

  The bond angles of methane, ammonia and water are correctly placed in decreasing order in which of the following ?

  • (a)H₂O > NH₃ > CH₄
  • (b)CH₄ > NH₃ > H₂O
  • (c)NH₃ > CH₄ > H₂O
  • (d)CH₄ > H₂O > NH₃

  Answer(b) CH₄ > NH₃ > H₂O — all three have four electron domains around the central atom, but methane has no lone pair and keeps the regular tetrahedral angle of 109.5°, ammonia has one lone pair which compresses the angle to about 107°, and water has two lone pairs which compress it further to about 104.5°. The order follows directly from the VSEPR rule that lone pairs repel bonding pairs more strongly than bonding pairs repel one another.