Arrange the bonds in order of increasing ionic character in the molecules. LiF, K_2O, N_2, SO_2 and ClF_3.

Solution & Explanation

### Core Logic The ionic character of a bond is directly proportional to the electronegativity difference (Delta EN) between the two bonded atoms. Larger Delta EN implies higher ionic character. ### Step 1: Assess Electronegativity Differences - N_2: Both atoms are Nitrogen. Delta EN = 0. Purely covalent. (Lowest ionic character) - SO_2: Bond between S and O. Moderate Delta EN. Covalent with some polarity. - ClF_3: Bond between Cl and F. Delta EN is higher than S-O as F is the most electronegative element. - K_2O: Bond between K (alkali metal, very low EN) and O. Very high Delta EN. Ionic. - LiF: Bond between Li (alkali metal) and F (highest EN). Maximum Delta EN possible among these options. Most ionic. ### Step 2: Order Derivation Increasing order of ionic character (or Delta EN): N_2 < SO_2 < ClF_3 < K_2O < LiF ### Pattern Recognition Homodiatomic (N_2) is always 0% ionic. Alkali metal + Halogen (LiF) represents the extreme of the ionic spectrum. Sorting non-metals by group distance yields the middle ranks. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

Reference Study Guides

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Q85 jee_main_2024_31_jan_morning Hybridization
The number of species from the following in which the central atom uses sp^3 hybrid orbitals in its bonding is NH_3, SO_2, SiO_2, BeCl_2, CO_2, H_2O, CH_4, BF_3
Numerical Answer. Answer: 4 to 4

Solution

### Core Logic Analyzing the hybridization of the central atom in each species: - NH_3: 3 bp + 1 lp = 4 electron domains rightarrow sp^3 - SO_2: 2 bp + 1 lp = 3 electron domains rightarrow sp^2 - SiO_2: A giant covalent network where each Si is bonded to 4 oxygens tetrahedrally rightarrow sp^3 - BeCl_2: 2 bp + 0 lp = 2 electron domains rightarrow sp - CO_2: 2 bp + 0 lp = 2 electron domains rightarrow sp - H_2O: 2 bp + 2 lp = 4 electron domains rightarrow sp^3 - CH_4: 4 bp + 0 lp = 4 electron domains rightarrow sp^3 - BF_3: 3 bp + 0 lp = 3 electron domains rightarrow sp^2 Total species with sp^3 hybridization: NH_3, SiO_2, H_2O, CH_4. Total count = 4. ### Evaluation Rubric / Model Answer null ### Chapter Mix Class 11 Chemistry: Chemical Bonding and Molecular Structure

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