In molecular geometry, match the compounds in List I (CH3F, HCHO, HCN and NH3) with the shapes of their molecules in List II (tetrahedral, trigonal planar, linear and trigonal pyramidal) and select the correct code given below.

Introduction / Context:
This question tests your understanding of basic molecular geometry, as predicted by the valence shell electron pair repulsion (VSEPR) theory. You are asked to match a set of simple molecules and ions, namely CH3F, HCHO, HCN and NH3, with their correct three dimensional shapes such as tetrahedral, trigonal planar, linear and trigonal pyramidal. Knowing how to assign shapes based on the number of electron domains and lone pairs around the central atom is a core skill in introductory chemistry and is frequently used to explain reactivity, polarity and physical properties.

Given Data / Assumptions:
In this question we consider the following points as given or standard knowledge.
CH3F is a molecule with carbon as the central atom bonded to three hydrogen atoms and one fluorine atom.
HCHO is methanal or formaldehyde, with carbon double bonded to oxygen and single bonded to two hydrogen atoms.
HCN is hydrogen cyanide, where carbon is between hydrogen and nitrogen and forms a triple bond with nitrogen.
NH3 is ammonia, with nitrogen bonded to three hydrogen atoms and having one lone pair of electrons.
We use standard VSEPR theory to predict shapes from electron domain geometry and lone pairs.

Concept / Approach:
According to VSEPR theory, the shape of a molecule is determined by the number of electron pairs (bonding plus lone pairs) around the central atom. The electron pair geometry considers all electron domains, while the molecular shape focuses on the positions of atoms. Four electron pairs with no lone pairs around the central atom give a tetrahedral shape. Three electron pairs with no lone pairs give a trigonal planar shape. Two electron pairs with no lone pairs give a linear shape. Three bonding pairs plus one lone pair give a trigonal pyramidal molecular shape. The task is to apply these ideas to each molecule and then write the correct code for the pairings between List I and List II.

Step-by-Step Solution:
Step 1: For CH3F, carbon is bonded to three hydrogen atoms and one fluorine atom. That means four bonding pairs and no lone pairs on carbon, so the shape around carbon is tetrahedral. Therefore CH3F corresponds to code 2 (tetrahedral). Step 2: For HCHO (formaldehyde), the carbon atom is bonded to two hydrogens and double bonded to oxygen. This gives three regions of electron density around carbon (two single bonds and one double bond) with no lone pairs, which corresponds to a trigonal planar shape. Therefore HCHO corresponds to code 1 (trigonal planar). Step 3: For HCN, the carbon is bonded linearly between hydrogen and nitrogen, and there are two regions of electron density (one single bond and one triple bond), which leads to a linear shape. Therefore HCN corresponds to code 4 (linear). Step 4: For NH3, nitrogen has three N−H bonds and one lone pair of electrons. This gives four electron pairs in total, but one is a lone pair. The electron pair geometry is tetrahedral, while the molecular shape is trigonal pyramidal. Therefore NH3 corresponds to code 3 (trigonal pyramidal). Step 5: Collecting the matches, we get A. CH3F → 2, B. HCHO → 1, C. HCN → 4, D. NH3 → 3, which produces the sequence 2 1 4 3.

Verification / Alternative check:
Another way to verify the shapes is to consider typical textbook examples. CH4 and CH3F are classic tetrahedral molecules with four sigma bonds around carbon. Formaldehyde HCHO is routinely cited as trigonal planar due to the sp2 hybridisation at carbon. Hydrogen cyanide HCN is used as a standard example of a linear molecule with a triple bond. Ammonia NH3 is a well known trigonal pyramidal molecule due to the lone pair on nitrogen. These standard shapes confirm that the mapping A → 2, B → 1, C → 4 and D → 3 is correct, reinforcing the earlier VSEPR based reasoning.

Why Other Options Are Wrong:
Option 2 4 1 3 would assign CH3F as tetrahedral but would incorrectly label HCHO as linear and HCN as trigonal planar, which contradicts VSEPR predictions and experimental data.
Option 3 4 1 2 mixes up the shapes even more, implying that CH3F is trigonal pyramidal and NH3 is tetrahedral without a lone pair, which is inconsistent with the known structure of ammonia.
Option 3 1 4 2 wrongly assigns NH3 as tetrahedral and CH3F as trigonal pyramidal, reversing the correct interpretation of lone pairs on nitrogen versus carbon in these molecules.

Common Pitfalls:
A common mistake is to confuse electron pair geometry with molecular shape and to forget that lone pairs change the molecular geometry but are not counted as atoms. Students also sometimes miscount double or triple bonds as multiple electron pairs, whereas VSEPR counts them as one domain of electron density. Another pitfall is to assume that any four bonded atoms around a central atom must always produce the same shape, without checking for the presence of lone pairs. Careful counting of electron domains and awareness of lone pairs help to avoid these errors.

Final Answer:
The correct matching of molecules with their shapes is CH3F tetrahedral, HCHO trigonal planar, HCN linear and NH3 trigonal pyramidal, which corresponds to the code 2 1 4 3.