Chemistry Test - 17

The structure of P₄S₃ is as follows:

Then, the total number of P - S bonds are 6.

Here Y is prop-2-enoic-acid as shown by below given reaction:

The preparation of amines by the reduction of the nitro compound is usually done by the catalytic hydrogenation of metal and acid, it can be used to carry out the reduction and the product formed will be acidic.

Hoffmann bromamide degradation is the reaction in which amines only primary amines are prepared.

The amine formed will have one carbon atom less than the amide.

In this reaction, there is the use of alkali as a strong base that attacks the amide which leads to the deprotonation and generation of an anion. The conversion of a primary amide to a primary amine with one less carbon is accomplished by heating the amide with a mixture of halogen, a strong base and hydrogen.

The mechanism involved is:

Mechanism:-

When Acetamide is treated with bromine and \(\mathrm{NaOH}\), methylamine is produced as potassium bromide, sodium carbonate and a water molecule is produced as a byproduct.

\(\mathrm{CH}_{3} \mathrm{CONH}_{2}+\mathrm{Br}_{2}+4 \mathrm{NaOH} \rightarrow \mathrm{CH}_{3} \mathrm{NH}_{2}+2 \mathrm{KBr}+\mathrm{Na}_{2} \mathrm{CO}_{3}+2 \mathrm{H}_{2} \mathrm{O}\)

The solutions which obey Raoult's law over the entire range of concentration are known as ideal solutions.

In option (D), the intermolecular forces are not the same because of hydrogen bonding between alcohol molecules. Hence, it is not an ideal solution.

The enthalpy of mixing of the pure components to form the solution is zero and the volume of mixing is also zero.

i.e. \(\Delta H _{\text {mix }}=0\) and \(\Delta V _{\text {mix }}=0\)

That means ideal solution is a mixture of solutions which have same intermolecular force of attractions.

In option, (A), (B), (C), the intermolecular forces are the same before and after the mixing.

Given:

\(A g^{+}+e^{-} \longrightarrow A g \quad (1 F\) is equal to \( 1 m o l\) of \(A g)\)

\(C u^{2+}+2 e \longrightarrow C u \quad(2 F\) is equal to \(1\) mol of \(C u)\)

\(F e^{3+}+3 e^{-} \longrightarrow F e \quad(3 F\) is equal to \(1 m o l\) of \(F e)\)

We know that:

Atomic weight of \(Ag = 108~g\)

Atomic weight of \(Cu = 158.8~g\)

Atomic weight of \(Fe = 56~g\)

As 5 Faraday of electricity is passed.

\(\therefore 5 \times 108 g=540 g\) of \(A g\)

\(\frac{5}{2} \times 63.5 g=158.8 g\) of \(C u\)

\(\frac{5}{3} \times 56 g=93.3 {~g}\) of \({Fe}\)

\(\left[ Co \left( NH _{3}\right)_{2} NO _{2}\right] Cl _{2}\)will exhibit linkage isomerism.

The coordination complexes that show linkage isomerism have ambidentate ligands.Ambidentate ligands are those which have two donor atoms, but only one of the donor atoms can bind to the central metal atom at a time.

\(NO_2\) is the ambidentate ligand in this coordination complex. This ligand is the reason the coordination complex shows linkage isomerism.When the central metal atom is bonded with nitrogen, the ligand is\(NO_2\).When the metal atom is bonded with oxygen, the ligand is \(ONO\).

\(\left[ Co \left( NH _{3}\right)_{2} NO _{2}\right] Cl _{2}\) will exhibit linkage isomerism as it contains nitrite ligand which has the same composition differing with the connectivity of the metal to a ligand. It can bind from \(N\) as well as from \(O\) side.

Above equilibrium can explain the formation of given product, hence reaction must proceed by E1 cb mechanism.

I. Starch is a polysaccharide. It is almost insoluble in cold water but relatively more soluble in boiling water. Thus, statement I is correct.

II. Starch solution gives blue colour with \(\mathrm{I}_{2}\) solution in cold. Amylose, a component of starch, gives this blue colour with iodine.Thus statement II is correct.

III. Starch is non-reducing sugar as the aldehyde group is not free.Hence, statement III is incorrect.

IV. Starch cannot form osazone as the aldehyde group is not free. Thus, statement IV is incorrect.

The exponential factor represents the fraction of "reactants" that have approached the activation energy hill and made it over per number of attempts.

The exponential factor reflects the percentage of reactants who have approached and crossed the activation energy hill in the number of tries.