As we know that if size of atomic radii increases, I.E decreases. But why it is said in a random topic that I.E increases down the group?
As we know that if size of atomic radii increases, I.E decreases. But why it is said in a random topic that I.E increases down the group?
I think u misundertood the concept
IE decreases down the group
Let me explain it The ionization energy is the lowest amount of energy required to remove one electron(the most loosely attached electron) from each atom in one mole of an element in gaseous state to form gaseous (positive) ion.
There are certain factors that have an impact on the ionization energy value.
To understand the answer to your question, you must understand the effect of all three
- Atomic radius (Inversely proportional to I.E. Value)
- Shielding effect of electrons (Inversely Proportional to I.E. Value)
- Nuclear Charge (Proportional to I.E. Value)
The Atomic radius is the distance between the to-be-removed electron and the nucleus. The greater the distance, lesser the force of attraction. Hence as this distance increases the electrons are easier to remove; lesser force is required to remove it from that attractive field of the nucleus. (This answers your question)
IE decreases down the group
Let me explain it
The ionization energy is the lowest amount of energy required to remove one electron(the most loosely attached electron) from each atom in one mole of an element in gaseous state to form gaseous (positive) ion.
There are certain factors that have an impact on the ionization energy value.
To understand the answer to your question, you must understand the effect of all three
- Atomic radius (Inversely proportional to I.E. Value)
- Shielding effect of electrons (Inversely Proportional to I.E. Value)
- Nuclear Charge (Proportional to I.E. Value)
The Atomic radius is the distance between the to-be-removed electron and the nucleus. The greater the distance, lesser the force of attraction. Hence as this distance increases the electrons are easier to remove; lesser force is required to remove it from that attractive field of the nucleus. (This answers your question)
