The option(s) in which at least three molecules follow Octet Rule is(are)
- CO2, C2H4, NO and HCl
- NO2, O3, HCl and H2SO4
- BCl3, NO, NO2 and H2SO4
- CO2, BCl3, O3 and C2H4
The Correct Option is A, D
Approach Solution - 1
To determine which option has at least three molecules that follow the octet rule, we need to evaluate each molecule in the given options. The octet rule states that atoms tend to gain, lose, or share electrons to achieve a stable configuration with 8 electrons in their valence shell (except for hydrogen, which follows the duet rule with 2 electrons).
Let’s analyze each option:
Option A: CO₂, C₂H₄, NO, and HCl
- CO₂ (Carbon Dioxide): Carbon is the central atom, double-bonded to two oxygen atoms (O=C=O). Carbon has 4 valence electrons and forms two double bonds (8 electrons shared), satisfying the octet rule. Each oxygen has 6 valence electrons, and with a double bond (4 electrons) plus 4 lone pair electrons, they also have 8 electrons. Follows octet rule.
- C₂H₄ (Ethylene): Each carbon is bonded to two hydrogens and double-bonded to the other carbon (H₂C=CH₂). Each carbon has 4 valence electrons, with 4 bonds (8 electrons shared), satisfying the octet rule. Hydrogens have 2 electrons (duet rule). Follows octet rule.
- NO (Nitric Oxide): Nitrogen has 5 valence electrons, oxygen has 6. In NO, there’s a double bond (sometimes represented with a triple bond in resonance structures), but nitrogen ends up with 7 electrons (5 + 2 from the bond) and oxygen with 8. Nitrogen does not satisfy the octet rule due to the odd number of electrons (11 total). Does not follow octet rule.
- HCl (Hydrogen Chloride): Hydrogen has 1 valence electron, chlorine has 7. They form a single bond, giving hydrogen 2 electrons (duet rule) and chlorine 8 electrons (1 bond + 6 lone pair electrons). Follows octet rule.
Result for Option A: CO₂, C₂H₄, and HCl follow the octet rule (3 molecules).
Option B: NO₂, O₃, HCl, and H₂SO₄
- NO₂ (Nitrogen Dioxide): Nitrogen has 5 valence electrons, each oxygen has 6. NO₂ has an odd number of electrons (17), so nitrogen cannot achieve an octet. Resonance structures show nitrogen with 7 electrons. Does not follow octet rule.
- O₃ (Ozone): Each oxygen has 6 valence electrons. In the resonance structures of O₃ (O–O=O ↔︎ O=O–O), the central oxygen has 8 electrons (double bond), and the terminal oxygens share electrons to achieve 8 electrons each. Follows octet rule.
- HCl: As analyzed in Option A, follows octet rule.
- H₂SO₄ (Sulfuric Acid): Sulfur is the central atom, bonded to four oxygens (two with double bonds, two with single bonds to OH groups). Sulfur has 6 valence electrons but forms 6 bonds (12 electrons in its valence shell), exceeding the octet rule (expanded octet, common for elements beyond the second period). Does not follow octet rule.
Result for Option B: O₃ and HCl follow the octet rule (2 molecules).
Option C: BCl₃, NO, NO₂, and H₂SO₄
- BCl₃ (Boron Trichloride): Boron has 3 valence electrons and forms 3 single bonds with chlorine, resulting in 6 electrons around boron. Does not follow octet rule.
- NO: As analyzed in Option A, does not follow octet rule.
- NO₂: As analyzed in Option B, does not follow octet rule.
- H₂SO₄: As analyzed in Option B, does not follow octet rule.
Result for Option C: No molecules follow the octet rule (0 molecules).
Option D: CO₂, BCl₃, O₃, and C₂H₄
- CO₂: As analyzed in Option A, follows octet rule.
- BCl₃: As analyzed in Option C, does not follow octet rule.
- O₃: As analyzed in Option B, follows octet rule.
- C₂H₄: As analyzed in Option A, follows octet rule.
Result for Option D: CO₂, O₃, and C₂H₄ follow the octet rule (3 molecules).
Conclusion
- Option A: 3 molecules follow the octet rule.
- Option B: 2 molecules follow the octet rule.
- Option C: 0 molecules follow the octet rule.
- Option D: 3 molecules follow the octet rule.
Final Answer: Both Options A and D have at least three molecules that follow the octet rule.
Approach Solution -2
To solve the problem, we verify which options contain at least three molecules following the octet rule.
1. Option 1: CO2, C2H4, NO, HCl
- CO2: Carbon follows octet.
- C2H4: Each carbon follows octet.
- NO: Odd-electron species, does not satisfy octet.
- HCl: Hydrogen (duet) and Cl (octet) satisfy octet.
Thus, 3 molecules follow octet rule.
Option 1 is correct.
2. Option 4: CO2, BCl3, O3, C2H4
- CO2: Octet satisfied.
- BCl3: Boron has 6 electrons, violates octet.
- O3: Octet satisfied.
- C2H4: Octet satisfied.
Thus, 3 molecules follow octet rule.
Option 4 is correct.
Final Answer:
Only options 1 and 4 have at least three molecules following the octet rule.
Top Questions on Chemical bonding and molecular structure
- The number of molecules having lone pair of electrons on central atom in the following is: BF$_3$, SF$_4$, SiCl$_4$, XeF$_4$, NCl$_3$, XeF$_6$, PCl$_5$, HgCl$_2$, SnCl$_2$
- AP EAPCET - 2025
- Chemistry
- Chemical bonding and molecular structure
- Identify the pair of molecules which have the same hybridisation as the hybridisation in Xenon (II) fluoride.
- AP EAPCET - 2025
- Chemistry
- Chemical bonding and molecular structure
- The correct order of increasing bond lengths of C--H, O--H, C--C, and H--H is
- AP EAPCET - 2025
- Chemistry
- Chemical bonding and molecular structure
How many of the following molecules / ions have a trigonal planar structure?
\( \text{BO}_3^{3-}, \, \text{NH}_3, \, \text{PCl}_3, \, \text{BCl}_3, \, \text{ClF}_3, \, \text{XeO}_3 \)- AP EAPCET - 2025
- Chemistry
- Chemical bonding and molecular structure
- Consider the following
Assertion (A): Dipole moment of NF\(_3\) is lesser than NH\(_3\)
Reason (R): In NF\(_3\), the orbital dipole due to lone pair of electrons is in the opposite direction to the resultant dipole moment of the three N-F bonds.
The correct answer is:- AP EAPCET - 2025
- Chemistry
- Chemical bonding and molecular structure
Questions Asked in JEE Advanced exam
- At 25$^\circ$C, the concentration of H$^+$ ions in $ 1.00 \times 10^{-3} \, \text{M} $ aqueous solution of a weak monobasic acid having acid dissociation constant $ K_a = 4.00 \times 10^{-11} $ is $ X \times 10^{-7} \, \text{M} $. The value of $ X $ is ____. {Use: Ionic product of water $ K_w = 1.00 \times 10^{-14} $ at 25$^\circ$C
- JEE Advanced - 2025
- Ionic Equilibrium
- Consider the vectors $$ \vec{x} = \hat{i} + 2\hat{j} + 3\hat{k},\quad \vec{y} = 2\hat{i} + 3\hat{j} + \hat{k},\quad \vec{z} = 3\hat{i} + \hat{j} + 2\hat{k}. $$ For two distinct positive real numbers $ \alpha $ and $ \beta $, define $$ \vec{X} = \alpha \vec{x} + \beta \vec{y} - \vec{z},\quad \vec{Y} = \alpha \vec{y} + \beta \vec{z} - \vec{x},\quad \vec{Z} = \alpha \vec{z} + \beta \vec{x} - \vec{y}. $$ If the vectors $ \vec{X}, \vec{Y}, \vec{Z} $ lie in a plane, then the value of $ \alpha + \beta - 3 $ is ________.
- Let $ x_0 $ be the real number such that $ e^{x_0} + x_0 = 0 $. For a given real number $ \alpha $, define $$ g(x) = \frac{3xe^x + 3x - \alpha e^x - \alpha x}{3(e^x + 1)} $$ for all real numbers $ x $. Then which one of the following statements is TRUE?
- JEE Advanced - 2025
- Fundamental Theorem of Calculus
Two identical concave mirrors each of focal length $ f $ are facing each other as shown. A glass slab of thickness $ t $ and refractive index $ n_0 $ is placed equidistant from both mirrors on the principal axis. A monochromatic point source $ S $ is placed at the center of the slab. For the image to be formed on $ S $ itself, which of the following distances between the two mirrors is/are correct:
- JEE Advanced - 2025
- Ray optics and optical instruments
- Let $$ \alpha = \frac{1}{\sin 60^\circ \sin 61^\circ} + \frac{1}{\sin 62^\circ \sin 63^\circ} + \cdots + \frac{1}{\sin 118^\circ \sin 119^\circ}. $$ Then the value of $$ \left( \frac{\csc 1^\circ}{\alpha} \right)^2 $$ is \rule{1cm}{0.15mm}.
- JEE Advanced - 2025
- Some Applications of Trigonometry
