In which of the following pairs, the ionisation energy of the first species is less than that of the second
(1) N, P
(2) Be+, Be
(3) N, N-
(4) Ne, Ne+
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Answers
Answer: Option (2)
The ionization energy of Be+ < Be.
Explanation:
Let us understand this bit in detail
Ionization energy or ionization enthalpy is the least possible energy required by an atom or molecule to loose its outermost electron, also known as valence electrons.
X + energy ---> (X+) + e-
here, X is atom or molecule, X+ is the atom or molecule after the removal of electron and e- is the electron removed from the atom or molecule.
As you can see from the figure attached below the ionization energy increases from left to right along the row and decreases from top to bottom along the column. Also the ionization energy increases with decrease in atomic size of the elements across the periodic table.
For N and P:
N + energy ---> (N+) + e-
P + energy ---> (P+) + e-
Since N lies above P in the periodic table so it can be said that the ionization energy of N is greater than P.
For Be+, Be, N, N-, Ne, Ne+:
Let’s see their electronic configuration
Be: 1s² 2s²
Be+:1s² 2s¹
N: 1s² 2s² 2p³
N- : 1s² 2s² 2p⁴
Ne : 1s² 2s² 2p⁶
Ne+ : 1s² 2s² 2p⁵
By studying the electronic configuration following comparisons based on ionization energy required to remove their valence electron can clearly be made:
Be+ < Be
N > N-
Ne > Ne+
Therefore, it is Be+ which has less ionization energy as compared to Be as in Be+ the electron from 2s¹ can easily be removed as compared to Be where the electron has to be removed from a fully filled 2s² orbital.
Answer:
Variation in the period among the representative elements:
Ionisation energies generally increase along the period because in moving left to right in a period the effective nuclear charge per outermost electron increases while the corresponding principal quantum number remains same.
However, due to half-filled configuration, P has higher IP than S.
Explanation: