Correct order of Ist IP among following elements Be, B, C, N, O is : -
A
B < Be < C < O < N
B
B < Be < C < N < O
C
Be < B < C < N < O
D
Be < B < C < O < N
The answer is A.
Solution
Ionisation enthalpy generally increases across a period, but there are two exceptions here: B (899 vs 801 kJ/mol) has a lower IE than Be, because B loses a 2p electron, which is higher in energy and shielded by the 2s electrons, while Be has a stable filled $2s^2$ configuration. O has a lower IE than N, because N has a stable half-filled $2p^3$ configuration, while in O ($2p^4$) the paired electrons repel each other, making one easier to remove. Values (kJ/mol): B 801, Be 899, C 1086, O 1314, N 1402. So the order is B $\lt$ Be $\lt$ C $\lt$ O $\lt$ N.
ConceptExceptions in second period
Ionisation enthalpy in past papers
4 questions from this chapter have appeared across 4 exam years.
