UPSC Civil Services (Mains), Chemistry optional · 2020 · Paper I
Q3(c). Linear SiH₃-NCS, bent CH₃-NCS
Question
Why SiH₃-NCS is linear, whereas CH₃-NCS is bent?
Answer in brief
In SiH₃NCS the lone pair on N is delocalised into empty acceptor orbitals of Si (π donation N → Si), so N is sp and Si–N–C is linear. Carbon has no such acceptor orbital, so in CH₃NCS the lone pair stays on N, which is sp², and C–N=C is bent.
Key points
- The N atom has three electron pairs to place: the σ bond to Si or C, the σ + π bonds to the C of NCS, and one lone pair. With the lone pair localised, VSEPR (three regions) gives a bent C–N=C arrangement around an sp² nitrogen.
- carbon of CH₃ is in Period 2 and has no low-lying vacant orbital. The lone pair remains on N and stays stereochemically active: the molecule is bent at N.
- silicon is in Period 3 and has vacant low-lying orbitals (3d in the older description; the Si–H σ* orbitals in the modern one). The filled N 2p lone-pair orbital overlaps them, giving N → Si π bonding (pπ–dπ).
- with its lone pair used in π bonding, N needs only two σ directions: it becomes sp hybridised and Si–N=C=S is linear. The Si–N bond gains partial double-bond character and is shorter than a single bond.
The full model answer
Written out in full, as it should go in the answer booklet. The solution page with its diagrams follows it.
- The N atom has three electron pairs to place: the σ bond to Si or C, the σ + π bonds to the C of NCS, and one lone pair. With the lone pair localised, VSEPR (three regions) gives a bent C–N=C arrangement around an sp² nitrogen.
- CH₃NCS: carbon of CH₃ is in Period 2 and has no low-lying vacant orbital. The lone pair remains on N and stays stereochemically active: the molecule is bent at N.
- SiH₃NCS: silicon is in Period 3 and has vacant low-lying orbitals (3d in the older description; the Si–H σ* orbitals in the modern one). The filled N 2p lone-pair orbital overlaps them, giving N → Si π bonding (pπ–dπ).
- Consequence: with its lone pair used in π bonding, N needs only two σ directions: it becomes sp hybridised and Si–N=C=S is linear. The Si–N bond gains partial double-bond character and is shorter than a single bond.
- The same effect explains planar trisilylamine N(SiH₃)₃ against pyramidal N(CH₃)₃, the weaker basicity of (H₃Si)₃N and (H₃Si)₂O, and the wide Si–O–Si angle in disiloxane.
The two molecules side by side
| CH₃-NCS | SiH₃-NCS | |
| Atom bonded to N | C, Period 2: no low-lying vacant orbital | Si, Period 3: low-lying vacant orbitals |
| Lone pair on N | localised, stereochemically active | delocalised into Si (N → Si π) |
| Hybridisation of N | sp² | sp |
| Shape at N | bent | linear, Si–N–C = 180° |
| X–N bond | single | partial double bond, shortened |
| Same pattern | N(CH₃)₃ pyramidal, stronger base | N(SiH₃)₃ planar, very weak base |
Application. silyl ethers and silylamines are weak Lewis bases for the same reason; in silicones and silicates the wide, flexible Si–O–Si angle gives the low glass-transition temperature and flexibility of silicone polymers.
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All parts of Question 3
- 3(a) Most probable distance of the 1s electron of hydrogen10 marks
- 3(b) Colourless Ce³⁺ (4f¹) and red Ce⁴⁺ (4f⁰)10 marks
- 3(c) Linear SiH₃-NCS, bent CH₃-NCS10 marks
- 3(d) Hybridisation and shapes of SO₃ and PF₅10 marks
- 3(e) Debye–Hückel limiting law for an ion i10 marks