There is an increase in ionization enthalpy from li to be .Why?
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The electronic configuration of Lithium is 1s2 2s1
This means that there's only one electron in the 2s orbital of Lithium
Now,
The electronic configuration of Beryllium is 1s2s2
Here, since there's 2 electrons in 2s orbital and we know that maximum 2 electrons can be put in s orbital
And, thus here since the orbital is fully filled. The atom gains more stability and thus it would resist giving it's electron easily.
And hence, The ionization enthalpy of Beryllium is greater than the ionization enthalpy of Lithium.
This means that there's only one electron in the 2s orbital of Lithium
Now,
The electronic configuration of Beryllium is 1s2s2
Here, since there's 2 electrons in 2s orbital and we know that maximum 2 electrons can be put in s orbital
And, thus here since the orbital is fully filled. The atom gains more stability and thus it would resist giving it's electron easily.
And hence, The ionization enthalpy of Beryllium is greater than the ionization enthalpy of Lithium.
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If sufficient energy is supplied ,electrons may be removed resulting in the formation of a positively charged ion.
The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so to convert it into gaseous cation is called ionisation enthalpy.
It is represented by Δ i H
This process may be represented as
M (g) + Δ i H ———–> M + (g) + e – (g)
where M (g) and M + (g) represent the gaseous atom and the resultant gaseous cation.
Ionisation enthalpy is also known as ionisation potential since it is the minimum potential difference required to remove the most loosely bound electrons from an isolated gaseous cation.
It is measured in units of electron volts (eV) per atom or kilo calorie per mole or kilo joules per mole.
One electron volt is the energy acquired by an electron while moving under a potential difference of 1 volt.
The energy required to remove the most loosely bound electrons from the isolated gaseous atom is called its first ionisation enthalpy and is denoted by Δ iH 1
M (g) + Δ i H 1 ———–> M + (g) + e – (g)
The energies required to knock out second and third electrons are called second and third ionisation energies.
M + (g) + Δ i H 2 ———–> M 2+ (g) + e –(g)
M 2+ (g) + Δ i H 3 ———–> M 3+ (g) + e –(g)
When one electron has been removed from the neutral gaseous atom the positively charged ions formed has 1 electrons less than the number of protons in the nucleus. As a result the electrostatic attraction between the nucleus and the remaining electrons in the cation increases i.e. effective nuclear charge increases. The positive ion holds its remaining electrons more firmly. Therefore ,the energy required to remove another electron from this positively charged Ion or second electron from the neutral atom must be higher than the first.
HOPE THIS WILL HELP YOU.........
The minimum amount of energy required to remove the most loosely bound electron from an isolated gaseous atom so to convert it into gaseous cation is called ionisation enthalpy.
It is represented by Δ i H
This process may be represented as
M (g) + Δ i H ———–> M + (g) + e – (g)
where M (g) and M + (g) represent the gaseous atom and the resultant gaseous cation.
Ionisation enthalpy is also known as ionisation potential since it is the minimum potential difference required to remove the most loosely bound electrons from an isolated gaseous cation.
It is measured in units of electron volts (eV) per atom or kilo calorie per mole or kilo joules per mole.
One electron volt is the energy acquired by an electron while moving under a potential difference of 1 volt.
The energy required to remove the most loosely bound electrons from the isolated gaseous atom is called its first ionisation enthalpy and is denoted by Δ iH 1
M (g) + Δ i H 1 ———–> M + (g) + e – (g)
The energies required to knock out second and third electrons are called second and third ionisation energies.
M + (g) + Δ i H 2 ———–> M 2+ (g) + e –(g)
M 2+ (g) + Δ i H 3 ———–> M 3+ (g) + e –(g)
When one electron has been removed from the neutral gaseous atom the positively charged ions formed has 1 electrons less than the number of protons in the nucleus. As a result the electrostatic attraction between the nucleus and the remaining electrons in the cation increases i.e. effective nuclear charge increases. The positive ion holds its remaining electrons more firmly. Therefore ,the energy required to remove another electron from this positively charged Ion or second electron from the neutral atom must be higher than the first.
HOPE THIS WILL HELP YOU.........
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