
For the second period elements the correct increasing order of first enthalpy is:
(A) Li < B < Be < C < N < O < F < Ne
(B) Li < B < Be < C < O < N < F < Ne
(C) Li < Be < B < C < N < O < F < Ne
(D) Li < B < Be < C < O < N < F < Ne
Answer
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Hint: According to the modern periodic table, the elements are arranged based on their order of increasing atomic number and periodic properties of elements. The periodic properties are in terms of ionization energy, electron affinity, atomic radius, electronegativity, and ionic radius. To predict the element’s physical, chemical, and atomic properties by using periodic law and table formation.
Complete step by step solution:
The minimum energy required to remove one electron from a neutral gaseous atom in its ground state to form a gaseous ion is called ionization energy or ionization enthalpy.
$X(g)\to {{X}^{+}}(g)+{{e}^{-}}$
The energy required for removing one electron is called first ionization potential and smaller ionization enthalpy which is easier to convert an atom into its positive ion.
From the above table, Ne has more ionization enthalpy than others due to its stable electronic configuration. ‘Be’ and ‘N’ have more stable valence subshells comparatively ‘B’ and ‘O’.
Hence, for the second-period elements, the correct increasing order of the first enthalpy is
Li < B < Be < C < O < N < F < Ne
The correct answer is option B.
Note: Variation of ionization enthalpy in the periodic table decreases in the group when the atomic number increases. In periods, ionization enthalpy increases with the atomic number from left to right. When the elements with half or filled electronic configuration have more ionization enthalpy.
Complete step by step solution:
The minimum energy required to remove one electron from a neutral gaseous atom in its ground state to form a gaseous ion is called ionization energy or ionization enthalpy.
$X(g)\to {{X}^{+}}(g)+{{e}^{-}}$
The energy required for removing one electron is called first ionization potential and smaller ionization enthalpy which is easier to convert an atom into its positive ion.
| Element | Atomic number | Electronic configuration |
| Lithium (Li) | 3 | $1{{s}^{2}}2{{s}^{1}}$ |
| Beryllium (Be) | 4 | $1{{s}^{2}}2{{s}^{2}}$ |
| Boron (B) | 5 | $1{{s}^{2}}2{{s}^{2}}2{{p}^{1}}$ |
| Carbon (C) | 6 | $1{{s}^{2}}2{{s}^{2}}2{{p}^{2}}$ |
| Nitrogen (N) | 7 | $1{{s}^{2}}2{{s}^{2}}2{{p}^{3}}$ |
| Oxygen (O) | 8 | $1{{s}^{2}}2{{s}^{2}}2{{p}^{4}}$ |
| Fluorine (F) | 9 | $1{{s}^{2}}2{{s}^{2}}2{{p}^{5}}$ |
| Neon (Ne) | 10 | $1{{s}^{2}}2{{s}^{2}}2{{p}^{6}}$ |
From the above table, Ne has more ionization enthalpy than others due to its stable electronic configuration. ‘Be’ and ‘N’ have more stable valence subshells comparatively ‘B’ and ‘O’.
Hence, for the second-period elements, the correct increasing order of the first enthalpy is
Li < B < Be < C < O < N < F < Ne
The correct answer is option B.
Note: Variation of ionization enthalpy in the periodic table decreases in the group when the atomic number increases. In periods, ionization enthalpy increases with the atomic number from left to right. When the elements with half or filled electronic configuration have more ionization enthalpy.
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