Chemistry Test - 22

The reaction given is a Sandmeyer's Reaction.

• The Sandmeyer reaction is used to prepare aryl halides from aryl diazonium salts. 
• It is named after the Swiss chemist Traugott Sandmeyer.
• The reaction is a method for substitution of an aromatic amino group via preparation of its diazonium salt followed by its displacement with a nucleophile, catalyzed by copper (I) salts. 

The reaction takes place as follows:

Chlorine atom in \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) is attached to \(\mathrm{sp}^{3}\) carbon.

The structure of \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) is as follows:

As from the structure, it can be seen that chlorine atom in \(\mathrm{CH}_{3} \mathrm{CH}_{2} \mathrm{Cl}\) is attached to a carbon with \(4 \sigma\) bond. 

Thus, it is attached to the \(\mathrm{sp}^{3}\) carbon atom.

Trigonal bipyramidal geometry is shown by \(XeO _{3} F _{2}\).

Trigonal bipyramidal geometry is seen in complexes that have hybridization of sp \(^{3} d\) and their coordination number is \(5\).

In the molecule \(XeO _{3} F _{2}\), the steric number of \(Xe\) is \(5\) as there are \(5\) sigma bonds and \(0\) lone pairs. Therefore \(5\) steric number implies that the hybridization is \(sp ^{3} d\) and therefore the geometry is trigonal bipyramidal.

Valence electrons in Xenon \(=8\)

Contribution of each \(F\) atom \(=2\)

\(\therefore\) Coordination number \(=(8+2) \div 2=5\)

\(2 \mathrm{C}(\mathrm{s})+\mathrm{O}_2(\mathrm{~g}) \rightarrow 2 \mathrm{CO}(\mathrm{g})\)

\(1 \mathrm{~mol} 1.5 \mathrm{~mol}\)

Limiting reagent is carbon. One mole carbon produces one mole CO. Hence, volume at STP is \(227 \times 10^{-1}\) litre

Freezing point of diluted milk \(=-0.2^{\circ} \mathrm{C}\)

\(\Delta \mathrm{T}_{\mathrm{f}}^{\prime}=0.2^{\circ} \mathrm{C}\)

Freezing point of pure milk \(=-0.5^{\circ} \mathrm{C}\)

\(\begin{aligned}& \Delta \mathrm{T}_{\mathrm{f}}=0.5^{\circ} \mathrm{C} \\& \frac{\Delta \mathrm{T}_{\mathrm{f}}}{\Delta \mathrm{T}_{\mathrm{f}}}=\frac{\mathrm{K}_{\mathrm{f}} \times \mathrm{m}}{\mathrm{K}_{\mathrm{f}} \times \mathrm{m}} ; \mathrm{m}=\frac{\text { moleof solute }}{\text { mass of solvent }(\mathrm{kg})}\end{aligned}\)

Moles of solute are same in both samples.

\(\begin{aligned}& \therefore \frac{0.5}{0.2}=\frac{W}{W} \\& \frac{W^{\prime}}{W}=\frac{5}{2} ; W^{\prime}=\frac{5}{2} W\end{aligned}\)

2 cups of pure milk is mixed with 3 cups of water to make 5 cups of diluted milk.

We know, \(\pi=1 \mathrm{iCRT} ; \pi_{\mathrm{xy}}=4 \pi_{\mathrm{BaCl}_2}\)

Since both are ionic solute, i.e.,

\(\mathrm{XY} \rightarrow \mathrm{X}^{+}+\mathrm{Y}^{-} \Rightarrow \mathrm{i}=2 \)

\( \mathrm{BaCl}_2 \rightarrow \mathrm{B}^2 \mathrm{a}^{+}+2 \mathrm{Cl}^{-} \Rightarrow \mathrm{i}=3\)

NOW,

\(2[\mathrm{XY}]=4 \times 3 \times[0.01] \)

\( {[\mathrm{XY}]=0.06} \)

\( =6 \times 10^{-2} \mathrm{~mol} / \mathrm{L}\)

Separating the oxidation and reduction reaction from the redox reaction:

\(3 \mathrm{~N}_{2} \mathrm{H}_{4}+2 \mathrm{BrO}_{3}^{-} \rightarrow 3 \mathrm{~N}_{2}+2 \mathrm{Br}^{-}+6 \mathrm{H}_{2} \mathrm{O}\)

Assigning the oxidation number on central atom (N and Br) in each molecules by considering oxidation number of \(\mathrm{H}=+1, \mathrm{O}=-2\),we get oxidation state as:

\(3\overset {-2}{N}_2H_4 + 2[\overset{+5}{Br}O_3]^- \rightarrow 3\overset{0}{N_2} + 2\overset{-1}{Br^-} +6H_2O\)

As the oxidation number of \(N\) changes from \(-2\) to 0 as:

\(\mathrm{N}_{2} \mathrm{H}_{4} \rightarrow N_{2}\)

Its a oxidation reaction where \(N_{2} H_{4}\) gets oxidized.

Similarly, as the oxidation number of \(B r\) changes from \(+5\) to \(-1\) as:

\(\mathrm{BrO}_{3}^{-} \rightarrow B r^{-}\)

Its a reduction reaction where \(\mathrm{BrO}_{3}^{-}\) gets reduced.

Given: Weight of Urea \(=12 {~g}\)

And, weight of sucrose \(=68.4 {~g}\)

We know that: Number of moles \(=\frac{\text{Weight}}{\text{Molar mass}}\)

Therefore, Moles of urea \(=\frac{12}{60}=0.2\)

(Molar mass of Urea is \(60 {gm}\). )

Moles of sucrose \(=\frac{68.4}{342}=0.2\)

(Molar mass of sucrose is \(342 {gm}\))

Since, there are equal moles of solute in equal volumes of water, the mole fraction is the same. As per Raoult's law, the relative lowering of pressure will also be the same.

Given,

Frequency, \(\nu=5 \times 10^{14}\) \(\mathrm{Hz}\)

As we know, Planck constant is given as,

\(h=6.626 \times 10^{-34} \mathrm{Js}\)

Energy of one photon is given by theexpression,

\(E=h \nu\)

\(E=\left(6.626 \times 10^{-34} \mathrm{~J} \mathrm{~s}\right) \times\left(5 \times 10^{14} \mathrm{~s}^{-1}\right)\)

\(=3.313 \times 10^{-19} \mathrm{~J}\)

As we know,

Avogadro's number \(= 6.022 × 10^{23} \mathrm{~mol}^{-1}\)

Energy of one mole of photons,

\(=\left(3.313 \times 10^{-19} \mathrm{~J}\right) \times\left(6.022 \times 10^{23} \mathrm{~mol}^{-1}\right)\)

\(=199.51 \mathrm{~kJ} \mathrm{~mol}^{-1}\)

The carbon atom is attached with one hydrogen atom and 3 chlorine atom.

It can be seen from structure that chloroform contains 4 single bonds. One single bond contains one sigma.  Therefore, it contains 4 sigma bond.