NEETChemistryClassification of Elements and Periodicity in Properties
Match List-I with List-II: List-I (Element) List-II (First ionization enthalpy in kJ mol⁻¹ ) (A) Boron (I) 1402 (B) Carbon (II) 1314 (C) Nitrogen (III) 1086 (D) Oxygen (IV) 801 Choose the correct answer from the options given below:
Options
- A(A)-(IV), (B)-(III), (C)-(II), (D)-(I)
- B(A)-(III), (B)-(IV), (C)-(I), (D)-(II)
- C(A)-(I), (B)-(II), (C)-(III), (D)-(IV)
- D(A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Correct answer
D. (A)-(IV), (B)-(III), (C)-(I), (D)-(II)
Step-by-step solution
The general trend across a period is an increase in first ionization enthalpy due to increasing effective nuclear charge. However, Nitrogen ( N , 2s^2 2p^3 ) has a higher first ionization enthalpy than Oxygen ( O , 2s^2 2p^4 ) because of the extra stability associated with its exactly half-filled 2p subshell. Thus, the correct increasing order of first ionization enthalpy is B Arranging the given numerical values in increasing order: 801 Matching the elements to these values: Boron ( B ) = 801 kJ mol⁻¹ (IV) Carbon