Chemistry Test-11

Since at \(n=1\) the population of electrons is maximum i.e., at the ground state. So, maximum excitation will take place from \(n=1\) to \(n=2\).

Thus, \(\mathrm{n}=2\) is the possible excited state.

Now, we have the formula for the energy of H-atom. \((\mathrm{E_n}) _{H}=-13.6 \frac{\mathrm{z}^{2}}{\mathrm{n}^{2}} \mathrm{eV}\), where \(Z\) = atomic number

\(\mathrm{Z}\) for \(\mathrm{H}\) -atom \(=1\) \(\therefore\left(\mathrm{E}_{n}\right)_{H}=13.6 \times \frac{1}{2^{2}} \mathrm{eV}\) \(=\frac{136}{4} \mathrm{eV}=-3.4 \mathrm{eV}\)

Given,

\(A_{2}(g)+B_{2}(g) \rightleftharpoons 2 A B(g)\)

1 mole of \(A_{2}\) reacts with 1 mole of \(B_{2}\) will produce 2 moles of \(A B\). Given, the value of equilibrium constant is \(K_{c}=4\). The \(K_{c}\) means the value of the compounds are taken in concentrations.

In the reaction given above, there are 2 moles of \(A_{2}\) and 4 moles of \(B_{2}\). Initially, the moles of \(A\) are 2 , moles of \(B\) are 4 , and mole of \(A B\) is 0 . After time \(t\), the moles of \(A\) will be \(2-x\), the moles of \(B\) will be \(4-x\), and the moles of \(A B\) will be \(2 x\). Given the volume is 1 liter of \(A_{2}\) and 3 liters of \(B_{2}\), so the total volume of the system will be \(3+1=4\) liters.

Total volume \(=1+3=4\) litre

\({[\mathrm{AB}]=\frac{2 x}{4}=\frac{x}{2} }\)

\({\left[\mathrm{~B}_{2}\right]=\frac{4-x}{4} }\)

\({\left[\mathrm{~A}_{2}\right]=\frac{2-x}{4} }\)

At equilibrium,

\(K_{c}=\frac{[A B]^{2}}{\left[A_{2}\right]\left[B_{2}\right]}\)

\(K_{\mathbf{c}}=\frac{\left(\frac{x}{2}\right)^{2}}{\left(\frac{4-x}{4}\right)\left(\frac{2-x}{4}\right)}\)

\(\Rightarrow 4=\frac{4 x^{2}}{(2-x)(4-x)}\)

\(\Rightarrow x^{2}=(4-x)(2-x)\)

\(\Rightarrow x^{2}=x^{2}-6 x+8\)

\(\Rightarrow 6 x=8\)

\(\Rightarrow x=\frac{4}{3}\)

\(\Rightarrow x=\frac{32}{24}=1.33\) mole

The concentration of \(A B\) is \(\frac{x}{2}\), putting the value of \(x\)

\([A B]=\frac{2}{3}=0.66 M\)

According to Fajan's rule, \(\mathrm{Li}^{+}\) shows the greatest polarising power. According to Fajan's rules of polarization, the more the size of anion, the more easily will it be polarized, and hence compound will be more non-polar. Less the size of the cation, more will be its polarizing power and hence compound will be more non-polar. Now, among all the ions given to us, the Li+ ion is the smallest in size because it has lost 3 electrons so it becomes very small as compared to other ions (the effective nuclear charge increases to a maximum amount). So, It will have maximum polarizing power. Smaller cations that have higher positive charges will have better polarizing power because the positive charge is distributed in a relatively small area. \(\mathrm{Li}^{+}\) ion has the highest polarising power among the alkali metal ions.

In this reaction:

\(\mathrm{Na}+\mathrm{NH}_{3} \rightarrow \underset{\text { Sodamide }(A)}{\mathrm{NaNH}_{2}}+\mathrm{H}_{2}\)

On reacting sodamide with water.

\(\mathrm{NaNH}_{2}+\mathrm{H}_{2} \mathrm{O} \rightarrow \mathrm{NaOH}+\mathrm{NH}_{3}\)

On reacting sodamide with N2O.

\(\mathrm{NaNH}_{2}+\mathrm{N}_{2} \mathrm{O} \rightarrow \underset{\text { Sodium Azide }(B)}{\mathrm{NaN}_{3}}+\mathrm{H}_{2} \mathrm{O}\)

Sodium Azide decomposes on heating to give Na and N2.

\(2 \mathrm{NaN}_{3} \stackrel{\text { Heat }}{\longrightarrow} 2 \mathrm{Na}+3 \mathrm{~N}_{2}(\) GasX \()\)

All monosaccharides whether aldoses or ketoses are reducing sugars. Disaccharides such as sucrose in which the two monosaccharide units are linked through their reducing centres i.e., aldehydic or ketonic groups are non-reducing. This is why sucrose is non-reducing sugar. The linkage between the glucose and fructose units in sucrose, which involves aldehyde and ketone groups, is responsible for the inability of sucrose to act as a reducing sugar.

Given,

\(\mathrm{C}+\mathrm{O}_{2} \longrightarrow \mathrm{CO}_{2} ; \quad \Delta \mathrm{G}=-393 \mathrm{~J} \ldots .\) (i)

\(2 \mathrm{Zn}+\mathrm{O}_{2} \longrightarrow 2 \mathrm{ZnO} ; \quad \Delta \mathrm{G}=-412 \mathrm{~J} \ldots .\) (ii)

By substracting equation (i) from equation (ii), we get

\(2 \mathrm{ZnO}+\mathrm{C} \rightarrow 2 \mathrm{Zn}+\mathrm{CO}_{2}\)

Then, \(\Delta G=-393+412=19\)

Since \(\Delta G\) is positive, therefore reduction of \(\mathrm{ZnO}\) by carbon is not feasible. So, Carbon cannot reduce \(\mathrm{ZnO}\) to \(\mathrm{Zn}\).

Gibb's free energy change for oxidation of both carbon and zinc is negative which implies that oxidation of carbon and zinc is feasible. Since Gibb's free energy of oxidation of zinc is more than that of carbon, oxidation of zinc is more feasible than carbon.

Polysubstitution is the drawback of the Friedel-Craft alkylation reaction. It is so because activating behaviour of the benzene ring increases with increase in number of alkyl group on the benzene ring. Friedel-Craft alkylation, the alkylated product obtained is more activated then reactant hence undergoes polysubstitution. The reaction is,

Adrenaline is produced under the conditions of stress which stimulate glycogenolysis in the liver of human beings. Adrenaline hormone helps to release fatty acids from fat and glucose from liver glycogen under the condition of stress. So, it is also called 'flight or fight hormone'. 

The element with atomic number 117 will be placed in the 17th group of the periodic table and the element with atomic number 120 will be placed in the 2nd group of the periodic table.

Electronic configuration of these elements are \([R n] 5 f^{14} 6 d^{10} 7 s^{2} 7 p^{5}\) and \([O g] 8 s^{2}\) respectively.

Now looking at the electronic configuration of an element with atomic number 117, we can say that its f, d and s orbitals are completely filled. Only the p orbital is lacking one electron to become stable, thus it is placed in the p block of elements.

As it is not stable, to calculate its group we have to add the number of electrons in p orbital and s orbital with 10, thus the group which it will be placed in is 17.

For the element having atomic number 120, the electronic configuration shows that its 8th s orbital is filled completely; thus, it is placed in the s block of the periodic table. And thus it is placed in the 2nd group of the periodic table.

Thus, the elements with atomic number 117 and 120 are placed in the 17th and 2nd group respectively.

Smog is classified into two types, classical smog and photochemical smog. Classical smog is a mixture containing smoke, fog and sulphur dioxide. It is a reducing mixture. So, it is known as reducing smog. It occurs in a cool, humid climate.

Positive Beilstein shows that halogens may be present. A positive Beilstein’s test for halogens does not always indicate the presence of halogen since some halogen-free compounds viz. urea, thiourea, amides etc. also respond this test. The reason being the fact that these halogen-free compounds form cuprous cyanide which is volatile and decomposes to copper which burns with green flame.

Blood is a colloidal solution containing a negatively charged colloidal particle (Albuminoid). Bleeding can be stopped by the use of alum or \(\mathrm{FeCl}{ }_{3}\) solution. The addition of \(\mathrm{Al}^{3+}\) or \(\mathrm{Fe}^{3+}\) causes coagulation of blood, so bleeding stops.

The polyacrylonitrile is a polymer of acrylonitrile. It is formed from addition reactions of acrylonitrile in the presence of peroxide as catalyst. The monomer is acrylonitrile. Thus, its commercial name is acrilan or orlon.

\(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\) has highest crystal field splitting energy \((\Delta)\). Crystal field splitting energy depends on the charge on the metal ion and the nature of the ligand. The oxidation state of \(\mathrm{Co}\) is \(+3\) in all the complexes so the strength of the ligand will be the deciding factor.

\(\mathrm{I}^{-}<\mathrm{Br}^{-}<\mathrm{S}^{2-}<\mathrm{SCN}^{-}<\mathrm{Cl}^{-}<\mathrm{NO}_{3}^{-}<\mathrm{N}_{3}^{-}<\mathrm{F}^{-}<\mathrm{OH}^{-}<\mathrm{C}_{2} \mathrm{O}_{4}^{2-}<\mathrm{H}_{2} \mathrm{O}<\mathrm{NCS}^{-}<\mathrm{CH}_{3} \mathrm{CN}<\mathrm{py}<\mathrm{NH}\)

\(<\) phen \(<\mathrm{NO}_{2}^{-}<\mathrm{PPh}_{3}<\mathrm{CN}^{-}<\mathrm{CO}<\mathrm{bipy}^{2}\)

As we can see from the spectrochemical series \(\mathrm{CN}^{-}\)is a stronger ligand and will cause more splitting and so crystal field splitting will be highest in \(\mathrm{K}_{3}\left[\mathrm{Co}(\mathrm{CN})_{6}\right]\).

Given,

\(\mathrm{Al}^{3+}(\mathrm{aq})+3 \mathrm{e}^{-} \rightarrow \mathrm{Al}(\mathrm{s}) ; \mathrm{E}^{\circ}=-1.66 \mathrm{~V}\)

\(\mathrm{Cu}^{2+}(\mathrm{aq})+2 \mathrm{e}^{-} \rightarrow \mathrm{Cu}(\mathrm{s}) ; \mathrm{E}^{\circ}=+0.34 \mathrm{~V}\)

The cell will be,

\(\mathrm{Al}(\mathrm{s})\left|\mathrm{Al}^{3+}(\mathrm{aq}) \| \mathrm{Cu}^{2+}(\mathrm{aq})\right| \mathrm{Cu}(\mathrm{s})\)

\(\mathrm{E}_{\text {cell }}^{o}=\mathrm{E}_{\text {cathode }}^{o}-\mathrm{E}_{\text {anode }}^{\circ}\)

\(=\mathrm{E}_{\mathrm{Cu}^{2+} / \mathrm{Cu}}^{\mathrm{o}}-\mathrm{E}_{\mathrm{Al}^{3+} / \mathrm{A}}^{\mathrm{o}}\)

\(=+0.34-(-1.66)\)

\(=+2.00 \mathrm{~V}\)