Classification of Elements 3 Flashcards
what is ionic radius
ionic radii is measured as the distance between an anion and cation in ionic crystals
a cation is always smaller than parent atom?
the cation is formed when an e- is removed from a neutral atom.
the cation is always smaller than a nauetral e- necause the atom has fewer e-s but the same no of protons in the nucleus so nuclear charge remains the same. so the protons can hold on and pull the e-s closer to the nucleus.
Na—–> Na⁺ + e-
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an anion is always greater than parent atom ? gr
gain of one or more e-s results in the formation of anion. The size of the anion will be larger than parents atom because of the increased interelectronic repulsion and the effective nuclear charge decreases. So the effect of nuclear charge of outer e-s is lesser, and the atom is bigger.
F + e- ——F-
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Consider a set of anions. out of the many anions, whcih one will have smaller and greater sizes.
For example,
O
2–, F–
, Na+
and Mg2+ have the same number of
electrons (10). Their radii would be different
because of their different nuclear charges.
The
cation with the greater positive charge will have
a smaller radius because of the greater attraction of the electrons to the nucleus. Cations with lesser positive charge have more bigger ions
Anion
with the greater negative charge will have the
larger radius. In this case, the net repulsion of
the electrons will outweigh the nuclear charge
and the ion will expand in size.
what is ionization enthalpy
A quantitative measure of the tendency of an
element to lose electron is given by its
Ionization Enthalpy. It represents the energy
required to remove an electron from an isolated
gaseous atom (X) in its ground state.
The ionization enthalpy is expressed in
units of kJ mol–1.
what is the 1st ionization enthalpy and 2nd delta iH
1st ionisation enthalpy is defined as the energy required to remove the first most loosely bound e- in an atom.
X—> X+ + e-
2nd ionisation enthalpy is defined as the energy reuired to remove the next most loosely bound e-
X+——> X2+ + e-
second ionization enthalpy is always greater than the 1st one why
The second
ionization enthalpy will be higher than the first
ionization enthalpy because it is more difficult
to remove an electron from a positively charged
ion than from a neutral atom. In the same way
the third ionization enthalpy will be higher than
the second and so on.
why is ΔiH always +ve
energy is always require dto remove an e- from an atom and hence it is always +ve
noble gases has highest IE while alkali metals have lowest IE why
Noble gases have fully filled shells and extremely stable eln confgs. SO a large amt of energy is required to pull an e- from the stable en confg. Have ns2np6 eln cng
Alkali metal on the other hand has fully filled shells and an ns shell with just one e-. So the ionization enthalpy is very low
how does ionzation enthalp vary across a period
The ionization enthalpi increases along he period.
The atomic radius of eleements increase along the period so do their nuclear charge due to increase in atomic number. but principal quantum number remains the same. Due to decreased size of the atom and stronger attraction to the nucleus, ithe valence e-s are held more tightly towards the nucleus.
successive electrons
are added to orbitals in the same principal
quantum level and the shielding of the nuclear
charge by the inner core of electrons does not
increase very much to compensate for the
increased attraction of the electron to the
nucleus. Thus, across a period, increasing
nuclear charge outweighs the shielding.
Consequently, the outermost electrons are held
more and more tightly and the ionization
enthalpy increases across a period.
how does ionization enthalpy vary down a group
The ionization enthalpy decreases as we go down a group.
As we go down a grp, the atomic radius increases. As the size increases, the valence e-s areplaced more farther away from the nucleus. Due to the shileding effect, the experience very little attraction and can be lost easily. So ionization enthalpy decrease
As we go
down a group, the outermost electron being
increasingly farther from the nucleus, there is
an increased shielding of the nuclear charge
by the electrons in the inner levels. In this case,
increase in shielding outweighs the increasing
nuclear charge and the removal of the
outermost electron requires less energy down
a group.
What are the factors affected ΔiH
a) Nuclear charge
as nuclear sharge increase IE increases because more nuclear charge more tightly held e-s by nucleus
b) Atomic radius
as atomic radius increases IE decreases because the valence e-s are placed farther away and hence can be removed easily
c) Pentetration efffect of e-s
how close the e-s are to the nucleus. The e-s in the su subshell eperience more penetration efect.
s>p>d>f
The ionization enthalpy increases with penetration efffect
d) Shielding effect of e-s,
greater the nof of iinner shell e-s, greater will be shielding and less effective nuclear charge so less IE
e) eln confg
There is extra stability associated with half filled or fully filled orbitals so it is more difficult to remove e-s , High value of IE
why is ionization enthalpy 1 of Al greater than ionisation enthalpy 1 of Magnesmi
Mg- [Ne]3s2
Al- [Ne] 3s2 3p1
In Al, the e- to be removed is a p-e-. It is easier to remove a p electron as it is placed further away from the nucleus and experiences lesser penetration effect. Hence it has low ionisation enthalpy. However the e- removed from Mg is a s-e-. The s-e- is placed more close to the nucleus and has a larger penetration effect. it is more difficult to remove an e- from it. High ih
The ionisation energy of Boron is slightly lesser than beryllium why
give electronic configuration of boron abd beryllium
we will notice that in boron, during ionization the p e- will be removed but in berylium an s e- has to removed. The penetration of a 2s e- is more than that of 2p e- Therefore 2p e- of boron is more shielded from the nucleus than the 2s electrons of beryllium. hence it is more attracted to the nucleus and more amount of energy is required to pull an e-
the ionisation energy of nitrogen is greater than oxygen
This is because the eln configuration of nitrogen is 1s2 2s2 2px 1 py 1 pz 1. This is half filled orbitals and hence as extra stability due to the symmetry. Whereas in oxygen it is 1s22s2 2px2 2py1 2pz1, due ot electron electron repulsion in 2px orbital, it ieasier to remove an e-
