Answer :
To determine the correct prediction about the properties of antimony (Sb) and iodine (I), let's analyze the given information step-by-step.
1. Ionization Energy: Ionization energy is the energy required to remove an electron from an atom in the gaseous phase.
- For iodine (I), the first ionization energy is provided as 1008 kJ/mol.
- For antimony (Sb), the ionization energy is not directly given.
2. Electron Affinity: Electron affinity is the energy change when an electron is added to an atom in the gaseous phase.
- The electron affinity of iodine (I) is -295 kJ/mol.
- The electron affinity of antimony (Sb) is -103 kJ/mol.
3. Electronegativity: Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons.
- The electronegativity of antimony (Sb) is 2.05.
- The electronegativity of iodine (I) is not given.
Comparing Trends:
- Typically, ionization energy and electronegativity increase across a period and decrease down a group in the periodic table.
- Iodine (I) is in Group 17 (Halogens) and Period 5.
- Antimony (Sb) is in Group 15 and Period 5, indicating that Sb is to the left of I in the same period.
Given these general trends, we can infer:
- Ionization Energy: Iodine (I) is expected to have a higher ionization energy than antimony (Sb) because ionization energy typically increases across a period. This means Sb has a lower ionization energy compared to I.
- Electronegativity: Iodine (I) is expected to have a higher electronegativity than antimony (Sb) as electronegativity also increases across a period. Therefore, Sb has a lower electronegativity compared to I.
Based on these interpretations:
- Sb has a lower ionization energy than I.
- Sb has a lower electronegativity than I.
These observations align with the following prediction:
"Sb has a lower ionization energy and a lower electronegativity than I."
Therefore, the correct answer is:
Sb has a lower ionization energy and a lower electronegativity than I.
This corresponds to:
Sb has a lower ionization energy and a lower electronegativity than I.
1. Ionization Energy: Ionization energy is the energy required to remove an electron from an atom in the gaseous phase.
- For iodine (I), the first ionization energy is provided as 1008 kJ/mol.
- For antimony (Sb), the ionization energy is not directly given.
2. Electron Affinity: Electron affinity is the energy change when an electron is added to an atom in the gaseous phase.
- The electron affinity of iodine (I) is -295 kJ/mol.
- The electron affinity of antimony (Sb) is -103 kJ/mol.
3. Electronegativity: Electronegativity is a measure of the tendency of an atom to attract a bonding pair of electrons.
- The electronegativity of antimony (Sb) is 2.05.
- The electronegativity of iodine (I) is not given.
Comparing Trends:
- Typically, ionization energy and electronegativity increase across a period and decrease down a group in the periodic table.
- Iodine (I) is in Group 17 (Halogens) and Period 5.
- Antimony (Sb) is in Group 15 and Period 5, indicating that Sb is to the left of I in the same period.
Given these general trends, we can infer:
- Ionization Energy: Iodine (I) is expected to have a higher ionization energy than antimony (Sb) because ionization energy typically increases across a period. This means Sb has a lower ionization energy compared to I.
- Electronegativity: Iodine (I) is expected to have a higher electronegativity than antimony (Sb) as electronegativity also increases across a period. Therefore, Sb has a lower electronegativity compared to I.
Based on these interpretations:
- Sb has a lower ionization energy than I.
- Sb has a lower electronegativity than I.
These observations align with the following prediction:
"Sb has a lower ionization energy and a lower electronegativity than I."
Therefore, the correct answer is:
Sb has a lower ionization energy and a lower electronegativity than I.
This corresponds to:
Sb has a lower ionization energy and a lower electronegativity than I.