Chemistry Test - 19

Basicity increases in the presence of \(+1\) groups and decreases in the presence of \(-1\) groups.\(\mathrm{CN}\) is electron withdrawing group (-I group) and \(\mathrm{CH}_{3}, \mathrm{NH}_{2}\) are electron releasing groups \((+1)\).

Thus, the correct order of basicity is (IV)<(I)<(III)<(II).

A solution of acetone in ethanol shows a positive deviation from Raoult's law.

• It isdue to miscibility of these two liquids with a difference of polarity and length of the hydrocarbon chain.
• Positive derivation occurs when vapour pressure of the component is greater than expected value.
• Acetone and ethanol both the components escape easily showing higher vapour pressure than the expected value.

Disproportionation reaction is a redox reaction in which a compound of intermediate oxidation state converts to two different compounds, one of higher and one of lower oxidation states.

(A) \(\stackrel{+4}{\mathrm{CH}_{4}}+2 \stackrel{0}{\mathrm{O}_{2}} \rightarrow \stackrel{+4-2}{\mathrm{CO}}_{2}+2 \mathrm{H}_{2} \mathrm{O}\): Not a disproportionate reaction.

(B) \(\stackrel{4+}{\mathrm{CH}_{4}}\stackrel{0}{+4 \mathrm{Cl}_{2}} \rightarrow \stackrel{+4-1}{\mathrm{CCl}_{4}}+4 \mathrm{HCl}\): Not a disproportionate reaction.

(C) \(\stackrel{0}{2 \mathrm{~F}_{2}}\stackrel{-2}{+2 \mathrm{OH}^{-}} \rightarrow \stackrel{-1}{2 \mathrm{~F^-}}+\stackrel{+2}{\mathrm{O}} \stackrel{-1}{~ \mathrm{F}_{2}} +\mathrm{H}_{2} \mathrm{O}\): Redox reaction but not disproportionate reaction.

(D) \(\stackrel{+4}{2 \mathrm{NO}_{2}}+2 \mathrm{OH}^{-} \rightarrow \stackrel{+3}{\mathrm{NO}_{2}^{-}}+\stackrel{+5}{\mathrm{NO}_{3}^{-}}+\mathrm{H}_{2} \mathrm{O}\): \(\mathrm{N}\) changes from \(+4\) to \(+3\) and \(+5\). Thus its a disproportionate reaction.

Since nitrogen atom have 5 electrons in its outermost shell, so higher electronegative elements can extend its oxidation state up to \(+5\), while in case of taking electrons it cannot go beyond 8 electrons. So, at most it can accept 3 electrons.

Formation of Ethanol:

\({ H _{2} C = CH _{2}}\) (Ethene) \(\underset{ H _{2} O }{\stackrel{ H _{2} SO _{4}}{\longrightarrow}} H _{3} C - CH _{2} OH\) (Ethanol)

\(\underset{\text { Glucose (Sugar) }}{ C _{6} H _{1} 2 O _{6}} \stackrel{\text { Zymase enzyme }}{\longrightarrow} \underset{\text { Ethanol }}{ C _{2} H _{5} OH }\)

So, Ethanol can be formed from Ethene and Fermentation of sugar.

Chemical kinetics is the study of the rates of chemical reactions, factors which are influential in the rates and the explanation of the rates with respect to the reaction mechanisms of chemical processes

The quenching of a reaction can be made by cooling as well as diluting the reaction mixture. The reaction is supposed to be completed if it is kept for a long time or strongly heated. Quenching a reaction is used to deactivate any unreacted reagents. It is also done by adding an antisolvent to induce precipitation, and collecting or removing the solids.

The compound that will react most readily with gaseous bromine has the formula \(\mathrm{C_{3} H_{6}}\).

Unsymmetrical alkenes generally are more reactive than the symmetrical alkenes, alkynes and alkanes. That is why, propene is more reactive than the other given compounds.

Given:

\(Z n^{2+}+2 e^{-} \rightarrow Z n\)

Charge = Current\(\times\) Time

Here,

Current \(=1.40\) ampere

and, Time\(=200\) seconds

We know that:

Faraday \(=\frac{\text { Charge }}{96500}\)

\(=\frac{1.40 \times 200}{96500}\)

\(=2.90 \times 10^{-3} {~mol}\)

\(\therefore Z n\) deposited \(=\frac{2.90 \times 10^{-3}}{2}=1.45 \times 10^{-3} {~mol}\)

Now, mol of \(Z n^{2+}\) initially \(=0.180 \times 0.5\)

Or, Molarity \(\times V=0.09\) mol

Therefore, mol of \(Z n^{2+}\) left after deposition \(=0.09-0.00145=0.08855 {~mol}\)

Molarity \(=\frac{\text { Moles }}{\text { Volume }}\)

\(=\frac{0.08855}{0.5}\)

\(=0.177 {M}\)

Methanol and ethanol are distinguished by heating with iodine and alkali.

Methanol and ethanol are distinguished by a type of test known as Iodoform. When ethanol is warmed with iodine in the presence of NaOH, it forms a yellow colored precipitate but methanol do not react positively to Iodoform test.

Mohr's salt is a compound known to contain two primary cations, namely the ammonium cation (denoted by \(\left.NH _{4}^{+}\right)\)and the ferrous cation (denoted by \(F e ^{2+}\)). Therefore, Mohr's salt can be categorized as a double salt of ammonium sulphate and ferrous sulphate.

All others are complexes.