Lewis Structures ā Octets, formal charge
Chemical Bonding
15
JEE Qs
8%
Hard
75
min
Master the systematic drawing of Lewis structures and precise formal charge calculation; it's fundamental for predicting molecular geometry and stability.
š§® Key Formulas
ā Key Points for JEE
- 1Follow a systematic procedure: count total valence electrons, identify central atom, form single bonds, complete octets for terminal atoms, then central atom. Use multiple bonds if necessary to satisfy octets or minimize formal charges.
- 2Understand and apply the octet rule (atoms tend to achieve 8 valence electrons), but also recognize its exceptions: incomplete octets (e.g., Be, B), expanded octets (hypervalent, for elements in Period 3 and beyond like P, S, Cl, Xe), and odd electron species.
- 3Calculate formal charge for each atom in a Lewis structure using FC = V - L - (B/2). The sum of formal charges in a molecule or ion must equal its net charge.
- 4The most stable Lewis structure is generally one where formal charges are closest to zero on all atoms, and any unavoidable negative formal charges reside on the more electronegative atoms, while positive formal charges reside on less electronegative atoms.
- 5To minimize formal charges, especially on the central atom for elements in Period 3 or higher, convert lone pairs into multiple bonds, ensuring octet rule (or expanded octet) is not violated.
ā ļø Common Mistakes
- āIncorrectly counting the total number of valence electrons for a molecule or polyatomic ion, leading to an incorrect distribution of electrons.
- āFailing to recognize or apply exceptions to the octet rule, particularly for elements from the third period and beyond (expanded octet) or for elements like B and Be (incomplete octet).
- āErrors in calculating formal charges, often by miscounting lone pair electrons or shared electrons, thus leading to incorrect assessment of structure stability.
- āNot using formal charge as a criterion to choose the most stable or dominant Lewis structure among several possibilities, especially in cases of resonance.
š Practice Questions
See allQ75.Total number of non bonded electrons present in NO2ā ion based on Lewis theory is 2025 (29 Jan Shift 2) JEE Main Previous Year Paper
Q71.The number of molecules/ions that show linear geometry among the following is ________ SO2, BeCl2, CO2, Nā3 , NO2, F2O, XeF2, NO+2 , Iā3 , O3
Q69.Arrange the following compounds in increasing order of their dipole moment : HBr, H2 S, NF3 and CHCl3 (1) H2 S < HBr < NF3 < CHCl3 (2) NF3 < HBr < H2 S < CHCl3 (3) HBr < H2 S < NF3 < CHCl3 (4) CHCl3 < NF3 < HBr < H2 S
Q64.Which of the following statement is true with respect to H2O, NH3 and CH4 ? A. The central atoms of all the molecules are sp3 hybridized. B. The H āO āH, H āN āH and H āC āH angles in the above molecules are 104.5ā, 107.5ā and 109.5ā , respectively. C. The increasing order of dipole moment is CH4 < NH3 < H2O. D. Both H2O and NH3 are Lewis acids and CH4 is a Lewis base. E. A solution of NH3 in H2O is basic. In this solution NH3 and H2O act as Lowry-Bronsted acid and base respectively. Choose the correct answer from the options given below: (1) A, B and C Only (2) A, D and E Only (3) C, D and E Only (4) A, B, C and E Only
Q70.Which of the following linear combination of atomic orbitals will lead to formation of molecular orbitals in homonuclear diatomic molecules [internuclear axis in z -direction] ? A. 2pz and 2px B. 2 s and 2px C. 3 dxy 2025 (24 Jan Shift 1) JEE Main Previous Year Paper and 3 dx2āy2 D. 2 s and 2pz E. 2pz and 3dx2 āy2 Choose the correct answer from the options given below: (1) A and B Only (2) D Only (3) E Only (4) C and D Only
Q53.Given below are two statements : Statement (I): Experimentally determined oxygen-oxygen bond lengths in the O3 are found to be same and the bond length is greater than that of a O = O (double bond) but less than 2025 (24 Jan Shift 2) JEE Main Previous Year Paper that of a single (O āO) bond. Statement (II) : The strong lone pair-lone pair repulsion between oxygen atoms is solely responsible for the fact that the bond length in ozone is smaller than that of a double bond (O = O) but more than that of a single bond (O āO). In the light of the above statements, choose the correct answer from the options given below : (1) Both Statement I and Statement II are false (2) Statement I is false but Statement II is true (3) Statement I is true but Statement II is false (4) Both Statement I and Statement II are true
NCERT Chapters
- Class 11 Chemistry Ch 4: Chemical Bonding and Molecular Structure