`Mg^(2+)O^(2-)` is formed since : [IE = Ionization energy, EA = Electron affinity]
`Mg^(2+)O^(2-)` is formed since : [IE = Ionization energy, EA = Electron affinity]
A
IE of Mg is low
B
EA of O is low
C
IE of Mg is high
D
IE of O is low
Text Solution
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The correct Answer is:
To understand why \( Mg^{2+} \) and \( O^{2-} \) ions are formed, we need to analyze the concepts of ionization energy (IE) and electron affinity (EA) for magnesium (Mg) and oxygen (O). Here’s a step-by-step breakdown of the reasoning:
### Step 1: Understanding Ionization Energy (IE)
- **Definition**: Ionization energy is the energy required to remove an electron from an atom.
- **For Magnesium (Mg)**: Magnesium is a metal located in Group 2 of the periodic table. Metals typically have a lower ionization energy because they tend to lose electrons easily.
- **Conclusion**: Since magnesium has a low ionization energy, it can easily lose two electrons to form \( Mg^{2+} \).
### Step 2: Understanding Electron Affinity (EA)
- **Definition**: Electron affinity is the energy change that occurs when an electron is added to a neutral atom. A higher electron affinity indicates a greater tendency to gain electrons.
- **For Oxygen (O)**: Oxygen is a non-metal and is highly electronegative. It has a high electron affinity because it readily gains electrons to achieve a stable electron configuration (octet).
- **Conclusion**: Oxygen can gain two electrons to form \( O^{2-} \) due to its high electron affinity.
### Step 3: Formation of Ionic Compounds
- **Ionic Bonding**: The formation of \( Mg^{2+} \) and \( O^{2-} \) leads to the creation of the ionic compound magnesium oxide (MgO).
- **Stability**: The transfer of electrons from magnesium to oxygen results in the formation of stable ions. The \( Mg^{2+} \) ion has lost two electrons, while the \( O^{2-} \) ion has gained two electrons.
### Step 4: Summary of Key Points
- **Low Ionization Energy of Mg**: This allows magnesium to lose electrons easily.
- **High Electron Affinity of O**: This allows oxygen to gain electrons easily.
- **Resulting Ions**: The combination of these two tendencies results in the formation of \( Mg^{2+} \) and \( O^{2-} \).
### Final Conclusion
The formation of \( Mg^{2+} \) and \( O^{2-} \) is primarily due to the low ionization energy of magnesium and the high electron affinity of oxygen, leading to the stable ionic compound \( MgO \).
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To understand why \( Mg^{2+} \) and \( O^{2-} \) ions are formed, we need to analyze the concepts of ionization energy (IE) and electron affinity (EA) for magnesium (Mg) and oxygen (O). Here’s a step-by-step breakdown of the reasoning:
### Step 1: Understanding Ionization Energy (IE)
- **Definition**: Ionization energy is the energy required to remove an electron from an atom.
- **For Magnesium (Mg)**: Magnesium is a metal located in Group 2 of the periodic table. Metals typically have a lower ionization energy because they tend to lose electrons easily.
- **Conclusion**: Since magnesium has a low ionization energy, it can easily lose two electrons to form \( Mg^{2+} \).
### Step 2: Understanding Electron Affinity (EA)
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Ionic compounds are formed most easily with (A) Low electron affinity, high ionization energy (B) High electron affinity, low ionization energy (C) Low electron affinity, low ionization energy (D) High electron affinity, high ionization energy
In 1931, Pauling defined the electronegativity of an atom as the tendency of the atom to attract electrons to itself when combined in a compound. The implication is that when a covalent bond is formed, the electrons used for bonding need not be shared equally by both atoms. If the bonding electrons spend more time around one atom, that atom will have a del- charge, and consequently the other atom will have a del+ charge. In the extreme case, where the bonding electrons are round one atom all of the time, the bond is ionic. Pauling and others have attempted to relate the electronegativity difference between two atoms to the amount of ionic character in the bond between them. Mulliken In 1934, Mulliken suggested an alternative approach to electronegativity based on the ionization energy and electron affinity of an atom. Consider two atoms A and B. If an electron is transferred from A to B, forming ions A^(+) and B^(-) , then the energy change is the ionization energy of atom A(I_(A)) minus the electron affinity of atom B(E_(B)) , that is I_(A)-E_(B) . Alternatively, if the electron was transferred the other way to give B^(+) and A^(-) ions, then the energy changed would be I_(B) - I_(A) . If A^(+) and B^(-) are actually formed, then this process require less energy , and (I_(A) - E_(B)) lt (I_(B) - E_(A)) Rerranging (I_(A)+E_(A)) lt (I_(A)-E_(B)) Now with respect to electronegativity for the same change, E.N_(A) lt E.N_(B) . Thus Mulliken suggested that electronegativity is proportional to I.E + E.A and could be regarded as the average of the ionization energy and the electron affinity of an atom. Electronegativity =((I+E))/(2) Mulliken used I and E values measured in electron volts, and the values were about 2.8 times alrger than the Pauling values. It is to be noted that I.E . and E.A values are defined for singular gaseous atoms. For a reaction A(g)+B(g) rar AB(S) the enthalpy change for the recation will be [assuming AB(s) to be an ionic compound] If I.E_(A)+E.A_(A) lt I.E_(B) +E.A_(B)
Mulliken defined the electronegativity of an atom as the arithmetic mean of its ionisation energy and electron affinity. X_(A)=(1)/(2)(I.P.+E.A.) One more relationship given by him, if the values are given in eV is X_(A)=("Ionisation potential"+ "Electron affinity")/(5.6) When there is pure covalent bond between A-B ((IP)_(A)+(EA)_(A))/(5.6)=((IP)_(B)+(EA)_(B))/(5.6) implies X_(A)=X_(B) Pauling's Electronegativity scale is based on
Mulliken defined the electronegativity of an atom as the arithmetic mean of its ionisation energy and electron affinity. X_(A)=(1)/(2)(I.P.+E.A.) One more relationship given by him, if the values are given in eV is X_(A)=("Ionisation potential"+ "Electron affinity")/(5.6) When there is pure covalent bond between A-B ((IP)_(A)+(EA)_(A))/(5.6)=((IP)_(B)+(EA)_(B))/(5.6) implies X_(A)=X_(B) When there is formation of overset(delta-)(A)-overset(delta+)(B) bond then condition will be
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