Exams › JEE Main › Chemistry
Find the total number of lone pairs of electrons present on the central atom, summed over the following species: [TeBr6]²-, [BrF2]+, SNF3 and [XeF3]- [Atomic numbers: N = 7, F = 9, S = 16, Br = 35, Te = 52, Xe = 54]
- 4
- 5
- 6
- 7
Correct answer: 6
Solution
Counting lone pairs on each central atom: Te = 1, Br = 2, S = 0, Xe = 3, giving a total of 6.
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⁻
- Atoms A and B have electronegativities of 1.20 and 4.0, respectively. What is the percentage ionic character of the A–B bond?
- 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?
⚔️ Practice JEE Main Chemistry free + battle 1v1 →