Chemistry Q125 – AP-EAMCET 2026

Home Chemistry Chemical BondingQuestion #209

The pair of molecules with same type of hybridisation is

Clue 1 of 3: VSEPR Theory and Hybridization

Hybridization of the central atom in a molecule is determined by the number of sigma bonds and lone pairs around it.

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Step 1: Determine Hybridization for Option 1:

H_{3} {BO}_{3}

and

H_{3} {PO}_{3}

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For $H_{3} {BO}_{3}$ (Boric acid), the central atom is Boron (B). Boron has 3 valence electrons and forms 3 single bonds with -OH groups. There are no lone pairs on Boron. Thus, the steric number for B is 3 (3 sigma bonds + 0 lone pairs). This corresponds to $s p^{2}$ hybridization.

For $H_{3} {PO}_{3}$ (Phosphorous acid), the central atom is Phosphorus (P). Phosphorus has 5 valence electrons. The structure is ${O=P(-OH)}_{2} H$ . P forms one double bond with O, two single bonds with -OH groups, and one single bond with H. This means P forms 4 sigma bonds (1 from P=O, 2 from P-OH, 1 from P-H) and has 0 lone pairs. Thus, the steric number for P is 4 (4 sigma bonds + 0 lone pairs). This corresponds to $s p^{3}$ hybridization. Since the hybridizations are different ( $s p^{2}$ vs $s p^{3}$ ), option 1 is incorrect.

💡 Teacher's Secret Hint

Remember that in oxoacids like $H_{3} {PO}_{3}$ , hydrogen atoms are often attached to oxygen atoms, but in phosphorous acid, one H atom is directly bonded to phosphorus.

Step 2: Determine Hybridization for Option 2:

{NO}_{2}

and

{SO}_{3}

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For ${NO}_{2}$ (Nitrogen dioxide), the central atom is Nitrogen (N). Nitrogen has 5 valence electrons. The molecule has an odd number of electrons (5 + 2*6 = 17). The structure involves resonance between O=N-O and O-N=O with an unpaired electron on N. In the VSEPR model for ${NO}_{2}$ , Nitrogen forms two sigma bonds (one to each Oxygen) and has one unpaired electron. These three regions of electron density (2 sigma bonds + 1 unpaired electron) correspond to a steric number of 3, leading to $s p^{2}$ hybridization.

For ${SO}_{3}$ (Sulfur trioxide), the central atom is Sulfur (S). Sulfur has 6 valence electrons. The structure shows Sulfur double-bonded to all three Oxygen atoms. Sulfur forms 3 sigma bonds (one to each O) and has 0 lone pairs. Thus, the steric number for S is 3 (3 sigma bonds + 0 lone pairs). This corresponds to $s p^{2}$ hybridization. Since both molecules have $s p^{2}$ hybridization, option 2 is correct.

💡 Teacher's Secret Hint

For molecules with odd electrons like ${NO}_{2}$ , the unpaired electron is often considered a region of electron density for hybridization purposes.

Step 3: Determine Hybridization for Option 3:

{XeO}_{3}

and

{BF}_{3}

○ Expand

For ${XeO}_{3}$ (Xenon trioxide), the central atom is Xenon (Xe). Xenon has 8 valence electrons. Xe forms double bonds with three Oxygen atoms and has one lone pair. Thus, Xe forms 3 sigma bonds and has 1 lone pair. The steric number for Xe is 4 (3 sigma bonds + 1 lone pair). This corresponds to $s p^{3}$ hybridization.

For ${BF}_{3}$ (Boron trifluoride), the central atom is Boron (B). Boron has 3 valence electrons and forms 3 single bonds with Fluorine atoms. There are no lone pairs on Boron. Thus, the steric number for B is 3 (3 sigma bonds + 0 lone pairs). This corresponds to $s p^{2}$ hybridization. Since the hybridizations are different ( $s p^{3}$ vs $s p^{2}$ ), option 3 is incorrect.

💡 Teacher's Secret Hint

Noble gases can form compounds, especially with highly electronegative elements like Fluorine and Oxygen, by expanding their octet.

Step 4: Determine Hybridization for Option 4:

{PCl}_{3}

and

{ClF}_{3}

○ Expand

For ${PCl}_{3}$ (Phosphorus trichloride), the central atom is Phosphorus (P). Phosphorus has 5 valence electrons. P forms 3 single bonds with Cl atoms and has 1 lone pair. Thus, the steric number for P is 4 (3 sigma bonds + 1 lone pair). This corresponds to $s p^{3}$ hybridization.

For ${ClF}_{3}$ (Chlorine trifluoride), the central atom is Chlorine (Cl). Chlorine has 7 valence electrons. Cl forms 3 single bonds with F atoms and has 2 lone pairs. Thus, the steric number for Cl is 5 (3 sigma bonds + 2 lone pairs). This corresponds to $s p^{3} d$ hybridization. Since the hybridizations are different ( $s p^{3}$ vs $s p^{3} d$ ), option 4 is incorrect.

💡 Teacher's Secret Hint

Elements from the third period and beyond can expand their octet, allowing for steric numbers greater than 4.

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Chemistry Question 125 – AP-EAMCET 2026

Clue 1 of 3: VSEPR Theory and Hybridization

🧠 Full Solution Path