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Academic session 2026-27, Chapter 3 in the current curriculum.
Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter Classification of Elements and Periodicity in Properties. This page offers a step-by-step solution to the specific question from Exercise 1, Question 16:
Among the second period elements the actual ionization enthalpies are in the order
Li < B < Be < C < O < N < F < Ne.
Explain why
(i) Be has higher ΔiH than B
(ii) O has lower ΔiH than N and F?
Among the second period elements the actual ionization enthalpies are in the order
Li < B < Be < C < O < N < F < Ne.
Explain why
(i) Be has higher ΔiH than B
(ii) O has lower ΔiH than N and F?
(i) Symmetry factor can be used for explaining higher ionization enthalpy of Be than B. The electronic configuration of Be is more symmetrical than that of B because both the occupied orbitals are filled in the case of Be while B has one half filled orbital in the 2p subshell. So during the process of ionization, the electron to be removed from beryllium atom is a 2s-electron, whereas the electron to be removed from boron atom is a 2p-electron. Now, 2s-electrons are more strongly attached to the nucleus than 2p-electrons and p orbital are also at slightly higher energy than the s orbital. Therefore, more energy is required to remove a 2s-electron of beryllium than that required to remove a 2p-electron of boron. Hence, beryllium has higher ΔiH than boron.
(ii) According to symmetry factor, nitrogen has all the three 2p orbitals half filled while oxygen has one filled and two half filled orbitals in 2p subshell. In nitrogen, the three 2p-electrons of nitrogen occupy three different atomic orbitals. However, in oxygen, two of the four 2p-electrons of oxygen occupy the same 2p-orbital. This results in increased electron-electron repulsion in oxygen atom. As a result, the energy required to remove the fourth 2p-electron from oxygen is less as compared to the energy required to remove one of the three 2p-electrons from nitrogen. Hence, oxygen has lower ΔiH than nitrogen.
Fluorine contains one electron and one proton more than oxygen. As the electron is being added to the same shell, the increase in nuclear attraction (due to the addition of a proton) is more than the increase in electronic repulsion (due to the addition of an electron). Therefore, the valence electrons in fluorine atom experience a more effective nuclear charge than that experienced by the electrons present in oxygen. As a result, more energy is required to remove an electron from fluorine atom than that required to remove an electron from oxygen atom. Hence, oxygen has lower ΔiH than fluorine.
The first (ΔiH1) and the second (ΔiH) ionization enthalpies (in kJ mol–1) and the (ΔegH) electron gain enthalpy (in kJ mol–1) of a few elements are given below:
| Elements | ΔiH1 | ΔiH | ΔegH |
| I | 520 | 7300 | -60 |
| II | 419 | 3051 | -48 |
| III | 1681 | 3374 | -328 |
| IV | 1008 | 1846 | -295 |
| V | 2372 | 5251 | +48 |
| VI | 738 | 1451 | -40 |
Which of the above elements is likely to be :
(a) the least reactive element.
(b) the most reactive metal.
(c) the most reactive non-metal.
(d) the least reactive non-metal.
(e) the metal which can form a stable binary halide of the formula MX2, (X=halogen).
(f) the metal which can form a predominantly stable covalent halide of the formula MX (X=halogen)?
NCERT questions are designed to test your understanding of the concepts and theories discussed in the chapter. Here are some tips to help you answer NCERT questions effectively:
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Welcome to the NCERT Solutions for Class 11 Chemistry - Chapter . This page offers a step-by-step solution to the specific question from Excercise 1 , Question 16: Among the second period elements the actual ionization enthalpies are in the order Li < B < B....
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Comments
Thank you for giving us Stasfied answer
I have question can anyone pls solve it: Why does BeCl2 exist but not BeH2?