Hybridization

*Intermixing of two or more pure atomic orbitals with different characteristics of same atom with almost same energy to give same number of identical and degenerate new type of orbitals is known as hybridization. The new orbitals formed are also known as hybrid orbitals.

E.g.
*If, one 's' and one 'p' orbitals of almost equal energy intermix to give two identical and degenerate hybrid orbitals, called sp hybrid orbitals.

*If, one 's' and two 'p' orbitals of almost equal energy intermix to give three identical and degenerate hybrid orbitals, called sp2 hybrid orbitals.

*If, one 's' and three 'p' orbitals of almost equal energy intermix to give four identical and degenerate hybrid orbitals, called sp3 hybrid orbitals.

  

*Hybridization is suitable for the pairing of electrons to form chemical bonds in valence bond theory.
*Provides a quantum mechanical insight to Lewis structures thus, during this process, the wave functions, Ψ of atomic orbitals of same atom are combined to give new wave functions corresponding to hybrid orbitals.

*The new orbitals have the same total electron capacity as the old ones.

*The properties and energies of the new, hybridized orbitals are an 'average' of the original unhybridized orbitals.

*Hybrid orbitals are very useful in the explanation of molecular geometry and atomic bonding properties. (Although sometimes it taught together with the valence shell electron-pair repulsion (VSEPR) theory but notice that valence bond and hybridization are in fact not related to the VSEPR model.)
*Hybrid orbitals cannot be used to interpret photoelectron spectra.
*Photoelectron spectra measure the energies of ionized states, identified with delocalized orbital energies using Koopmans' theorem (The first ionization energy of a molecular system is equal to the negative of the orbital energy of the highest occupied molecular orbital, HOMO, or for closed-shell systems, the negative of the HOMO energy is the ionization potential i.e. the energy required to form the cation provided that the ionization process is adequately represented by the removal of an electron from an orbital without change in the wave-functions of the other electrons.)
*Structure and hybridization: Click here.

Base induced E2 eliminations:

Base induced E2 eliminations of alcohols may be achieved by their sulfonate ester derivatives. This has the advantage of avoiding strong acids, which may cause molecular rearrangement and or double bond migration in some cases.

*Since 3º-sulfonate derivatives are sometimes unstable, this procedure is best used with 1º and 2º-mesylates or tosylates.

Application of this reaction sequence is shown here for 2-butanol.

The Zaitsev Rule favors formation of 2-butene (cis + trans) over 1-butene.

*Eliminations often favor the more stable trans-product over the cis-product (stereoselectivity).

*The E2 elimination of 3º-alcohols under relatively non-acidic conditions may be accomplished by treatment with phosphorous oxychloride (POCl₃) in pyridine.

*This procedure is also effective with hindered 2º-alcohols, but for unhindered and 1º-alcohols an S_N2 chloride ion substitution of the chlorophosphate intermediate competes with elimination.

*Some examples of these and related reactions are:

*The first equation shows the dehydration of a 3º-alcohol. The predominance of the non-Zaitsev product (less substituted double bond) is presumed due to steric hindrance of the methylene group hydrogens, which interferes with the approach of base at that site.

*The second example shows two elimination procedures applied to the same 2º-alcohol.

The first uses the single step POCl₃ method, which works well in this case because SN² substitution is retarded by steric hindrance.

*The second method is another example in which an intermediate sulfonate ester confers halogen-like reactivity on an alcohol. In every case the anionic leaving group is the conjugate base of a strong acid.

Elimination Reactions of Alcohols

In alkyl halide reactions  2º and 3º-alkyl halides experienced rapid E2 elimination when treated with strong bases, such as hydroxide and alkoxides. But-

*Alcohols do not undergo such base-induced elimination reactions and are, in fact, often used as solvents for such reactions. This is a example of how leaving group stability often influences the rate of a reaction.

*When an alcohol is treated with sodium hydroxide, the following acid-base equilibrium occurs. Most alcohols are slightly weaker acids than water (why?) so the left side is favored.

R–O–H   +   Na⁽⁺⁾ OH^(–)   <=====>  R–O^(–) Na⁽⁺⁾   +   H–OH

*This type of elimination in which water released from an alcohol is called dehydration. 

*It is important to note that-

(a) "Water is a much better leaving group than hydroxide ion, it is sensible to use acid-catalysis rather than base-catalysis to achieve such reactions".

(b) "Hydrohalic acids (HX) are not normally used as catalysts because their conjugate bases are good            nucleophiles and may give substitution products"

(c) "The conjugate bases of sulfuric and phosphoric acids are not good nucleophiles and do not give substitution  under the usual conditions of their use.

(d) "The acid-catalyzed dehydrations are the reverse of the acid-catalyzed hydration reactions of alkenes.

(e) "The laws of thermodynamics require that the mechanism in both directions proceed by the same reaction path. This is known as the principle of microscopic reversibility.  

Reactions:

The first two examples (top row) are typical, and the more facile elimination of the 3º-alcohol suggests predominant E1 character for the reaction. This agrees with the tendency of branched 1º and 2º-alcohols to give rearrangement products, as shown in the last example. The last two reactions also demonstrate that the Zaitsev Rule applies to alcohol dehydrations as well as alkyl halide eliminations. Thus the more highly-substituted double bond isomer is favored among the products.

E1 - Mechanism:

To illustrate, the following diagram lists the three steps in each transformation.

The dehydration reaction is shown by the blue arrows; the hydration reaction by magenta arrows. The intermediates in these reactions are common to both, and common transition states are involved.

Periodic properties

Atomic radii:

“The distance from the center of the nucleus to the outermost shell containing electrons”.  Since absolute value of atomic size cannot be determined, instead, it is determined in terms of operational definition such as covalent radius, van der Waals radius etc.

Along the period; Normally, covalent radius and van der Waals radius decreases from left to right because left to right shells remain constants and the relative or effective nuclear charge is increased.
 *Alkali metals have the largest size in a period.                                                                                             

*Atomic size order in; Inert gas > Alkali metals > other group elements > halogen > Nitrogen.

*After nitrogen the atomic size for oxygen increases for oxygen and then decreases for F (why?) because in nitrogen no inter electronic repulsion is observed due to three unpaired electrons in different orbitals but in case of oxygen two paired electron repel each other hence due to this inter electronic repulsion the atomic size increases from N to O. in case of F, the enhanced nuclear charge predominates over inter electronic repulsions.

*The size of the atoms of inert gases are larger than others, (why?) because here van der Waals radius is considered since they do not form covalent bond and van der Wall Radius  is greater than  covalent radius always.

*Periodic order of atomic radii; s block > d block > p block.

*The size of the atoms of inert gases are, however, larger than those of the preceding halogens because here van der Wall Radius (> metallic radius > covalent radius) is considered not covalent radius.

*d and f – blocks elements show exceptional properties.

*d – block; the covalent radii of the elements decrease from left to half midway (to Cr) and the radii of the elements, from Cr to Cu are very close to one another until near the end where the size increase slightly, across a period; (Why?) because on passing from left to right, extra protons are placed in the nucleus and extra orbital electrons are added. The orbital electrons shield the nuclear charge incompletely (because shielding effect order s > p > d > f). Because of this poor shielding effect by d - electrons, the nuclear charge attracts all of the electrons more strongly: hence a contraction in size occurs.¹

            The radii of the elements, from Cr to Cu are very close to one another; this is due to the fact that the successive addition of d – electrons screens the outer electron (4s) from the inward pull of the nucleus. As a result of this, the size of the atom does not alter much in moving from Cr to Cu.²

            Near the end of the period, there is a slight increase in the atomic radii; this is due to the fact that near the end of the series, the electron - electron repulsions between added electrons in the same orbitals are greater than the attractive forces due to the increased nuclear charge. This result in the expansion of the electron cloud and thus the atomic radius increases.

Sc 1.44

Ti 1.32

V 1.22

Cr 1.17

Mn 1.17

Fe 1.17

Co 1.16

Ni 1.15

Cu 1.16

Zn 1.25

*The elements in the 3^rd group of the d-block show the expected increase in size Sc ----Y-----La.

*The elements in the 4^th to 12^th group of d – block the atomic radii of the first and second transition elements show expected increases but  radii of the second  and third transition series are almost same;  (why?) this is due to the fact in the atoms of the second transition series, the number of shells increases, their atomic radii are larger than those of the elements of the first transition series but in second and third transition metals between  lanthanum and hafnium there are  14 lanthanide elements are present, in which the antepenultimate 4f shell of electrons is filled. There is a gradual decrease in size of the 14- lanthanide elements from Ce to Lu. This is called lanthanide contraction. Thus the lanthanide contraction cancels almost exactly the normal size increases on descending a group of transition elements.

*Along the group; Normally, size is increased from top to bottom (why?) because top to bottom number of shells is increased and the relative or effective nuclear charge is decreased.

Ionic Radii:-

“The effective distance from the center of the nucleus of the ion upto which it exerts its influence on its electronic cloud is called ionic radii.”  Cation always has smaller radius and an anion is always larger than it parent atom.

*In case of isoelectronic ions, i.e., ions having same number of electrons but different nuclear charge, ionic radius decreases with increase in nuclear charge.

*Ionic radii follow the same trend as the atomic radii.

*KU 2012; How does the atomic size of elements vary in the periodic table and why?

Answer; Normally, decreases from left to right because left to right shells remain constants, the relative or effective nuclear charge is increased.

*Periodic order of atomic radii; s block > d block > p block.

*d and f – blocks elements show exceptional properties.

*d – block; the covalent radii of the elements decrease from left to half midway (to Cr) and the radii of the elements, from Cr to Cu are very close to one another until near the end where the size increase slightly, across a period; ), because of poor shielding effect by d – electrons.

KU 2005; Why does the size of an anion is greater than its parent atoms?

KU 2011; Which has the smallest size?

(a)    Cl

(b)   Cl^-

(c)    Br

(d)   Br^-

Answer; Cl < Br < Cl^- < Br^-

KU 2011; Why is the size of Na⁺ ion less than of Na atom?

KU 2010; Which following has largest size?

(a)    Al

(b)   Al⁺

(c)    Al⁺²

(d)   Al³⁺

Answer; Al > Al⁺ > Al²⁺ > Al³⁺

KU 2009, 2006; Which following has largest size?

(a)    Li²⁺

(b)   Be²⁺

(c)    O^2-

(d)   N^3-

Answer; N^3- > O^2- > Be²⁺ > Li²⁺

Hint; O and N are isoelectronic, Li and Be follows normal order

KU 2009; Why is the size of Na⁺ ion less than of Na atom and that of Cl^- is bigger than Cl atom?

*Arrange the following ions in order of their increasing ionic radii:

Li⁺, Mg²⁺, K⁺, Al³⁺.

Answer;    Al³⁺ < Li⁺ < Mg²⁺ < K⁺

Hint;   Al³⁺and Mg² are isoelectronic.  Li⁺ and K⁺ follow normal trend.

*Arrange the following ions in the order of increasing size:

Be²⁺, Cl^-, S^2-, Na⁺, Mg²⁺, Br^-.

Answer;  Be²⁺ < Mg² < Na⁺ < Cl^- < S^2- < Br^-.

*Arrange the following in decreasing order of their van der Waals radii:

Cl, H, O, N.

Answer; Cl > N > O > H

Hint; The van der Waals radii increases s the number of energy shells increases and decreases as the nuclear charges increases.

*Atomic size of N < O but van der Waals radii is N > O. (why?)  Because both N and O have two energy shells but the nuclear charge on O (+8) is higher than that on N (+7).

*Among the elements B, Al, C and Si which has the largest radii?

Answer;  Al > B > Si > C.

 Some useful information for you!!!

1.      For Rajya sabha how many members are nominated by president? 12

2.      Chankya was the chief advisor of which ruler? Chandragupta Maurya.

3.      What is the direction of rotation of earth on its axis? West to East

4.      In Computer, One byte equals to how many bits? 8

5.      Which is the highest civilian award in India? Bharat Ratna.

Chemistry Notes

1. The Mole

Don’t forget that individual atoms are far too small to weigh!

ü  Chemist’s ‘dozen’ is the mole

ü  One ‘dozen’ means 12 bananas.

ü  One ‘moles’ means 6.02 x 10²³ of any particles,called (Avogadro’s Number )

ü  A mole = 22.4 litres = molecular mass of gas or liquid.

ü  A mole = relative atomic mass of an element.

1 mole of zinc atoms (ANY particles!) would have a mass of 65.4g (This is the mass number printed on your periodic table) and so on.

1 mole of zinc atoms (I repeat ANY particles!) contains 6.02 x 10²³ atoms.

1 mole of methane molecules (I repeat again ANY particles!) contains 6.02 x 10²³ molecules.

Notice the following:

Average weight: If 75% of all chlorine atoms are Cl-35 and 25% are Cl-37, then the weighted average is:-

(0.75 × 35) + (0.25 × 37) = 35.5.

The mole is a number that answers the question, "How many?", in the same way a meter answers the question, "How far?", or a kilogram answers the question, "How much mass?"

For example, sodium chloride has the formula NaCl. When 6.02 x 10²³ sodium ions and 6.02 x 10²³ chloride ions combine together we have 1 mole of sodium chloride. This means that:

a.      One mole of NaCl contains one mole of sodium ions and one mole of chloride ions.

b.      One mole of Na is the same amount as 22.9898 grams.

c.      One mole of Cl is the same amount as 35.453 grams.

d.      One mole of NaCl is the same amount as 22.9898 + 35.453 grams, or 58.4418 grams. It follows that:

§  1/2 mole of NaCl is 29.2209 grams.

§  2 mole of NaCl is 104.8836 grams.

§  10 mole of NaCl is 584.418 grams.

Note: The mass of an electron is negligible compared to the rest of the atom. Compare this to another ionic compound, K₂O. This compound contains two moles of potassium ions, K⁺, and one mole of oxide ions, O^2-, in one mole of potassium oxide.

*The formula of ionic compound does not represent its molecule. It simply represents a unit of lattice.

*One mole NaCl = 6.022 X 10²³ units of NaCl  i.e. each units of NaCl = 6.022 X 10²³ particles of Na⁺ and 6.022 X 10²³ particles of Cl^- .

At last for you!

How can I Convert mole into weight which I need for my experiment?

Use this relation:

     Mole = Given weight / Molecular weight

 

Do you know? That – We retain information at these rates:

ü  10 percent of what we read.

ü  20 percent of what we hear.

ü  30 percent of what we see.

ü  50 percent of what we see and hear.

ü  70 percent of what we see and discuss.

ü  90 percent of what we do.

Another way to think about how we retain information is this edge:

Tell me and I will probably forget,

Show me and I might remember,

Involve me and I will learn.

Ha ha ha ! I am going to involve you to solve my problems!

Problems related to moles:

A – Theory:

Short answers type questions:

ü  How is the mass of an element related to the number of atoms present in it?

ü  Which does have more atoms 1.0 g of hydrogen or 1.0 g of oxygen?

ü  What is Avogadro’s number and what is its significance?

ü  How is the mole concept useful in calculations based on chemical equations?

Is it true or false?

ü  One mole of K₂SO₄ contains 6.02 X 10²³ potassium ions.

ü  One gram mole of a monoatomic gas occupies 22.4 L at S.T.P.

ü  A chemical equation tells the actual number of moles of reactants and products.

ü  The number of atoms present in 1 g of hydrogen is the same as that in 1 g of oxygen.

ü  The ratio of atoms can vary in differently prepared samples of a compound.

ü  How many moles of chromium ions and oxide ions are present in one mole of Chromium (III) oxide, Cr₂O₃?

Fill in the blanks:

ü  One gram mole, one gram molecules, and one gram molecular mass have the …………..meaning.

ü  A mole is defined as the amount of the substance that contains as many specified elementary particles as the number of atoms in ……………..g of C¹².

ü  Total numbers of atoms in 90 g of water are…………

ü  The weight of 1 X 10²² molecules of CuSO₄.5H₂O is………….

ü  The volume occupied by the 64 g of SO₂ at S.T.P. is………………