its because of the size difference. s orbital is very compared to other orbitals therefore
the decrease in electropositivity is due to poor shielding effect of s and d orbitals
The reduction in the force of attraction between the nucleus and outer most electron is known as shielding effect
yes, increases from left to right
It had most effect on outer shell electrons.
its because of the size difference. s orbital is very compared to other orbitals therefore
the correct order is-4s>4p>4d>4f this is because of the shape of the orbitals
the decrease in electropositivity is due to poor shielding effect of s and d orbitals
An s orbital is closer to the nucleus than a p orbital, so it shields outer electrons more than a p orbital does. Therefore, it's penetration effect is greater than the p orbital's. The penetration effect is the tendency of orbitals closer to the nucleus shielding outer electrons.
Na have higher shielding effect than Li *According to my chemistry book
YES
in metals due to shielding effect ionization value is low
The reduction in the force of attraction between the nucleus and outer most electron is known as shielding effect
The shielding effect reduces the ionization energy and so makes cation formation easier.
When a period of elements are considered, the element in group 18 has the highest shielding effect.
yes, increases from left to right
It had most effect on outer shell electrons.