Chemistry Test - 50

Physical adsorption decreases with increase in temperature.

Physical adsorption is non-directional, reversible, multilayers exothermic process where adsorbate molecules are held on surface of adsorbent by physical forces such as van der Waals forces.

Van der Waals interactions cause these surface atoms to be reactive, causing them to attract liquids, vapors and gases in order to satisfy their atomic force imbalance. The attracted molecules fill in the pores on the surface of the solid when adsorption occurs.

Temporary hardness may be removed from water by adding\(Ca ( OH )_{2}\).

The temporary hardness of water is due to the presence of bicarbonates of calcium and magnesium.The temporary hardness of water can be removed by the addition of a calculated quantity of milk of lime which converts soluble bicarbonates into insoluble carbonates which can be removed.

\(Ca \left( HCO _{3}\right)_{2}+ Ca ( OH )_{2} \rightarrow 2 CaCO _{3} \downarrow+2 H _{2} O\)

Reaction involved-

\(\mathrm{NH} 3+\mathrm{H} 2 \mathrm{O} \rightarrow \mathrm{NH} 4++\mathrm{OH}\)

Given- \(-\) \(\mathrm{NH} 4+=\mathrm{OH}-=1.33 \times 10^{-3} \mathrm{M}\)

\(\mathrm{NH} 3\) reacted \(=\left(0.1-1.33 \times 10^{-3}\right) \mathrm{M}\)

Condition at equilibrium. \(=\left(0.1-1.33 \times 10^{-3}\right) \mathrm{M}\)

\(=0.1-0.00133 \mathrm{M}\)

\(=0.09867 \mathrm{M}\)

\(\mathrm{Kb}\)

\(=\frac{[\mathrm{NH}_{4}^{+}][\mathrm{OH}^{-}]}{[\mathrm{NH} 3]}\)

\(=\frac{(1.33 \times 10^{-3} \times 1.33 \times 10^{-3})} { 0.09867}\)

\(=1.79 \times 10^{-5}\)

The ionic radii of elements exhibit the same trend as the atomic radii. The atomic size generally decreases across a period for the elements of the second period. It is because, within the period, the outer electrons are in the same valence shell and the effective nuclear charge increases as the atomic number increases resulting in the increased attraction of electrons to the nucleus. The size of an anion will be larger than that of the parent atom because the addition of one or more electrons would result in increased repulsion among the electrons and a decrease in effective nuclear charge.

In given question option \(\mathrm{A}\) and \(\mathrm{C}\) are cations in the second period. So they will have less ionic radius than that of the anions o same period. oxygen falls left to the Fluorine and also has one extra negative charge \(\left(\mathrm{O}_2^{-2}\right)\) than Fluorine ion \(\left(\mathrm{F}^{-}\right)\)

Gas, \(\mathrm{X}\) tums acidified dichromate \(\left(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\right)\) green that means it is a reducing agent.

\(\mathrm{Cr}(\mathrm{VI})\) from compound \(\mathrm{K}_{22} \mathrm{Cr}_2 \mathrm{O}_7\) on reduction changes its colour from orange to green which is the colour of \(\mathrm{Cr}\) (III) compound \(\mathrm{Y}\). \(\mathrm{SO}_2\) can show both reducing and oxidising properties whereas \(\mathrm{SO}_3\) cannot show reducing property because of highest group number 16 , oxidation state (+6) of sulphur in \(\mathrm{SO}_3\).

So, \(\mathrm{X}\) is \(\mathrm{SO}_2\).

As, \(\mathrm{X}\) reacts with \(\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7\) in dil. \(\mathrm{H}_2 \mathrm{SO}_4\) medium, the green coloured \(\mathrm{Cr}\) (III) compound, \(\mathrm{Y}\) must be \(\mathrm{Cr}\) (III) sulphate or \(\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3\).

So, \(\mathrm{Y}\) is \(\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3\).

The overall reactions can be shown as :

\(\underset{\text { Water soluble } \mathrm{SO}_3^{-} \text {salt }}{(\text { Compound) }} \xrightarrow[\mathrm{H}_2 \mathrm{SO}_4]{\text { Dii. }} \underset{(\mathrm{X})}{\mathrm{SO}_2(\mathrm{~g})} \xrightarrow[\text { dil. } \mathrm{H}_2 \mathrm{SO}_4]{\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7} \underset{\substack{(\mathrm{Y}) \\ \text { Green }}}{\mathrm{Cr}_2\left(\mathrm{SO}_4\right)_3}\)

(i) \(\mathrm{SO}_3{ }^{2-}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \mathrm{SO}_2 \uparrow+\mathrm{H}_2 \mathrm{O}+\mathrm{SO}_4{ }^{2-}\)

(ii) \(\underset{\text{(Orange)}}{\mathrm{K}_2 \stackrel{+6}{\mathrm{Cr}_2} \mathrm{O}_7}+\stackrel{+4}{\mathrm{SO}_2}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow \underset{\substack{(\mathrm{Y}) \\ \text { Green }}}{\stackrel{+3}{\mathrm{Cr}_2}\left(\mathrm{SO}_4\right)_3}+\mathrm{K}_2 \stackrel{+6}{\mathrm{SO}_4}+\mathrm{H}_2 \mathrm{O}\)

Methyl isocyanide on reduction using \(\mathrm{LiAlH}_{4}\) gives dimethyl amine.

\(\mathrm{CH}_{3}-\mathrm{N}^{+} \equiv \mathrm{C}^{-} \stackrel{\mathrm{LiAlH}_{4}}{\longrightarrow} \mathrm{CH}_{3}-\mathrm{NH}-\mathrm{CH}_{3}\)

Methyl isocyanide \(\quad \quad \quad \quad \)Di methyl amine

Two hydrogen molecules are added across \(\mathrm{N} \equiv \mathrm{C}\) triple bond.

Chlorotoluenes can be obtained by passing dry chlorine through toluene using iron or ferric chloride as catalyst. The substitution products are ortho and para as methyl is an Electron releasing group.

The temperature of the reaction is about 310 -320K.

\({I}_{2}\) or \({AlCl}_{3}\) can also be used as a halogen carrier. The function of \({FeCl}_{3}\) is to generate \({Cl}^{+}\)which then attacks the benzene ring to form an intermediate carbocation.

The carbocation then loses a proton to form Chlorotoluene. Formation of the intermediate carbocation is the slow rate-determining step and the halogenation of toluene is an example of electrophilic substitution reaction. The reaction should be carried in absence of light to prevent the formation of addition products.

The products formed are ortho and para chloro toluenes.

K₂[Hgl₄](potassium mercuric iodide) also known as Nessler's reagent detects NH₄⁺ion.Here potassium reacts with ammonia (NH₃) in basic media of KOH to make a complex which is known as Nessler's reagent.

The reaction taking place is:

\(2 \mathrm{~K}_{2}\left[\mathrm{HgI}_{4}\right]+3 \mathrm{KOH}+\mathrm{NH}_{3} \rightarrow\left[\mathrm{OHg}_{2} \cdot \mathrm{NH}_{2}\right] \mathrm{I}+7 \mathrm{KI}+2 \mathrm{H}_{2} \mathrm{O}\)

The colour of nesseler's reagent changes to brown in basic medium.

According to the Aufbau principle, electron enters into the lowest energy level (orbital), before entering into a higher energy level. Hence, according to this principle, the electronic configuration of \(\mathrm{Ca}^{20}\) is \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 4 \mathrm{~s}^{2}\), as differentiating electron enters into the s-orbital,\(\mathrm{Ca}\) belongs to s-block.

If we don't follow the Aufbau rule, the electronic configuration of \(\mathrm{Ca}^{20}\) would be \(1 \mathrm{~s}^{2} 2 \mathrm{~s}^{2} 2 \mathrm{p}^{6} 3 \mathrm{~s}^{2} 3 \mathrm{p}^{6} 3 \mathrm{~d}^{2}\).

So, the differentiating electron is entering in the d orbital, so it would belong to the d-block element.

Given,

Wavelength of violet light \(=400 \mathrm{~nm} =400 \times 10^{-9} \mathrm{~m} \)

Wavelength of red light \(=750 \mathrm{~nm}=750 \times 10^{-9} \mathrm{~m}\)

As we know,

The speed of light,

\(c= 3 \times 10^8\) m/s

Frequency of violet light is given as,

\(\nu=\frac{c}{\lambda}\)

\(=\frac{3.00 \times 10^{8}}{400 \times 10^{-9}}\)

\(=7.50 \times 10^{14}\) Hz

Frequency of red light is given as,

\(\nu=\frac{c}{\lambda}\)

\(=\frac{3.00 \times 10^{8}}{750 \times 10^{-9}}\)

\(=4 \times 10^{14}\) Hz