Chemistry Test - 13

Redox reaction involves both oxidation and reduction reactions which are complimentary to each other.

In the reaction given in option (C), there is no change in the oxidation states of any atom. Therefore, it is not redox reaction. This reaction does not involve either oxidation or reduction. The type of this reaction is called as decomposition.

\(\mathrm{CaCO}_{3} \rightarrow \mathrm{CaO}^{+2+4-2}+{\mathrm{CO}_{2}}^{+4-2}\)

The other name for branched chain alkanes is Isoparaffins.

Linear and branched chain alkanes have difference in their physical properties and hence they are given different prefix like n- and iso-respectively.

Here, the\(\mathrm{K}_{\mathrm{b}}\)for\(\mathrm{B}^{-}\)is \(10^{-10}\)

\(\therefore \mathrm{pK}_{\mathrm{b}}=-\log _{10} \mathrm{~K}_{\mathrm{b}}=-\log _{10}\left(10^{-10}\right)=10\)

We know that,\(\mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}+\log [\) salt \(] /[\) acid \(]\)

Here, the concentration of the salt and the acid is same\(\therefore \log _{10} 1=0\)

\(\therefore \mathrm{pOH}=\mathrm{pK}_{\mathrm{b}}=10\) and

\(\mathrm{pH}=14-\mathrm{pOH}\)

\(=14-10\)

\(=4\)

\(\mathrm{P}=\) Can be prepased by Gabriel phthalimide synthesis it should be i-amine

\(\mathrm{Q}=\) React with Hinsberg's reagent and insoluble in \(\mathrm{NaOH}\) it should be \(2^{\circ}\)-amine

\(\mathrm{R}=\) React with \(\mathrm{HNO}_2\) followed by B-Napthol in \(\mathrm{NaOH}\) it give red dye it must be Aromatic Amine

\(\begin{aligned} & \mathrm{Ca}\left(\mathrm{HCO}_3\right)_2 \xrightarrow{\Delta} \mathrm{CaCO}_3 \downarrow+\mathrm{H}_2 \mathrm{O}+\mathrm{CO}_2 \uparrow \\ & \mathrm{Mg}\left(\mathrm{HCO}_3\right)_2 \xrightarrow{\mathrm{A}} \mathrm{Mg}(\mathrm{OH})_2+2 \mathrm{CO}_2 \uparrow\end{aligned}\)

\(\begin{aligned} & \mathrm{CaCO}_3(\mathrm{~s}) \xrightarrow{\Delta} \mathrm{CaO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g}) \\ & \mathrm{MgCO}_3(\mathrm{~s}) \xrightarrow{\Delta} \mathrm{MgO}(\mathrm{s})+\mathrm{CO}_2(\mathrm{~g})\end{aligned}\)

Let the weight of \(\mathrm{CaCO}_3\) be \(\mathrm{xgm}\)

\(\therefore\) weight of \(\mathrm{MgCO}_3=(2.21-\mathrm{x}) \mathrm{gm}\)

Moles of \(\mathrm{CaCO}_3\) decomposed \(=\) moles of \(\mathrm{CaO}\) formed

\(\frac{\mathrm{x}}{100}=\) moles of \(\mathrm{CaO}\) formed

\(\therefore\) weight of \(\mathrm{CaO}\) formed \(=\frac{\mathrm{x}}{100} \times 56\)

Moles of \(\mathrm{MgCO}_3\) decomposed = moles of \(\mathrm{MgO}\) formed

\(\frac{(2.21-\mathrm{x})}{84}=\) moles of \(\mathrm{MgO}\) formed

\(\therefore\) weight of \(\mathrm{MgO}\) formed \(=\frac{2.21-\mathrm{x}}{84} \times 40\)

\(\Rightarrow \frac{2.21-\mathrm{x}}{84} \times 40+\frac{\mathrm{x}}{100} \times 56=1.152\)

\(\therefore \mathrm{x}=1.1886 \mathrm{~g}=\) weight of \(\mathrm{CaCO}_3\)

& weight of \(\mathrm{MgCO}_3=1.0214 \mathrm{~g}\)

We know,

\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{k} \mathrm{p}^{1 / \mathrm{n}}\)

or, \(\log \frac{\mathrm{x}}{\mathrm{m}} =\log \mathrm{k}+\frac{1}{\mathrm{n}} \log \mathrm{p}\)

\(\therefore\) Plot of \(\log \frac{\mathrm{x}}{\mathrm{m}} \mathrm{vs} \cdot \log \mathrm{p}\) is linear

with slope \(=\frac{1}{\mathrm{n}}\) and intercept \(=\log \mathrm{k}\)

Hence, \(\frac{1}{\mathrm{n}}=\tan \theta=\tan 45^{\circ}=1\) i.e., \(\mathrm{n}=1\)

and \(\log \mathrm{k}=0.3010\) or \(\mathrm{k}=\operatorname{antilog}(0.3010)=2\)

At \(\mathrm{p}=0.5 \mathrm{~atm}\), 

\(\frac{\mathrm{x}}{\mathrm{m}}=\mathrm{kp}^{1 / \mathrm{n}}=2 \times(0.5)^{1}=1.0\)

In the mixture of nitrite and nitrate ions, the nitrite ions can be removed by heating with urea and supluric acid.

The reaction takes place as follows:

\(2 \mathrm{NaNO}_{2}+\mathrm{H}_{2} \mathrm{SO}_{4} \longrightarrow \mathrm{Na}_{2} \mathrm{SO}_{4}+2 \mathrm{HNO}_{2}\)

\(2 \mathrm{HNO}_{2}+\mathrm{CO}\left(\mathrm{NH}_{2}\right)_{2} \longrightarrow \mathrm{CO}_{2}+\mathrm{N}_{2}+3 \mathrm{H}_{2} \mathrm{O}\)

In the above reaction, we can see that the products, nitrogen and carbon dioxide are in gaseous form and thus, do not stay in the solution. This reaction is exothermic and thus, will occur spontaneously without the supply of any extra energy.

Given,

The energy of a hydrogen atom in the ground state \(=-13.6 \mathrm{eV}\)

Atomic number of helium \(Z=2\)

\(n=2\)

The energy of \(\mathrm{He}^+\) ion in the first excited state is,

\(\mathrm{E}_{\mathrm{n}}=-13.6\left(\frac{\mathrm{Z}^{2}}{\mathrm{n}^{2}}\right)\)

\(=(-13.6)\left(\frac{4}{4}\right)\)

\(=-13.6 \mathrm{eV}\)

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 \(=(8+2) \div 2=5\)