Chemistry Q124 – AP-EAMCET 2025

Home Chemistry Chemical BondingQuestion #210

Observe the following molecules.

{SO}_{2}

H_{2} O

{HgCl}_{2}

{BeCl}_{2}

{XeF}_{2}

O_{3}

{NO}_{2}

{PbCl}_{2}

The number of molecules with same geometry as that of

{SnCl}_{2}

Clue 1 of 3: VSEPR Theory

Molecular geometry is determined by the repulsion between electron pairs (both bonding and lone pairs) around the central atom.

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Step 1: Determine the molecular geometry of

{SnCl}_{2}

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Tin (Sn) is the central atom. It belongs to Group 14, so it has 4 valence electrons. Each Chlorine (Cl) atom forms a single bond with Sn. So, 2 bonding pairs are formed. The number of electrons used in bonding is $2 \times 1 = 2$ . The number of remaining valence electrons on Sn is $4 - 2 = 2$ . These 2 electrons form 1 lone pair. Total electron domains around Sn = 2 (bonding pairs) + 1 (lone pair) = 3. According to VSEPR theory, 3 electron domains with 1 lone pair result in a **bent** (or V-shaped) molecular geometry. The electron geometry is trigonal planar.

💡 Teacher's Secret Hint

Remember that lone pairs exert more repulsion than bonding pairs, which compresses the bond angles in molecules like ${SnCl}_{2}$ and $H_{2} O$ .

Step 2: Determine the molecular geometry of each given molecule○ Expand

We will now analyze each molecule to determine its molecular geometry based on VSEPR theory:

- ${SO}_{2}$ : Sulfur (S) is central, 6 valence electrons. It forms 2 bonds with oxygen atoms (one double, one single in resonance structures, but counts as 2 bonding domains) and has 1 lone pair. Total 3 electron domains, 1 lone pair $\to$ **Bent**.

- $H_{2} O$ : Oxygen (O) is central, 6 valence electrons. It forms 2 single bonds with hydrogen atoms and has 2 lone pairs. Total 4 electron domains, 2 lone pairs $\to$ **Bent**.

- ${HgCl}_{2}$ : Mercury (Hg) is central, 2 valence electrons. It forms 2 single bonds with chlorine atoms and has 0 lone pairs. Total 2 electron domains, 0 lone pairs $\to$ **Linear**.

- ${BeCl}_{2}$ : Beryllium (Be) is central, 2 valence electrons. It forms 2 single bonds with chlorine atoms and has 0 lone pairs. Total 2 electron domains, 0 lone pairs $\to$ **Linear**.

- ${XeF}_{2}$ : Xenon (Xe) is central, 8 valence electrons. It forms 2 single bonds with fluorine atoms and has 3 lone pairs. Total 5 electron domains, 3 lone pairs $\to$ **Linear**.

- $O_{3}$ : Central Oxygen (O) has 6 valence electrons. It forms 2 bonds with other oxygen atoms (one double, one single in resonance structures, but counts as 2 bonding domains) and has 1 lone pair. Total 3 electron domains, 1 lone pair $\to$ **Bent**.

- ${NO}_{2}$ : Nitrogen (N) is central, 5 valence electrons. It forms 2 bonds with oxygen atoms (one double, one single) and has 1 unpaired electron. The unpaired electron contributes to repulsion, resulting in a **Bent** geometry.

- ${PbCl}_{2}$ : Lead (Pb) is central, 4 valence electrons. It forms 2 single bonds with chlorine atoms and has 1 lone pair. Total 3 electron domains, 1 lone pair $\to$ **Bent**.

💡 Teacher's Secret Hint

For molecules with odd electrons like ${NO}_{2}$ , the unpaired electron still occupies a region of space and influences the molecular geometry, often leading to a bent shape.

Step 3: Count the molecules with bent geometry○ Expand

The molecules that have a bent molecular geometry, similar to ${SnCl}_{2}$ , are:

1. ${SO}_{2}$ 2. $H_{2} O$ 3. $O_{3}$ 4. ${NO}_{2}$ 5. ${PbCl}_{2}$

There are 5 molecules that have the same bent geometry as ${SnCl}_{2}$ .

💡 Teacher's Secret Hint

Double-check your Lewis structures and electron domain counts for each molecule to ensure accuracy.

Step 4: Select the correct option○ Expand

The count is 5, which corresponds to option 2.

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Chemistry Question 124 – AP-EAMCET 2025

Clue 1 of 3: VSEPR Theory

🧠 Full Solution Path