## Innovative and Time Economic Pedagogical Views in Chemical Education - A Review Article

**Arijit Das**^{1,}, **R. Sanjeev**^{2}, **V. Jagannadham**^{3}

^{1}Department of Chemistry, Ramthakur College, Agartala, Tripura, India

^{2}Department of Pharmacy, School of Health Science, University of KwaZulu-Natal, Durban, South Africa

^{3}Department of Chemistry, Osmania University, Hyderabad, India

2. Innovative Thinking on the Prediction of Hybridization State

3. New Dimension for the Prediction of Bond-Order

4. Isolation of Magnetic Behavior of Homo and Hetero Nuclear Diatomic Molecules or Ions

5. Evaluation of Spin Multiplicity Value

6. Identification of Aromatic and Anti-Aromatic Nature of Organic Compounds

7. Calculation of π-bonds, σ-bonds, Single and Double Bonds in Alkenes and Cycloalkenes

8. Calculation of π-bonds, σ-bonds, single and Triple Bonds in Alkyne and Cycloalkyne

### Abstract

Context based learning approaches have been presented in this article as a way to enhance student’s interest in, as well as time economic learning outcomes from chemical education. In this article 16 years research based time economic study has been presented on hydrocarbon, aromaticity and chemical bonding to cultivate the interest of teaching and learning in organic and inorganic chemistry when solving context-based chemistry problems.** **In this pedagogical survey, we have tried to hub twelve (12) innovative and time economic methodologies by including thirty two (32) completely new formulae in the field of Chemical Education. Concepts that usually confuse the students in the examination hall have been explained here by some innovative methods based on applied mathematics. In this context, current trends in chemistry have been highlighted by some technical methods. The review explores the results and gives implications for context-based teaching, learning and assessment.

### At a glance: Figures

**Keywords:** general public, high school, graduate student, chemical education research, organic and inorganic chemistry, problem solving, M.O. theory, alkenes, alkynes, aromatic compounds

*World Journal of Chemical Education*, 2014 2 (3),
pp 29-38.

DOI: 10.12691/wjce-2-3-1

Received June 20, 2014; Revised July 03, 2014; Accepted July 18, 2014

**Copyright**© 2013 Science and Education Publishing. All Rights Reserved.

### Cite this article:

- Das, Arijit, R. Sanjeev, and V. Jagannadham. "Innovative and Time Economic Pedagogical Views in Chemical Education - A Review Article."
*World Journal of Chemical Education*2.3 (2014): 29-38.

- Das, A. , Sanjeev, R. , & Jagannadham, V. (2014). Innovative and Time Economic Pedagogical Views in Chemical Education - A Review Article.
*World Journal of Chemical Education*,*2*(3), 29-38.

- Das, Arijit, R. Sanjeev, and V. Jagannadham. "Innovative and Time Economic Pedagogical Views in Chemical Education - A Review Article."
*World Journal of Chemical Education*2, no. 3 (2014): 29-38.

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### 1. Introduction

In this article, we have introduced some time economic innovative methods [1-10]^{[1]} comprising prediction of hybridization state, bond order for diatomic homo / hetero nuclear molecules or ions having (1-20) electrons, bond-order of oxide based acid radicals, prediction of spin state using spin multiplicity value, aromatic and anti-aromatic behavior of organic compounds, evaluation of magnetic behavior of homo and hetero nuclear diatomic molecules or ions having 1-20 electrons, calculation of the number of π-bonds, σ-bonds, single and double bonds in aliphatic unsaturated open chain and cyclic olefinic hydrocarbons and alkynes etc. on conventional methods [11-23]^{[11]} to make chemistry time economic and more interesting. Here, we have tried to discuss them abruptly.

### 2. Innovative Thinking on the Prediction of Hybridization State

Hybridization state generally considered an operative empirical for excusing the structures of organic and inorganic compounds and their related problems.

**2.1. sp, sp**

^{2}, sp^{3}HybridizationWe know that, hybridization is nothing but the mixing of orbital’s in different ratio to form some newly synthesized orbitals called hybrid orbitals. The mixing pattern is as follows:

Formula for determination of hybridization state like sp, sp^{2}, sp^{3} followed the following method:

Power of the hybridization state of the central atom = (Total no of σ bonds around each central atom -1)

All single (-) bonds are σ bond, in double bond (=) there is one σ and 1π, in triple bond there is one σ and 2π. In addition to these each lone pair (i.e.no of electrons in the outermost orbit which should not take part in bond formation) and co-ordinate bond can be treated as one σ bond.

Eg.

a. In NH_{3}, central atom N is surrounded by three N-H single bonds i.e. three sigma (σ) bonds and one lone pair (LP) i.e. one additional σ bond. So, in NH_{3} there is a total of four σ bonds [3 bond pairs (BPs) + 1 lone pair (LP)] around central atom N. Therefore, in this case power of the hybridization state of N = 4-1 = 3 i.e. hybridization state = sp^{3}.

b. In H_{2}O, central atom O is surrounded by two O-H single bonds i.e. two sigma (σ) bonds and two lone pairs i.e. two additional σ bonds. So, altogether in H_{2}O there are four σ bonds (2 bond pairs + 2 lone pairs) around central atom O, So, in this case power of the hybridization state of O = 4-1 =3 i.e. hybridization state of O in H_{2}O = sp^{3}.

c. In H_{3}BO_{3}_{ }:- B has 3σ bonds (3BPs but no LPs) and oxygen has 4σ bonds (2BPs & 2LPs) so, in this case power of the hybridization state of B = 3-1 = 2 i.e. B is sp^{2} hybridized in H_{3}BO_{3}. On the other hand, power of the hybridization state of O = 4-1= 3 i.e. hybridization state of O in H_{3}BO_{3} is sp^{3}.

d. In I-Cl, I and Cl both have 4**σ*** *bonds and 3LPs, so, in this case power of the hybridization state of both I and Cl = 4 - 1 = 3 i.e. hybridization state of I and Cl both are sp^{3}.

e. In CH_{2}=CH_{2}, each carbon is attached with 2 C-H single bonds (2 **σ*** *bonds) and one C=C bond (1**σ*** *bond), so, altogether there are 3 sigma bonds. So, in this case, power of the hybridization state of both C = 3-1 = 2 i.e. hybridization state of both C’s are sp^{2}.

**2.2. sp**

^{3}d, sp^{3}d^{2}, sp^{3}d^{3}HybridizationIn case of sp^{3}d, sp^{3}d^{2} and sp^{3}d^{3} hybridization state there is a common term sp^{3} for which 4 sigma bonds are responsible. So, in addition to 4 sigma bonds, for each additional sigma, added one d orbital gradually as follows:-

5σ bonds = 4σ bonds + 1 additional σ bond = sp^{3}d hybridization

6σ bonds = 4σ bonds + 2 additional σ bonds = sp^{3}d^{2} hybridization

7σ bonds = 4σ bonds + 3 additional σ bonds = sp^{3}d^{3 }hybridization

Eg:-

a. IF_{4}^{-}** **

**:** I has 7 e^{-}s in its outermost shell, so, in this case, subtract one e^{-} from 7 i.e. 7 – 1 = 6. So, out of 6 electrons, 4 electrons form 4 I-F bonds i.e. 4 sigma bonds and there is one LP. So, altogether there are 5 σ bonds. So, 5σ bonds = 4 σ bonds + 1 additional σ bond = sp^{3}d hybridization

b.IF_{7}

7 I-F single bonds i.e. 7σ bonds = 4σ bonds + 3 additional σ bonds = sp^{3}d^{3} hybridization.

c. ICl_{2}^{-} - I has 7 e^{-}s in its outermost shell, so, in this case, add one e^{-} with 7(overall charge on the compound) i.e. 07+1= 08. So, out of 08 electrons, 02 electrons form 02 I-Cl bonds i.e. 02 sigma bonds and there is 03 LPs. So, altogether there are 05σ bonds. So, 5σ bonds = 04 σ bonds + 01 additional σ bond = sp^{3}d hybridization.

d. XeF_{4}-

Xe, an inert gas, consider 8 e^{-}s in its outermost shell, 04 of which form 04 Xe-F sigma bonds and there is two LPs, i.e. altogether there is 06 σ bonds = 04 σ bonds + 02 additional σ bonds = sp^{3}d^{2} hybridization.

In case of determination of the hybridization state by using the above method, one must have a clear idea about the outermost electrons of different family members in the periodic table as follows:

And in case of cationic species you must remove electron from the outermost orbit of the central atom and incase of anionic species you must add electron with the outermost electrons of the central atom.

### 3. New Dimension for the Prediction of Bond-Order

**3.1. For Diatomic Homo and Hetero Nuclear Molecules or Ions Having (1-20)e**

^{-}s:**Fig**

**ure**

**1**

**.**

**(B.O. vs number of electrons)**

The graphical representation presented in Figure 1 shows that bond-order gradually increases to 1 in the range (0-2) electrons then it falls to zero in the range (2-4) electrons then it further rises to 1 for (4-6) electrons and once again falls to zero for (6-8) electrons then again rises to 3 in the range (8-14) electrons and then finally falls to zero for (14-20) electrons. For total no of electrons 2, 6 and 14, we use multiple formulae, because they fall in the overlapping region in which they intersect with each other.

First of all we classify the molecules or ions into four (4) types based on the total number of electrons.

i) Molecules and ions having total no of electrons within the range (1-2)

In such case Bond order = n/2; [Where n = Total no of electrons]

Eg. H_{2} (Total es = 2), Therefore B.O. = n/2 = 2/2 = 1

ii) Molecules and ions having total no of electrons within the range (2-6).

In such case Bond order = I 4- n I/2;

[Where n = Total no of electrons, ‘I I’ indicates Mod function i.e. the value of bond order is always positive]

Eg. Li_{2}^{+}(5e^{-}s) Therefore B.O. = I 4-5 I / 2 = 1/2 = 0.5.

iii) Molecules and ions having total no of electrons within the range (6-14).

In such case Bond order = I 8-n I / 2

Eg: CO (Total es = 6+8=14), Therefore B.O.= I 8-14 I / 2 = 3

iv)Molecules and ions having total no of electrons within the range (14-20).

In such case Bond order = (20-n) / 2; [Where n = Total no of electrons]

Eg. NO (Total es = 15), Therefore B.O. = 20-15/2 = 2.5

**3.2. Bond-Order Prediction for Oxide Based Acid Radicals:**

In case of Acid Radicals

Eg.

**Lewis structure of ClO**_{4}^{-}

**ClO**_{4}^{-} ; (Oxidation number of one Peripheral atom i.e. Oxygen = **2**, Charge on acid radical = -1, Total Number of Peripheral atoms = 04), Therefore B.O. = 2 + (-1/4) = 1.75;

**Lewis structure of ClO**_{3}^{-}

**ClO**_{3}^{-}; (Oxidation number of one Peripheral atom i.e. Oxygen = 2, Charge on acid radical = -1, Total Number of Peripheral atoms = 03), Therefore B.O. = 2 + (-1/3) = 1.66;

**Lewis structure of ClO**_{2}^{-}

**ClO**_{2}^{-}; (Oxidation number of one Peripheral atom i.e. Oxygen = 2, Charge on acid radical = -1, Total Number of Peripheral atoms = 02), Therefore B.O. = 2 + (-1/2) = 1.5;

**AsO**_{4}^{3-} ; (Oxidation number of one Peripheral atom i.e. Oxygen = 2, Charge on acid radical = -3, Total Number of Peripheral atoms = 04), Therefore B.O. = 2 + (-3/4) = 1.25,

**AsO**_{3}^{3-}; (Oxidation number of one Peripheral atom i.e. Oxygen = 2, Charge on acid radical = -3, Total Number of Peripheral atoms = 03), Therefore B.O. = 2 + (-3/3) = 1.0;

**SO**_{4}^{2-}^{ }; (Oxidation number of Peripheral atom i.e. Oxygen = 2, Charge on acid radical = -2, Number of Peripheral atoms = 04), Therefore B.O. = 2 + (-2/4) = 1.5;

**SO**_{3}^{2-}^{ }; (Oxidation number of Peripheral atom i.e. Oxygen = 2, Charge on acid radical = -2, Number of Peripheral atoms = 03), Therefore B.O. = 2 + (-2/3) = 1.33

**PO**_{4}^{3-}^{ }; (Oxidation number of Peripheral atom i.e. Oxygen = 2, Charge on acid radical = -3, Number of Peripheral atoms = 04), Therefore B.O. = 2 + (-3/4) = 1.25

**BO**_{3}^{3-}^{ }; (Oxidation number of Peripheral atom i.e. Oxygen = 2, Charge on acid radical = -3, Number of Peripheral atoms = 03), Therefore B.O. = 2 + (-3/3) = 1

**CO**_{3}^{2-}^{ }; (Oxidation number of Peripheral atom i.e. Oxygen = 2, Charge on acid radical = -2, Number of Peripheral atoms = 03), Therefore B.O. = 2 + (-2/3) = 1.33

**SiO**_{4}^{4-}**; **(Oxidation number of Peripheral atom i.e. Oxygen = 2, Charge on acid radical = - 4, Number of Peripheral atoms = 04), Therefore B.O. = 2 + (- 4/4) = 1

Relation of Bond order with Bond length, Bond Strength, Bond energy, Thermal stability and Reactivity:

B.O. α 1 / Bond length or Bond distance;

B.O. α Bond strength;

B.O. α Bond Energy;

B.O. α Thermal Stability; B.O. α 1 / Reactivity

**Correlation among / between Literature values of bond-distances of some oxide based acid radicals with their predicted bond order values**:

Literature values of the Cl-O average bond lengths in ClO_{4}^{-},ClO_{3}^{-} and ClO_{2}^{-} ; As-O bond lengths in AsO_{4}^{3-} and AsO_{3}^{3-}^{ }with respect to their Bond order suggest that with increasing bond-order M-O bond length (Where M = Cl, As etc.) decreases which is shown in Table 1.

### 4. Isolation of Magnetic Behavior of Homo and Hetero Nuclear Diatomic Molecules or Ions

The present study involves three new formulae by just manipulating the number of unpaired electrons (n) using mod function (based on Applied Mathematics) and by means of these n values one can easily stumble the magnetic moment values in Bohr-Magneton using spin only formula μ_{s }= √n (n+2) B.M., where B.M. = Bohr Magneton = Unit of Magnetic Moment, n = number of unpaired electrons.

First of all we classify the molecules or ions depending on the total number of electrons present in them in the following three (03) sets.

**Set-1: Molecules or ions having (1-3)e**^{-}**s, (3-5)e**^{-}**s, (5-7)e**^{-}**s, (7-10)e**^{-}**s, (13-16)e**^{-}**s; **

**Set-2: Molecules or ions having (10-13)e**^{-}**s and (16-19)e**^{-}**s**

**Set-3: Molecules or ions having 20 e**^{-}**s**

Then for different set we have to use three different formulae to calculate the number of unpaired electrons which have been presented in Table 2 and thus magnetic moment (μ_{s }in B.M.) can be evaluated in the following way.

**4.1. F-1(For Set-1) - for the Determination of Number of Unpaired Electrons (n) of Molecules or Ions Having Total Number of Electrons (1-3),(3-5),(5-7),(7-10) and (13-16)e**

^{-}s:In this case, the number of unpaired electrons n = [ I (ND - total e^{-}s) I ]

Here, ND = next digit i.e. digit next to minimum digit and ‘I I’ indicates Mod Function.

Eg:Molecules or ions having (1-3)e^{-}s, in this case ND = 2 because here minimum digit is 1.

Eg. He_{2}^{+} (3e^{-}s), the total number of electrons will be 3, ND = 2, Hence, unpaired electron n = I (ND - total e^{-}s) I = I (2-3) I = 1. Hence, Magnetic Moment μ_{s }= √n(n+2) B.M. = √ 1(1+2) BM = √3 BM = 1.73BM.

For the molecules or ions containing (3-5)e^{-}s, (5-7)e^{-}s, (7-10)e^{-}s, and (13-16)e^{-}s the ND value will be 4, 6, 8 and 14 respectively.

Hence, the value of n = [ I (4-total e^{-}s) I ]; [ I (6- total e^{-}s) I] [I (8- total e^{-}s) I ] and [ I (14- total e^{-}s) I ] respectively.

**4.2. F-2(For Set-2) - for the Determination of Number of Unpaired Electrons (n) of Molecules or Ions Having Total Number of Electrons (10-13) and (16-19):**

In this case, the number of unpaired electrons n = [ I (PD - total e^{-}s) I ]

Here, PD = Penultimate electron digit (i.e. before last electron).

Eg: for C_{2}^{-} (13e^{-}s), the total number of electrons will be 13, PD = 12

Hence, unpaired electron n = I (12 - total e^{-}s) I = I (12-13) I = 1

Hence, Magnetic Moment μ_{s }= √n(n+2) B.M. = √ 1(1+2) BM = √3 BM = 1.73BM

For F_{2} (18e^{-}s), the total number of electrons will be 18, PD = 18

Hence, unpaired electron n = I (18 - total e^{-}s) I = I (18-18) I = 0

Hence, Magnetic Moment μ_{s }= √n(n+2) B.M. = √ 0(0+2) BM = 0 BM = Diamagnetic in nature.

**4.3. F-3(For Set-3) - for the Determination of Number of Unpaired Electrons (n) of Molecules or Ions Having Total Number of Electrons 20:**

In this case, the number of unpaired electrons n = [ I (20 - total e^{-}s) I ]

Eg: for Ne_{2} (20e^{-}s), the total number of electrons will be 20,

Hence, unpaired electron n = I (20 - total e^{-}s) I = I (20-20) I = 0

Hence, Magnetic Moment μ_{s }= √n(n+2) B.M. = √ 0(0+2) BM = 0 BM = Diamagnetic in nature.

### 5. Evaluation of Spin Multiplicity Value

The formula which is generally used for the prediction of spin multiplicity value is [(2S+1), where S = Σs = total spin quantum no]. Here another innovative method has to be introduced for calculation of spin-multiplicity value and thus spin state, shown in Table 3.

First of all we should classify the species (atoms, molecules, ions or complexes) for which spin multiplicity should be evaluated into three types based on the nature of alignment of unpaired electrons present in them.

**5.1. Species Having Unpaired Electrons in Upward Alignment (↑):**

In this case, spin multiplicity = (n+1); where n = number of unpaired electrons

Spin multiplicity = (n +1) = (1+1) = 2 (spin state = doublet); (2+1) = 3 (spin state = triplet) and (3 + 1) = 4 (spin state = quartet) respectively.

Spin multiplicity = (n +1) = (2 + 1) = 3 (in this case ignore paired electrons) (spin state = triplet) and (1 + 1) = 2 (spin state = doublet)

Spin multiplicity = (n +1) = (0 + 1) = 1 (spin state = singlet).

**5.2. Species Having Unpaired Electrons in Downward Alignment (↓):**

In this case spin multiplicity = (-n+1)

Here (-ve) sign indicate downward arrow.

Spin multiplicity = (-n +1) = (-1 + 1) = 0; (-2 + 1) = -1 and (-3 + 1) = -2 respectively.

Spin multiplicity = (-n + 1) = (-2 + 1) = -1( ignore paired electrons) and (-1 + 1) = 0 respectively.

**5.3. Species Having Unpaired Electrons in Both Upward and Downward Alignment**

In this case spin multiplicity = [(+n) + (-n) + 1]

where, n = number of unpaired electrons in each alignment. Here, (+ve) sign and

(–ve) sign indicate upward and downward alignment respectively.

Here total no of unpaired electrons = 2 in which one having upward direction (+1) and other having downward mode (-1).

Hence Spin multiplicity = [(+n) + (-n) +1] = [(+1) + (-1) + 1] = 1 (spin state = singlet)

Here the total no of unpaired electrons = 3 in which two unpaired electrons lie in upward (+2) and one unpaired electrons lie in downward (-1).

Hence Spin multiplicity = [(+n) + (-n) + 1] = [(+2) + (-1) + 1] = 2 (spin state = doublet)

Here the total no of unpaired electrons = 5 in which three unpaired electrons lie upward (+3) and two unpaired electrons lie downward (-2).

Hence Spin multiplicity = [(+n) + (-n) + 1] = [(+3) + (-2) +1] = 2 (spin state = doublet).

### 6. Identification of Aromatic and Anti-Aromatic Nature of Organic Compounds

The present study will be an innovative method involving two formulae by just manipulating the no of π bonds within the ring system and delocalized electron pair (excluding π electron pair within the ring system) with one (01).

**6.1. Prediction of Aromatic Behavior:**

In the first case, the compound must be cyclic, planar (i.e. all the carbon atoms having same state of hybridization) with even number of A value, **where [A = πb+e**^{-}**p+1(constant)]**, here πb = number of π bonds with in the ring system and e^{-}p = number of electron pair outside or adjacent to the ring system i.e. if the ring contains hetero atoms (atoms containing lone pair of electrons) which can undergo delocalization and each negative charge if present may be treated as one pair of electrons.

If the value of ‘A’, for a certain organic compound comes out as even number then this compound will be treated as aromatic compound.

**6.2. Prediction of Anti-aromatic Behavior**

In the second case, the compound must be cyclic, planar (i.e. all the carbon atoms having same state of hybridization) with odd number of A value, where **[A = πb+e**^{-}**p+1(constant)]**, here πb = number of π bonds with in the ring system and e^{-}p = number of electron pair outside or adjacent to the ring system i.e. if the ring contains hetero atoms which can undergo delocalization and each negative charge if present, may be treated as one pair of electrons.

If the value of ‘A’,** **for a certain organic compound comes out as odd number then this compound will treat as anti-aromatic compound.

**6.3. In General Condition for Non-aromatic Behavior of Organic Compounds**

Any compound that lacks one or more of the above features i.e. it may be acyclic / non-planar, is to be treated as non aromatic. But in this case, ‘A’ value may be even or odd number.

It is always to be noted that if the ring contains hetero atom like N, O, S etc, in this case we must** **count that electron pair in the evaluation of ‘A’ value which can undergo delocalization. We never count localized electron pair. Examples have been illustrated in Table 4.

There are some compounds which do not follow the above rule. Huckel’s also cannot explain the aromatic or non aromatic behavior of these compounds. These compounds have been represented in the Table 5.

If we easily predict the nature of organic compound i.e. aromatic, anti aromatic or non aromatic then we can resolve different kind of problems regarding stability, reactivity, acidity etc. by using the following supposition.

1. Order of stability is aromatic > non aromatic > anti aromatic

2. Order of reactivity just follows the reverse order of stability as follows

Anti-aromatic > non aromatic > aromatic

3. Acidity: Stability of Conjugate base α acidity

Eg: cyclopentadienyl anion (aromatic) > cyclopentadiene (non-aromatic) > cyclopentadienyl cation (anti aromatic). Hence, cyclopentadiene (its conjugate base i.e. Cyclopentadienyl anion is aromatic in nature) is much more acidic than cycloheptatriene (its conjugate base i.e. Cycloheptatrienyl anion is anti-aromatic in nature).

### 7. Calculation of π-bonds, σ-bonds, Single and Double Bonds in Alkenes and Cycloalkenes

The molecular formula which defines a very large number of chemical structure, in this particular case, it is a herculean_{ }task to calculate the nature and number of bonds. Earlier Badertscher *et al*^{ }studied a novel formalism to characterize the degree of unsaturation of organic molecules. But no such work has not been taken till now to calculate the number and types of bonds in open chain olefinic system having complex molecular formulae like C_{176}H_{250}, C_{2000}H_{2000}.

Keeping this in view, a rapid method has been proposed for the calculation of number of π-bonds, σ-bonds, single and double bonds with the help of following 06 (six) completely new formulae for certain aliphatic unsaturated open chain and cyclic olefinic hydrocarbons.

**7.1. For Open Chain Aliphatic Hydrocarbons:**

**7.1.1. Calculation of π-bonds and Double Bonds (P):**

In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in a given unsaturated hydrocarbon containing double bonds.

The formula to calculate the number of π bonds or double bonds for an aliphatic straight chain olefin is **P= [(2X-Y)/2] + 1****.**

Where, X = number of carbon atoms; Y = number of hydrogen atoms and P = number of π bonds/double bonds. E.g.: In C_{176}H_{250}, X = 176, Y = 250, therefore P = (2 x 176 – 250)/2 +1 = 51 + 1 = 52 number of π bonds or double bonds.

**7.1.2. Calculation of σ-bonds (S):**

In this case, first we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in the given unsaturated hydrocarbon containing double bonds.

The formula to calculate the number of σ bonds for an aliphatic straight chain olefin is** S = [X + Y - 1]**

Where, X = number of carbon atoms; Y = number of hydrogen atoms and S = number of sigma bonds (σ-bonds).

E.g.: In C_{176}H_{250}, X = 176, Y = 250, therefore P = 176 + 250 -1 = 425 σ bonds.

**7.1.3. Calculation of Single Bonds (A):**

The total number of single bond for an aliphatic straight chain olefin is **A = [(3Y/2)-2]**

where A = number of single bonds and Y is number of hydrogen atoms.

E.g.: In C_{176}H_{250}, Y = 250, therefore **A = **[(3 x 250)/2] = 375 -2 = 373 single bonds. Examples have been illustrated in Table 6.

**7.2. For Cyclic Aliphatic Olefinic Hydrocarbons:**

**7.2.1. Calculation of π-bonds and Double Bonds (P**

_{c}):In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in the given unsaturated cyclic olefinic hydrocarbons.

The formula to calculate the number of π bonds or double bonds for an aliphatic cyclic olefin is **P**_{c}**= [(2X-Y)/2]**

Where, X = number of carbon atoms; Y = number of hydrogen atoms and P_{c} = number of π bonds or double bonds in the cyclic olefinic system.

E.g.: In cyclooctatetraene (C_{8}H_{8}), X = Y = 8, therefore P_{c} = 16-8/2 = 4 number of π bonds or double bonds.

**7.2.2. Calculation of σ-bonds (S**

_{c}):In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in the given unsaturated cyclic olefinic hydrocarbons.

The formula to calculate the number of σ bonds for an aliphatic cyclic olefin is ** S**_{c}** = [X + Y]**

Where, X = number of carbon atoms; Y = number of hydrogen atoms and S_{c} = number of sigma bonds (σ-bonds) in cyclic olefinic system.

Eg: In cyclooctatetraene (C_{8}H_{8}), X = Y = 8, therefore S_{c} = 8+8 = 16 number of σ bonds.

**7.2.3. Calculation of Single Bonds (A**

_{c}):The total number of single bonds in aliphatic cyclic olefin can be calculated by using the formula **A**_{c}** = [3Y/2] **where A_{c} = number of single bonds and y is number of hydrogen atoms in aliphatic cyclic olefin.

E.g.: In cyclooctatetraene (C_{8}H_{8}), Y = 8, therefore A_{c} = 24/2 = 12 number of single bonds. Examples have been illustrated in Table 7.

### 8. Calculation of π-bonds, σ-bonds, single and Triple Bonds in Alkyne and Cycloalkyne

The number and types of bonds in open chain and cycloalkynes having complex molecular formula is a Herculean task. Keeping this in view, a rapid method has been proposed for the calculation of number of π-bonds, σ-bonds, single and triple bonds with the help of following 08 (eight) completely new formulae by just manipulating the number of carbon and hydrogen atoms by using some factors for certain aliphatic unsaturated open chain and cycloalkynes.

**8.1. Open Chain Aliphatic Alkynes**

**8.1.1. Calculation of π-bonds (P)**

In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in a given unsaturated hydrocarbon containing triple bonds. The formula to calculate the number of π bonds for an aliphatic open chain alkyne, where there is one or more than one triple bonds is **P= ****[{(2X-Y)/2} + 1]**

Where, X = number of carbon atoms; Y = number of hydrogen atoms and P = number of π bonds.

E.g.: In C_{16}H_{30}, X = 16, Y = 30, therefore P = [{(2X-Y)/2} + 1] = [{(2 x 16 – 30)/2} +1] = 1 + 1

= 2 number of π bonds.

**8.1.2. Calculation of σ-bonds (S):**

In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in a given unsaturated hydrocarbon containing triple bonds. The formula to calculate the number of σ bonds for an aliphatic open chain alkyne, where there is one or more than one triple bonds is **S= ****[X+Y-1]**

Where, X = number of carbon atoms; Y = number of hydrogen atoms and S = number of σ bonds.

E.g.: In C_{16}H_{30}, X = 16, Y = 30, therefore, S= [X+Y-1] = [16+30-1] = 45 numbers of σ bonds.

**8.1.3. Calculation of Single Bonds (A)**

The total number of single bond for an aliphatic open chain alkyne, where there is one or more than one triple bonds is **A = ****[{(2X+5Y)/2} - 3]/2**

Where, A = number of single bonds, X = number of carbon atoms and Y = number of hydrogen atoms.

E.g.: In C_{16}H_{30}, X = 16, Y = 30, therefore, A = [{(2X+5Y) / 2} - 3]/2

= [{(2x16+5x30)/2}-3]/2 = [91-3]/2 = 44 numbers of single bonds.

**8.1.4. Calculation of Triple bonds (T)**

In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in a given unsaturated hydrocarbon containing triple bonds. The formula to calculate the number of triple bonds for an aliphatic open chain alkyne, where there is one or more than one triple bonds is **T= ****[{(2X-Y)/2} + 1]/2**

Where, X = number of carbon atoms; Y = number of hydrogen atoms and T = number of triple bonds.

E.g.: In C_{16}H_{30}, X = 16, Y = 30, therefore, T = [{(2X-Y)/2} + 1]/2 = [{(2 x 16 – 30)/2} +1]/2

= 2/2 = 1 triple bond.

Examples have been illustrated in Table 8.

**8.2. Cycloalkynes**

**8.2.1. Calculation of π-bonds (P**

_{c}):In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in the given unsaturated cycloalkyne. The formula to calculate the number of π bonds for an aliphatic cycloalkyne is **P**_{c}**= [(2X-Y)/2]**

Where, X = number of carbon atoms; Y = number of hydrogen atoms and P_{c} = number of π bonds in the cycloalkyne system.

E.g.: In Cycloheptyne (C_{7}H_{10}), X =7, Y = 10, therefore P_{c} = (2x7-10)/2 = 2 number of π bonds.

**8.2.2. Calculation of σ-bonds (S**

_{c}):In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in the given unsaturated cycloalkyne. The formula to calculate the number of σ bonds for an aliphatic cycloalkyne is **S**_{c}** = [X + Y]**

Where, X = number of carbon atoms; Y = number of hydrogen atoms and S_{c} = number of sigma bonds (σ-bonds) in cyclo alkyne system.

E.g.: In Cycloheptyne (C_{7}H_{10}), X =7, Y = 10, therefore S_{c} = (7+10) = 17 number of σ bonds.

**8.2.3. Calculation of Single Bonds (A**

_{c}):The total number of single bond for an aliphatic cyclo alkyne is **A**_{c}** = ****[{(2X+5Y)/2}]/2; **Where, A_{c} = number of single bonds in cycloalkyne, X = number of carbon atoms and Y = number of hydrogen atoms.

E.g.: In Cycloheptyne (C_{7}H_{10}), X = 7, Y = 10, therefore, A_{c} = [{(2X+5Y)/2}]/2

= [{(2x7+5x10)/2}]/2=32/2 = 16 numbers of single bonds.

**8.2.4. Calculation of Triple bonds (T):**

In the first case, we have to count the number of carbon atoms (X) and the number of hydrogen atoms (Y) in a given unsaturated cyclo system containing triple bond. The formula to calculate the number of triple bond is **T**_{c}**= ****[{(2X-Y)/2}]/2****.**

Where, X = number of carbon atoms; Y = number of hydrogen atoms and T_{c} = number of triple bond.

E.g.: In Cycloheptyne (C_{7}H_{10}), X =7, Y = 10, therefore, T_{c} = [{(2X-Y)/2}]/2 = [{(2 x 7 – 10)/2}]/2 = 2/2 = 1 triple bond.

### 9. Conclusions

It may be expected that these innovative methods would go a long way to help to the students of chemistry at Undergraduate, Senior Undergraduate and Post-Graduate level who would choose the subject as their career. Experiment *in vitro* on 150 students at Undergraduate, Senior Undergraduate and Post-Graduate level showed that by using these new innovative methods students can save up to 30-40 mins time in the examination hall. On the basis of this, we can strongly recommend to use these new time economic interesting pedagogies.

### Acknowledgement

The corresponding author Dr. Arijit Das, wants to dedicate this review article to his student Late Uddipan Debborma (17yrs), who unfortunately breathed his last on 9^{th} June 2014 due to heart attack.

The corresponding author Dr. Arijit Das, would be is grateful to the SERB, DST, New Delhi, Govt. of India, for their financial assistance (Sanction no – SERB / F / 5537 / 2013-14 dated 27/11/2013 and D.O. No. SB / EMEQ - 014 / 2013).

The corresponding author would be also grateful subsequently to Prof. R. N. Mukherjee, Director, IISER, Kolkata, Prof. P. K. Chattaraj, Convenor, centre for Theoretical studies, Deptt. of Chemistry, IIT Kharagpur, India,; Prof. Samar Kumar Das, School of Chemistry, University of Hyderabad, Prof. Partha Sarathi Mukherjee, Dept. of Chemistry, Indian Institute of Science, Bangalore, Prof. A. T. Khan, Head, IIT Patna, Dr.Satish Nimse, Dept. of Chemistry, Hyllym University, South Korea, Prof. G.N.Mukherjee, Dept. of Chemistry, Calcutta University, Prof. A.K.Das, Ex Vice-Chancellor of Kalyani University, Prof. Nilashis Nandi, Dept. of Chemistry, Kalyani University, W.B., India, Prof.Md.Ali, Deptt. of Chemistry, Jadavpur University, Prof. R. A. Lal, Head, Dept. of Chemistry, NEHU, Shillong, Prof.M.K.Singh and Prof.R.K.Nath, Deptt. of Chemistry, Tripura Central University for their appreciation and recognition in this regard.

Further, the corresponding author, Dr. Arijit Das, give his cordial thanks to Prof.(Dr.) Debabrata Goswami, Principal, Ramthakur College, Agartala, Tripura(w), Tripura, India, for giving him this opportunity to carry out the research work in the college. He also give his cordial thanks to other faculty members of different departments in the college for providing him valuable supporting mentality. Finally, Dr.Das would be grateful to the Honble Chief Minister of the State,Mr. Manik Sarkar and Honble Ex-Higher Education Minister, Mr. Anil Sarkar for their felicitation with ‘Certificate of Honour’ on behalf of the State ‘Tripura’, Mr.Tapan Chakraborty, Education & Higher Education Minister, Govt. of Tripura,Mr. Sekhar Datta, Chairman, tripurainfo.com and Senior Correspondent, ‘The Telegraph’ newspaper, Mr. Goutam Das, Editor, ‘Daily Desher Katha’ news Paper, Mr.Bijan Dhar, CPIM Party Secretary of Tripura, and Mr.Jayanta Debnath, MD, Tripurainfo.com for providing their valuable sustaining mentality in this regard.

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