Periodic Table & Periodicity PDF

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B.S. Degree in Chemistry....

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LEC LECTURE TURE # 1

Introduce Modern Periodic Table to the students for Few Minutes :

D I AG O N A L R E L AT ATII O N S H I P : Some elements of certain groups of 2nd period resemble much in properties with the elements of third period of next group i.e. elements of second and third period are diagonally related in properties. This phenomenon is known as diagonal relationship. 2nd period

Li

Be

B

C

3rd period

Na

Mg

Al

Si

Diagonal relationship arises because of (i) similar size of atom and ions (Li = 1.23 ≈ & Mg = 1.36 ≈ ; Li+ = 0.76 ≈ & Mg2+ = 0.72 ≈) (ii) similar polarising powers (charge to radius ratio) (iii) similarity in electronegativity values (Li = 1.0 & Mg = 1.2 ; Be = 1.5 & Al = 1.5)

T H E PE R IO D I C I T Y OF ATO ATOM M IC PR O P E R T I E S : (I) E F F E C T I V E NU C L E A R CH A R G E : Between the outer most valence electrons and the nucleus of an atom, there exists finite number of shells containing electrons. Due to the presence of these intervening electrons, the valence electrons are unable to experience the attractive pull of the actual number of protons in the nucleus. These intervening electrons act as shield between the valence electrons and protons in the nucleus. Thus, the presence of intervening (shielding) electrons reduces the electrostatic attraction between the protons in the nucleus and the valenece electrons because intervening electrons repel the valence electrons. The concept of effective nuclear charge allows us to account for the effects of shielding on periodic properties. The effective nuclear charge (Zeff) is the charge felt by the valence electron. Zeff is given by Zeff = Z ñ . Where Z is the actual nuclear charge (atomic number of the element) and  is the shielding (screening) constant. There are some simple rules for determining the degree to which electrons in the various types of orbitals shield other electrons from the nucleus & hence for determining the Zeff experienced by other electrons. Write the full electronic configuration. If the electron in question resides in an s or p orbital, 1. All electrons in principal shells higher than the electron in question contribute zero to . 2. Each electron (other than in the question) in the same principal shell contributes 0.35 to . 3. Electrons in the (n ñ 1) shell each contribute 0.85 to . 4. Electrons in deeper shells each contribute 1.00 to . If the electron in question resides in a d or f orbital, 1. All electrons in principal shells higher than electron in question contribute zero to . 2. Each electron (other than in the question) in the same principal shell contributes 0.35 to . 3. All inner-shell electrons [i.e., (n ñ 1) and lower] uniformly contribute 1.00 to . Sum the shielding constants from steps 1ñ4 for s or p orbital or from steps 1ñ3 for d or f orbital and subtract them from the actual nuclear charge (Z) of the atom in question to obtain the Zeff felt by the electron in question. (Source : Basic Inorganic Chemistry (Third Edition) By Cotton Wilkinson and Gaus) The relative extent to which the various orbital penetrate the electron clouds of of other orbitals is s > p > d > f. Thus for any given principal quantum number n, an electron will experience the greatest effective nuclear charge when in s-orbital than p-orbital and so on. ____________________________________________________________________________________________________

What is the effective nuclear charge at the periphery of nitrogen atom when a extra electron is added during the formation of an anlon. Compare the vlaue of Zeff when the atom is ionized.

____________________________________________________________________________________________________

Sol. Ground state electron configuration of N(Z = 7) = 1s2 2s2 2p3 Electron configuration of Nñ = (1s2) (2s2 2p4) Shielding constant for the last 2p electron,  = [(5 ◊ 0.35) + (2 ◊ 0.85)] = 3.45  Zeff = Z ñ Zeff for last p electron on Nñ = 7 ñ 3.45 = 3.55 Electron configuration of N+ = (1s2) (2s2 2p2) Shielding constant for the last 2p electron,  = [(3 ◊ 0.35) + (2 ◊ 0.85)] = 2.75 Zeff for last electron on N+ = 7 ñ 2.75 = 4.25

2.

AT O M I C RA D I U S :

(A)

Covalent radius : It is one-half of the distance between the centres of two nuclei (of like atoms) bonded by a single covalent bond. Covalent radius is generally used for non-metals.

Single Bond Covalent Radius, SBCR (bond length) (a) For homodiatomic molecules 

dAñA = r A + r A or 2rA

so,

rA =

dA A 2

(b) For heterodiatomic molecules in which electronegativity remains approximately same. dA ñ B = rA + rB Do not tell this formula in class. Only for reference of faculty.



For heteronuclear diatomic molecule, AñB, where difference between the electronegativity values of atom A and atom B is relatively larger, dA ñ B = rA + rB ñ 9.0  Electronegativity values are given in pauling units and radius in picometers.  = XA ñ XB where XA and XB are electronegativity values of high electronegative element A and less electronegative element B. This formula is given by Stevenson & Schomaker.

LEC LECTURE TURE # 2 (B)

Van der Waals radius (Collision radius) : It is one-half of the internuclear distance between two adjacent atoms in two nearest neighbouring molecules of the substance in solid state.  Van der Waalís radius does not apply to metals. Its magnitude depends upon the packing of the atoms when the element is in the solid state. Co mp a r i sio n of c ov a le n t r a di u s an and d v a n d e r W a a l í s r a di u s

(i)

(ii)

The van der Waalís force of attractions are weak, therefore, their internuclear distances in case of atoms held by van der Waalís forces are much larger than those of between covalently bonded atoms. Therefore van der Waalís radii are always larger than covalent radii. A covalent bond is formed by the overlaping of two half-filled atomic orbitals, a part of the orbital becomes common. Therefore, covalent radii are always smaller than the van der Waals radii. For example, Elements

H

O

F

S

Br

Covalent radius (≈)

0.37

0.66

0.64

1.04

1.11

van der Waal's radius (≈)

1.20

1.40

1.35

1.85

1.95

(C)

Metallic radius (Crystal radius) : It is one-half of the distance between the nuclei of two adjacent metal atoms in the metallic crystal lattice.  Metallic radius of an element is always greater than its covalent radius. It is due to the fact that metallic bond (electrical attraction between positive charge of an atom and mobile electrons) is weaker than covalent bond and hence the internuclear distance between the two adjacent atoms in a metallic crystal is longer than the internuclear distance between the covalently bonded atom. For example :



rcovalent < rcrystal

Metallic radius K 231 pm Na 186 pm < rvander Walls

Covalent radius 203 pm 154 pm

Variation in a period: On moving left to right due to increased nuclear charge the size decreases. Variation in a group: On moving top to bottom due extra addition of a shell the size increases. --------------------------------------------------------------------------------------------------------------------------------------------------------------------Note : This portion is not properly discussed so do explain all these a bit slow Some Irregularities (a)

The atomic radius of inert gas (zero group) is given largest in a period because it is represented as van der Waalsís radius which is generally larger than the covalent radius. The van der Waalís radius of inert gases also increases from top to bottom in a group. Talk about general trend. Do not stress on exception. Need not memorize.

(b)

In the transition series (e.g. in first transition series), the covalent radii of the elements decrease from left to right across a row until near the end when the size increases slightly. On moving from left to right, extra protons are placed in the nucleus and the extra electron are added. The orbital electrons shield the nuclear charge incompletely. Thus the nuclear charge attracts all the electrons more strongly, hence a contraction in size occurs. The radii of the elements from Cr to Cu, are very close to one another because the extra electron being added increasses the repulsion between the electrons and counter balances the increased nuclear charge on the outer electrons (4s). As a result of this, the size of the atom does not change much in moving from Cr to Cu and for zinc this repulsion even dominates the nuclear charge so size slightly increases. Element Sc Ti V Cr Mn Fe Co Ni Cu Zn Atomic radius (≈) 1.44 1.32 1.22 1.18 1.17 1.17 1.16 1.15 1.17 1.25

(c)

rAl  rGa because of d orbital contraction.

(d)

4d 5d (Zr ñ Hf) ( lanthanide contraction) The lanthanide contraction counter balances almost exactly the normal size increase on descending a group of transition elements. Thus covalent and ionic radii of Nb (5th peroid) and Ta (6th period) are almost same due to poor shielding of f-orbitals electrons. Define lanthanide and transition metal contraction.

Note:

I O N I C RA D I U S : 



The sizes of ions increases as we go down a group (considering the ions of same charge). For example : Li+ < Na+ < K+ < Rb+ Be2+ < Mg2+ < Ca2+ < Sr2+ Fñ < Clñ < Br ñ < ñ Cation is smaller than parent atom but anion is bigger than parent atom.

 The species containing the same number of electrons but differ in the magnitude of their nuclear charges are called as isoelectronic species. For example, N3ñ , O2ñ, Fñ, Ne, Na+ , Mg2+ and Al3+ are all isoelectronic species with same number of electrons (i.e 10) but different nuclear charges of +7, +8, +9, +10, +11, +12 and +13 respectively.

Within a series of isoelectronic species as the nuclear charge increases, the force of attraction by the nucleus on the electrons also increases. As a result, the ionic radii of isoelectronic species decrease with increases in the magnitude of nuclear charges. For example, Al3+

Mg2+

Na+

Fñ

O2ñ

N3ñ

Ionic radii increase As effective nuclear charge decrease.  Zero group elements should not be considered while comparing the size or ionic radii as their atomic radii are expressed as van der Wallís radii.

LECTURE # 3 Ionisation Energy ( Ionisation potential) Min amount of energy required to remove an electron from an isolaled gaseous atom / species.

(I.E.)1

M(g)  M+(g) + eñ

(I.E.)2

M+(g)  M2+ (g) + eñ

(I.E.)3 > (I.E)2 > (I.E.)1

H = +ve

ñ Difficult to remove an electron from a positively charged ion than from a neutral atom.

Factors affecting I.E. Size

Size   I.E. 

(b)

Zeff

Zeff   I.E. 

(c)

Electronic configuration ñ For a stable electronic configuration i.e. half filled or fully filled configurations (stable because of symmetry) I.E. will be larger The relative extent to which the various orbital penetrate the electron clouds of of other orbitals is s > p > d > f. Thus for any given principal quantum number n, an electron will experience the greatest effective nuclear charge when in s-orbital than p-orbital and so on. Hence the order of I.E. is as follows : s > p > d > f

(a)

(d)

but

  (i.e. Zeff  ) E 

Teach following portion a bit slowly Periodicity LñR

I.E. 

(Zeff  )

TñB

I.E. 

(Size  )

(a)

 rregularities In  period ( also in  period )

(b)

Be > B, N > O, Ne >> F .E.Ga> .E.Al

(c)

size same extra nuclear charge  5d > 3d > 4d (lanthanide contraction) Note : I.E. can be correlated with the reactivity e.g., (a) Noble gases are inert towards chemical reactivity.

(b) low I.E. of alkalimetals correlate with their high reactivity. ---------------------------------------------------------------------------------------------------------------------------------------------------------------------

Puts some examples of ionization energy here Electron Affinity (E.A) The energy released when an electron is added to an isolated gaseous atom to produce a monovalent anion is known as e.a. and enthalpy change of this process is known as electron gain enthalphy  Heg   ve energy released E.A.   ve

M(g) + eñ  Mñ(g)

 Heg   ve energy absorbed E.A.   ve

(E.A.)2 is always +ve ñ because of electrostatic repulsion between anion and electron (having same charge). Heg2 = + ve

Mñ(g) + eñ  M2ñ(g) Factors affecting E.A. (a)

Size

Size   E.A. 

(b)

Zeff

Zeff   E.A. 

(c)

E.C.

Stable E.C. will have smaller or ñve E.A

   E.A. 

Periodicity L R

Zeff   E.A. 

TB

Size   E.A.  (Except 3rd period elements)

Irregularities (a)

 nd period

(b)

E.A. 3P > 2P E.A (low) B C N O F (added electron goes to the samller n = 2 and suffer significant repulsion from the other electrons present in this level.) E.A.(high)

Al

Si

P

S

Cl

(added electron goes to the bigger n = 3 i.e. occupies larger region of space and the electron-electron repulsion is much less.) (c)

Noble gases have large positive electron gain enthalpies because the electron has to enter the next higher principal quantum level leading to a very unstable electronic configuration. e.g.H in kJ / mol of groups 16th and 17th S ñ200

Se ñ195

Te ñ190

Po ñ174

O ñ141

Cl ñ349

F ñ328

Br ñ325

I ñ295

At ñ270

LEC LECTURE TURE # 4 Electonegativity. A qualitative measure of the ability of an atom in a chemical compound to attract shared electrons to itself iscalled electronegativity not a measurable quantity. A number of numerical scales of electronegativity have been proposed. Differences in E.A. & E.N.

(a) (b)

E.A. is defined in isolated gaseous state while E.N. is defined in bonded state E.A. is a absobule term, it has proper units while E.N. is a relative concept has no units

Different Scales of Measurement of Electonegativity. (a)

Paulingís scale : Linus Pauling developed a method for calculating relative electronegativities of most elements. According to Pauling (a)

 = XA ñ XB = O.208

E.A B  E A  A  EB B

(All bond energies are in kcal / mol)

EA-B = Bond enthalpy/ Bond energy of A ñ B bond. EA - A = Bond energy of A ñ A bond EB ñB = Bond energy of B ñ B bond (b)

 = XA ñ XB = O.1017

E.A B  E A  A  EB B

(All bond energies are in kJ / mol.) (b)

Mullikenís scale : Electronegativity  (chi) can be regarded as the average of the ionisation energy (IE) and the electron affinity (EA) of an atom (both expressed in electron volts). M =

IE  EA 2

Only For Faculty refrence. Not to be told in class. Paulingsís electronegativity P isrelated to Mullikenís electronegativity M as given below. P = 1.35 (M)1/2 ñ 1.37 Mullikenís values were about 2.8 times larger than the Paulingís values. Factors affecting Electronegativity. (a)

Size

Size   EN 

(b)

Zeff

Zeff   EN 

(c)

Charge on cationic species A3+ > A2+ > A+1 (for the same element) Greater the charge on cation, greater E.N value Charge on anionic species

(d) (e)

   EN 

A3ñ < A2ñ < Añ State of hybridization Greater the %s character  greater the attraction on the shared pair  Greater will be E.N sp > sp2 > sp3

Electronegativity of some elements according to Pauling scale Elements Electronegativy

H

Li

Be

B

C

N

O

F

Na

P

S

Cl

Br

I

2.1

1.0

1.5

2.0

2.5

3.0

3.5

4.0

0.9

2.1

2.5

3.0

2.8

2.5

Periodicity L ñ R  Zeff  E.N 

T ñ P  Size  E.N 

--------------------------------------------------------------------------------------------------------------------------------------------------------------------Note : BE SLOW AND EXPLAIN PROPERLY

Applications Prediction of nature of bond (a) XA = XB

 pure covalent bond

 partly ionic + covalent

(b) XA  XB

Only for faculty refrence (Debatable theory) Not to be done in class Acc. to pauling, if XA ñ XB = 1.7 50% ionic  < 1.7 more covalent less ionic  > 1.7 more ionic less covalent Henny smith formula % ionic character Bond to be 50% ionic  = 2.1 Q.

= 16 + 3.5 2   = | XA ñ XB|

Calculate % ionic character of bond formed between the most electropositive element Cs (x = .7) and most electronegative element F (x = 4.0). According to Hanny Smith formula % ionic character = 16(3.3) + 3.5 (3.3)2 = 52.8 + 38 .115 = 91.915 %

LEC LECTUR TUR TURE E#5 (a)

Inert pair effect : The outer shell ësí electrons (ns2) penetrate to (n ñ 1) d electrons and thus become closer to nucleus and are more effectively pulled towards the nucleus. This results in less availability of ns2 electron pair for bonding of ns2 electron pair becomes inert. The inert pair effect begins after n  4 and increases with increases value of n. (i) Dominance of lower oxidation state on moving down the group in boron, carbon & nitrogen family may also be explained by the inert pair effect. (ii) The inert pair effect is not the explanation of why monovalency , bivalency and trivalency in group 13, 14 & 15. It merely describe what happens, i.e. two electrons do not participate in bonding. The reason that they do not take part in bonding in energy. Order of stability of oxidation state according to Inert pair effect :

Tl   Tl 3 Pb 2  Pb 4 Bi 3  Bi 5 (b)

Sn 2  Pb  2

Ga  In   Tl  Ga  3  In 3  Tl  3

Sn

4

 Pb

4

Sb 3  Bi 3   Sb 5  Bi 5

Note: Discussion of all sheets

LEC LECTUR TUR TURE E#6 (Shift to chemical bonding) Types of Oxides and acidic / basic nature of oxyacids and hydra acid : (a) Ex.

Acidic oxides (a) Solution in water will be acidic in nature H2 O SO2   H2SO3 Sulphurous acid



(b) will react with base but not with an acid H 2O SO3   H2SO4 sulphsic acid

H2 O CO2   H2CO3 Carbonic acid

generally non-metallic oxider are acidic oxides (b) Ex.

Basic oxides (a) Solution in water will be basic in nature

(b) will react with an acid but not with a base

H2 O Na2O   2NaOH

H2 O CaO    Ca(OH)2
...