Question on Nitrogen Oxides

Which oxide of nitrogen is produced on heating a mixture of ammonium sulphate and sodium nitrate?

Answer:

It is N₂O.

(NH₄)₂SO₄ + 2NaNO₃ = Na₂SO₄ + 2NH₄NO₃

NH₄NO₃ on heating gives N₂O and H₂O

Question on Nucleophilic Strength

Out of Fluoride ion and Iodide ion which one is a better nucleophile and why?

Answer:

Out of Fluoride and Iodide ions, Iodide ion is less electronegative and has high tendency to release electron density. Moreover it is large and hence can easily be polarized. Also, because of the larger size Iodide ion is soft and prefers to form bonds with soft species like carbocations. So, Iodide ion is a strong nucleophile out of the two.

Question on Chemical Bonding

How many pi-bonds are present in a molecule of tetracyano ethene?

Answer:

There are 9 pi-bonds in a molecule of tetracyano ethene, 8 between C and N and 1 between C and C.

Question on Basic Strength

In the molecule given below which of the three labelled nitrogen atoms will get protonated first when treated with an acid?

Answer:

Nitrogen atom labelled 2 will get protonated first because its conjugate acid will be stabilized by identical resonating structures.

Nitrogen atom labelled 3 is the least basic as the lone pair on that nitrogen atom will be used in making the 5 membered ring aromatic and will be unavailable for an acid.

Question on Acid Base Equilibrium

In the acid – base equilibrium equilibrium NaCN + Na₂HPO₄ = Na₃PO₄ + HCN, if pK_a of Na₂HPO₄ and HCN are 12.3 and 9.3 respectively which side (forward / backward) of the equilibrium is favoured?

Answer:

From the pK_a it is evident that HCN is a stronger acid than Na₂HPO₄. As Na₂HPO₄ is a weaker acid, its conjugate base Na₃PO₄ will be stronger than NaCN (conjugate base of HCN). Both the species on the product side are stronger and hence are relatively unstable than the reactants. So the equilibrium is favoured in the backward direction. 

Question on Acidic Strength

Out of Chloroform and Fluoroform which one is a stronger acid and why?

Answer:

While comparing two compounds for acidic strength, we can check for the stability of their conjugate bases. The one having the most stable (weak) conjugate base is the strong acid.

The conjugate bases of Chloroform and Fluoroform are ^-CCl₃ and ^-CF₃ respectively.

The negative charge in ^-CF₃ is stabilized by the inductive effect of the electronegative fluorine atoms while the negative charge in ^-CCl₃ is stabilized by the resonance effect in which the negative charge is delocalized in the vacant d-orbitals of Cl atoms. F can’t stabilize the negative charge by resonance effect as it does not have d-orbitals. Resonance effect stabilizes charges much better than inductive effect as inductive effect operated through sigma bonds, electron density of which is difficult to perturb, while resonance effect operates through the labile pi bonds and lone pairs. So, ^-CCl₃ is more stable than ^-CF₃ .

Based on the above argument we can conclude that Chloroform is a stronger acid than Fluoroform.

Question on Chemical Equilibrium

K_p for an equilibrium at some temperature is “x” atm. 

(1) K_p = K_c × (RT)^-1                  

(2) K_p = K_c × (RT)                              

(3) K_p = K_c

Out of the three options given above which option shows the correct relation between K_p and K_c at the same temperature?

Answer:

The relation between K_pand K_c is K_p = K_c × (RT)ⁿ where n is the change in number of gas moles in the equilibrium.

K_p for the equilibrium given is “x” atm and the units show that the value of n = 1. So the relation between K_p and K_c is (2) i.e. K_p = K_c  × (RT)

Question on Balancing Chemical Equations

Mn²⁺ + BiO₃^- + H⁺ = Bi³⁺ + MnO₄^- + H₂O

What will the coefficient of H⁺ be when the above equation is balanced with whole number coefficients?

Answer:

In the above reaction Mn gets oxidized from +2 state to +7 state by losing 5 moles of electrons per mole of it. Similarly Bi is getting reduced from +5 state to +3 state by gaining 2 moles of electrons per mole of it. In a redox reaction the number of electrons lost by the reducing agent should be equal to the number of electrons gained by the oxidising agent. So, 2 moles of Mn²⁺ will have to react with 5 moles BiO₃^- in order to have same number of electrons (in this case 10 electrons) lost and gained in this redox reaction. Note that the number of moles of Mn²⁺ & BiO₃^- used are the inverse ratio of the electrons exchanged by them in this reaction. To balance the atoms on the product side we need 5 moles of Bi³⁺ and 2 moles of MnO₄^-. This brings 15 moles and 8 moles of O atoms on the reactant side and product side respectively. To balance the remaining O atoms on the product side we need 7 moles of H₂O. This brings 14 moles of H atoms on the product side and to balance it we need 14 moles of  H⁺ on the reactant side. Finally the balanced equation is

2Mn²⁺ + 5BiO₃^- + 14H⁺ = 5Bi³⁺ + 2MnO₄^- + 7H₂O

Question on pH of a solution

1mL of 0.1M HCl solution is diluted 1000 times. What will be the pH of the diluted solution?

Answer:

Millimoles of HCl taken = Millimoles of H⁺ taken = 1mL × 0.1M = 0.1

Concentration of the diluted HCl solution = (0.1mmol / 1000mL) = 0.0001M

Concentration of H⁺ in diluted solution = 0.0001M

pH = -log [H⁺]

So, pH = -log (0.0001) = 4

Question on Organic Synthesis

C₂H₅MgBr reacts with ethanal followed by acid work-up to form a substance D, which exists as a pair of optical isomers. D is heated with a solution of potassium dichromate in dilute sulphuric acid. What will be the colour of the solution remaining after this reaction?

Answer:

Substance D is butan-2-ol which when heated with potassium dichromate in dilute sulphuric acid gets oxidised to butan-2-one. During this reaction potassium dichromate will reduce to chromic sulphate which is green in colour. So the solution remaining after this reaction will be green in colour.

Question on Quantum Numbers

What is the correct set of quantum numbers for the unpaired electron in a chlorine atom?

Answer:

The atomic number of chlorine is 17. Its electronic configuration is 1s² 2s²2p⁶3s²3p⁵. From the electronic configuration we can see that is has an unpaired electron in a 3p sub-shell. So, the quantum numbers would be n = 3; l = 1; m = + 1; s = +/- ½

Question on Physical States

Why is CO₂ a gas while SiO₂ is a solid at room temperature?

Answer:

C can form multiple bonds with O as the orbitals used of the lateral overlap are of same size and energy, thus satisfying its valancies by forming 2 s bonds and 2 p bonds with O. Si cannot form such multiple bonds with O as the orbitals overlapping laterally will be of different size and energy. So, Si satisfies all its valancies by forming 4 s bonds with O. This results in CO₂ being monomeric and SiO₂ being polymeric and hence CO₂ is a gas and SiO₂ is a solid at room temperature.

Question on Ionization Potential

In spite of having the highest Ionization Potential among the alkali metals, why is Lithium the strongest reducing agent?

Answer:

The hydration energy of Li⁺ is the highest among the alkali metals as it has very high charge density due to very small size. The energy released in the process of hydration is more than the amount of energy spent on ionizing Li.

If the energy spent in ionizing Li to Li⁺ is E₁ and the energy recovered in hydrating Li⁺ to Li⁺_(aq) is E₂, then E₂ > E₁. The overall process of converting Li_(g) to Li⁺_(aq)is thus energetically favoured. Hence Li acts as a strong reducing agent.

Question on Carbocation Stability

Will a tertiary carbocation always be more stable than a primary carbocation?

Answer:

No. A primary carbocation with resonance stabilization will be more stable than a tertiary carbocation without resonance stabilization. Moreover, it also depends on the nature of the groups connected to the carbon atom carrying the positive charge.