Chemistry Test - 32

Given reaction is:

\(\mathrm{Cl}_{2}+\mathrm{OH}^{-} \rightarrow \mathrm{Cl}^{-}+\mathrm{ClO}_{3}+\mathrm{H}_{2} O\)

The balanced reaction will be:

\(3 \mathrm{Cl}_{2}+6 \mathrm{OH}^{-} \longrightarrow 5 \mathrm{Cl}^{-}+\mathrm{ClO}_{3}+3 \mathrm{H}_{2} \mathrm{O}\)

Oxidation number of \(C l_{2}=0\)

Oxidation number \(C l\) in \(C l^{-}=-1\)

Oxidation number of\(\mathrm{Cl}\) in \(\mathrm{ClO}_{3}:\)

\(\Rightarrow x+3 \times(-2) =-1\)

\(\Rightarrow x=+5\)

So, \(\mathrm{Cl}_{2}\) is oxidised as well as reduced.

Chromatography is based on physical adsorption.

Chromatography is based on the principle of distributing the components of a mixture of organic compounds between two phases. Now since the process does not involve formation of chemical covalent bond but involves the formation of weak wander Waals fond. Thus, chromatography is physical adsorbtion.

H₂gas is used in the hydrogenation of oils in presence of nickel as a catalyst to give vanaspati.

Hydrogenation is the addition of hydrogen. So, hydrogen gas is used in the hydrogenation of oils in the presence of a Nickel catalyst to give vanaspati. By hydrogenation of oil, the unsaturated oils are converted to saturated vanaspati. Apart from that hydrogenation improves taste, odour, and preservation.

Here, \(\mathrm{K}_{3}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) with dissociation constant \(1.9 \times 10^{17}\) is more stable than \(\mathrm{K}_{4}\left[\mathrm{Fe}(\mathrm{CN})_{6}\right]\) with dissociation constant \(2.6 \times 10^{37}\).

Higher is the dissociation constant lesser will be the stability of complex. So\(\mathrm{B}\) is more stable than \(\mathrm{A}\).

\(-\mathrm{Cl}\) and \(-\mathrm{CH}_3\) groups are o and \(\mathrm{p}\) directing. They are electron releasing due to \(+\mathrm{M}\) and hyperconjugation effects. Further since such groups increase electron density in the nucleus, they facilitate further electrophilic substitutionand hence known as activating group. The activating effect of these groups is in order of \(-\mathrm{CH}_3>-\mathrm{X}\) but chlorine exceptionally deactive the ring due to strong - I effect. Hence, it is difficult to carry out substitution in chlorobenzene thanin benzene. Further \(-\mathrm{NO}_2\) is a deactivating group, hence deactivates the benzene nucleus, i.e. hinder the further substitution. Thus nitrobenzene undergo electrophilic substitution with a great difficulty, hence the correct order will be

'Arylamines are generally less basic than alkylamines due to delocalisation of lone pair of electrons in the benzene ring.'is true for the basicity of amines.

The basicity of amines varies by molecule and largely depends on: the presence of the lone pair of nitrogen electrons.The electronic properties of the substituent groups attached (e.g., alkyl groups increase the basicity, aryl groups decrease it, etc.) the degree of solvation of the protonated amine, which mostly depends on the solvent used in the reaction.

\(\mathrm{XeF}_6+\mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Partial }} \mathrm{XeOF}_4+2 \mathrm{HF} \)

\(\mathrm{XeF}_6+2 \mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Partial }} \mathrm{XeO}_2 \mathrm{~F}_2+4 \mathrm{HF} \)

\(\mathrm{XeF}_6+3 \mathrm{H}_2 \mathrm{O} \xrightarrow{\text { Complete }} \mathrm{XeO}_3+6 \mathrm{HF}\)

Oxygen gas can be prepared from solid KMnO₄

250°C

2KMnO₄→ KMnO₄+ MnO₂+ O₂

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 violetlight 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

Here, (B), (C) and (D) are p-block elements of the 2 nd period.

(D) \(\rightarrow p^1\) configuration \(\rightarrow B\) of group 13

(C) \(\rightarrow \mathrm{p}^3\) configuration \(\rightarrow \mathrm{N}\) of group 15

(B) \(\rightarrow p^4\) configuration \(\rightarrow 0\) of group 16

We know, ionisation enthalpy \(\left(\Delta_i H\right) \propto\) stability of the subshell concerned.

Therefore, half-filled subshell is more stable than partially filled.

\(\therefore \Delta_i \mathrm{H}\) order is \(\mathrm{c}>\mathrm{b}>>\mathrm{d}\)

(A) is a s-block element (group 2) of 2 nd period with \(s^2\) -

configuration \(\rightarrow\) Be of group 2 [fully-filled; stable]

So, the correct order of \(\mathrm{IE}\), or \(\Delta_{\mathrm{i}} \mathrm{H}_1\) (in \(\mathrm{kJ} \mathrm{mol}^{-1}\) ) of the \(2 \mathrm{nd}\) period elements will be

The correct matching is (A)-(ii), (B)-(iii), (C)-(iv), (D)-(i)

Note The order of IE, or \(\Delta_1 \mathrm{H}_1\) of \(2 \mathrm{nd}\) period elements is

\(\mathrm{Li}<\mathrm{B}<\mathrm{Be}<\mathrm{C}<\mathrm{O}<\mathrm{N}<\mathrm{F}<\mathrm{Ne}\)