6. Explain in detail why the ground state electronic configuration does not expl

Question

6. Explain in detail why the ground state electronic configuration does not explain the bonding in methane, ethene (or ethylene) and ehyne (or acetylene)!!

7. For the reactionHCN+KClHCl+KCN , answer the following questions:

a. Complete the above reaction by placingthe “reaction” arrow in the appropriate direction. To earn full credit, you must fully explain your answer.

b. Based on the direction of the arrow in a., label the acid, base, conjugate acid, and conjugate base.

c. Based on the direction of the arrow in 7a., write the equilibrium constant expression, K.

Explanation / Answer

Ans: 6)

Atomic number of carbon is 6,

Ground state electronic configuration of 6C : 1S2 2s2 2Px1 2py1 2pz0.

As we can see there are 2 unpaired electrons present in valence shell of C , expected valency of C in its compounds is 2 and could form 2 bonds with neighbouring atom(s)

But in methane, ethene and ethyne valency of C is 4 i.e. it forms 4 bonds with neighbouring atoms.

CH4, H2C=CH2 and H-CC-H.

Hence Ground state electronic configuration of C do not explain tetravalency of C in its compounds.

Tetravalency of C can be explained on the basis of hybridization approach.

By transition one of the two 2S electron into the 2Pz0 sub-orbital C achieve excited state electronic configuration as,

Excited state electronic configuration of 6C : 1S2 2s1 2Px1 2py1 2pz1.

Then as per need requisite number of orbitals undergo mixing and recasting of energy (what we call as hybridization) and forms exactly same number of Hybrid orbitals.

e.g.

i)In Methane case C is sp3 hybridized. 4 atomic orbitals (AOs) 2s1 2Px1 2py1 2pz1 with 1 unpaired electron each undergo hhybridization and forms 4 sp3 hybrid orbital (H.O.). Each sp3 H.O. forms bond with 1s orbital of H and 4 bonds formed i.e. CH4 forms.

ii) In ethene case 3 AOs 2s1 2Px1 2py1 undergo hybridation called sp2 hybridization and forms 3SP2 H.O. which forms 3 sigma bond with 2H and 1C

and unhbridized 2pz1 orbital of 2 C used in pi bonding.(H2C=CH2)

iii) In Ethyne case 2 AOs 2s1 2Px1 hybridizes to form 2 sp H.O used in sigma bonding with 1H and 1C and 2 unhybridized 2py1 2pz1.used in 2 pi bonding.

Hence the explanation.

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7)

a) HCl is far more stronger acid than HCN and hence arrow will point towards the formation of HCN hence we can write,

KCN + HCl ----------> HCN + KCl

Arrow shifted more towards HCN.

b) KCN and KCl are salts and hence are strong electrolytes moreover HCl acid is also a strong electrolyte and hence they will exist in ionized state in aqueous medium and hence we can write the same reaction as,

K+ (aq.) + CN– (aq.) + H+ (aq.) + Cl– (aq.) -------------> HCN + K+ (aq.) + Cl– (aq.)

Let us cancel out common ions from both sides of reaction, then,

CN– (aq.) + H+ (aq.) <-----------> HCN

In Water as solvent we can write,

                 CN– (aq.) + H3O+ (aq.) <-----------> HCN + H2O

Or,

               HCN + H2O <-----------> CN– (aq.) +                  H3O+ (aq)

        Acid               Base              Conjugate base        conjugate acid.

Arrow more towards HCN unionized form.

c) we have equilibrium expression as

KCN + HCl <----------> HCN + KCl

Expression for equilibrium constant in the specified direction is,

K = [Products]/[Reactants]

K = [KCl][HCN] / [KCN][HCl]

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