NEET Chemistry Solved Question Paper 2013

The formula of dichlorotetraqua chromium (lll) chloride is [Cr(H₂O)₄Cl₂] Cl.
On ionisation it generates only one Cl^- ion

        $$\begin{gathered} \left[\mathrm{Cr}\right({\mathrm{H}}_2\mathrm{O}{)}_4{\mathrm{Cl}}_2]\mathrm{Cl} \to [\mathrm{Cr}({\mathrm{H}}_2\mathrm{O}{)}_4{\mathrm{Cl}}_2{]}^{+} + {\mathrm{Cl}}^{-} \\ \mathrm{Intial} 100\times 0.01 0 0 \\ \mathrm{mmol} = 1\mathrm{mmol} \\ \mathrm{After} 0 1\mathrm{mmol}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}\hspace{0.17em} 1\mathrm{mmol} \\ \mathrm{inonisation}\hspace{0.17em} \end{gathered}$$

One mole of Cl^- ions react with on only 1 mole of AgNO₃, molecule to produce 1 mole of AgCl.
.:. 1 mmol or 1 x 10^-3 mole reacts with AgNO₃ to give AgCl

               =$\frac{1 \times 1\times {10}^{-3}}{1}= {10}^{-3} \mathrm{or} 0.001 \mathrm{mol} \mathrm{AgCl}$ 

$$\begin{gathered} \mathrm{If} \mathrm{n} = 1 \\ {\mathrm{E}}_1 = -2. 178 \times {10}^{-18 }{\mathrm{Z}}^2\mathrm{J} \\ \mathrm{If} \mathrm{n} = 6 \\ {\mathrm{E}}_{6 }= \frac{- 2.178 \times {10}^{-18} {\mathrm{Z}}^2}{36}\mathrm{J} \\ = 6.05 \times {10}^{-20} {\mathrm{Z}}^2 \mathrm{J} \end{gathered}$$

From the above calculation, it is obvious that electron has a more negative energy than it does for n=6. It means that electron is more strongly bound in the smallest allowed orbit.

Electron rich species are called Lewis base. Among the given, BF₃ is an electron deficient species, so have a capacity of electron accepting instead of donating that's why it is least likely to act as a Lewis base. It is a Lewis acid.

Easily liquifiable gases like NH₃, SO₂ etc. exhibit maximum deviation from ideal gas as for them Z <<<1.
CH₄ also exhibits deviation but it is less as compared to NH₃.

The orbital of the electron having n= 3, l=1 and m = -1 is 3p_z (as nl_m) and an orbital can have a maximum of two electrons with opposite spins.
:. 3p_z orbital contains only two electrons or only 2 electrons are associated with n= 3 , l = 1, m =-1.

$$\begin{gathered} \mathrm{Given}, \mathrm{molairty} \mathrm{of} \mathrm{solution} = 2 \\ \mathrm{Volume} \mathrm{of} \mathrm{solution} = 250 \mathrm{mL} =\frac{250}{1000}=\frac{1}{4}\mathrm{L} \\ \mathrm{Molar} \mathrm{mass} \mathrm{of} \\ {\mathrm{HNO}}_3= 1 +\hspace{0.17em}14+ 3 \times 16 = 63 \mathrm{g} {\mathrm{mol}}^{-1} \\ \mathrm{therefore}, \mathrm{Molarity} \\ = \frac{\mathrm{weight} \mathrm{of} {\mathrm{HNO}}_3}{\mathrm{mass} \mathrm{of} {\mathrm{HNO}}_3\times \mathrm{volume} \mathrm{of} \mathrm{solution} \left(\mathrm{L}\right)} \\ \mathrm{therefore}, \\ \mathrm{weight} \mathrm{of} {\mathrm{HNO}}_3 = \mathrm{molarity} \times \mathrm{mol}. \mathrm{mass} \times \mathrm{volume}\left(\mathrm{L}\right) \\ = 2 \times 63\times \frac{1}{4}\mathrm{g} = 31.5 \mathrm{g} \\ \mathrm{It} \mathrm{is} \mathrm{the} \mathrm{weight} \mathrm{of} 100\% {\mathrm{HNO}}_3. \\ \mathrm{But} \mathrm{the} \mathrm{given} \mathrm{acid} \mathrm{is} 70\% {\mathrm{HNO}}_3 \\ \mathrm{therefore}, \\ \mathrm{Its} \mathrm{weight} = 31.5 \times \frac{100}{70}\mathrm{g} = 45 \mathrm{g} \end{gathered}$$

Paramagnetic species contains unpaired electrons in their molecular orbital electronic configuration.
Molecular orbital configuration of the given species is as

  $$\begin{gathered} \mathrm{CO}(6+8 =14) \\ =\mathrm{\sigma }1{\mathrm{s}}^2,\stackrel{*}{\mathrm{\sigma }}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2,\stackrel{*}{ \mathrm{\sigma }}2{\mathrm{s}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2 \simeq \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2, \mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2 \end{gathered}$$ 

(All the electrons are paired so it is diamagneitc.)

 $$\begin{gathered} O_2^{-}(8+8+1=17) \\ = \mathrm{\sigma }1{\mathrm{s}}^2,\stackrel{*}{ \mathrm{\sigma }}1{\mathrm{s}}^2, \mathrm{\sigma }2{\mathrm{s}}^2,\stackrel{*}{\mathrm{\sigma }}2{\mathrm{s}}^2,\mathrm{\sigma }2{\mathrm{p}}_{\mathrm{z}}^2, \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^2, \approx \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^2,\stackrel{*}{\mathrm{\pi }}2{\mathrm{p}}_{\mathrm{x}}^2\approx \stackrel{*}{\mathrm{\pi }}2{\mathrm{p}}_{\mathrm{y}}^1 \end{gathered}$$

It contains one unpaired electron so it is paramagnetic.
CN^- (6+ 74 1= 14)= same as CO
NO⁺ (7+ 8- 1)= same as CO
Thus, among the given species only ${\mathrm{O}}_2^{-}$ is paramagnetic.

Species having the same number of bond pairs and lone pairs are isostructural (have same structure).

Thus, XeF₂ is isostructural with ICl₂^- and BaCl₂.

Dipole-induced dipole interaction are present in the pair in which the first species is polar and the other is non-polar.H₂O and alcohol both are polar molecules, so there exists dipole-dipole interactions in between them. Cl₂ and CCl₄ both are non-polar so there exists induced dipole-induced dipole interactions in between them. Similar is true for SiCl₄ and He atoms pair. HCl is a polar molecule, whereas He atoms are non-polar, so in between them dipole-induced dipole interactions exist.

Given, Plank's constant: h= 6.63 x 10^-34 Js

Speed of light, c= 3 x 10¹⁷ nms^-1

Frequency of quanta, v= 6 x 10¹⁵s^-1

Wavelength, $\mathrm{\lambda }$ = ?

We know that, $$\begin{gathered} \mathrm{v}=\frac{\mathrm{\lambda }}{\mathrm{c}} \\ \mathrm{or} \mathrm{\lambda }= \frac{\mathrm{c}}{\mathrm{v}} \\ = \frac{3 \times {10}^{17}}{6 \times {10}^{15}} \\ = 0.5 \times {10}^2 \mathrm{nm} \\ = 50\mathrm{nm} \end{gathered}$$