Exams › JEE Advanced › Chemistry

White phosphorus is a tetra-atomic solid P4(s) at room temperature. Calculate the P-P bond enthalpy in kJ/mol. Given: enthalpy of sublimation of P4(s) = 61 kJ/mol; enthalpy of atomization of P4(s) = 1321 kJ/mol. Report the answer as the repeated digit-sum (sum of digits, then sum those digits, continuing until a single digit remains).

1. 1
2. 2
3. 3
4. 6

Correct answer: 3

Solution

The energy to atomize P4 from solid involves sublimation (solid->gas P4) plus breaking all 6 P-P bonds. Bond enthalpy = (1321 - 61)/6 = 210 kJ/mol. Digit sum: 2+1+0 = 3, which is already a single digit.

Related JEE Advanced Chemistry questions

• When N₂ is converted to N₂⁺, what happens to the bond dissociation energy of the N–N bond, and when O₂ is converted to O₂⁺, what happens to the bond dissociation energy of the O–O bond?
• In which of these ions is pπ–dπ bonding absent?
• What is the VSEPR formula for the chlorine trifluoride molecule?
• Consider the following molecules: CO2, SO2, and H2O. Arrange them in ascending order of their dipole moments.
• Which of the following occurrences is primarily influenced by hydrogen bonding?
• When two 1s atomic orbitals (one from each atom) combine to form molecular orbitals, which of the following statements is correct?

⚔️ Practice JEE Advanced Chemistry free + battle 1v1 →