Exams › JEE Main › Chemistry
Atoms A and B have electronegativities of 1.20 and 4.0, respectively. What is the percentage ionic character of the A–B bond?
- 50%
- 72.24%
- 55.3%
- 43%
Correct answer: 72.24%
Solution
The percentage ionic character of a bond can be estimated using the difference in electronegativities between the two atoms. In this case, the significant difference of 2.8 between atoms A and B indicates a high ionic character, which is calculated to be approximately 72.24%.
Related JEE Main Chemistry questions
- Although the electronegativity gap between N and F is larger than that between N and H, ammonia has a dipole moment of 1.5 D while nitrogen trifluoride has only 0.2 D. The reason is that
- When N₂ is converted into N₂⁺, the dissociation energy of the N–N bond ______, and when O₂ is converted into O₂⁺, the dissociation energy of the O–O bond ______.
- From the ions listed below, which pair has geometries that can be accounted for by the same type of orbital hybridization? NO₂⁻, NO₃⁻, NH₂⁻, NH₄⁺, SCN⁻
- In the phosphate ion, PO₄³⁻, what is the formal charge on an oxygen atom that is singly bonded to phosphorus in a P–O bond?
- Among the following species, which one has no unpaired electrons?
- Which pair of ions has the same total number of electrons and the same molecular geometry?
⚔️ Practice JEE Main Chemistry free + battle 1v1 →