NEET Chemistry Solved Question Paper 2014

$$\begin{gathered} \because \mathrm{KE} = \frac{3}{2}\mathrm{RT} \\ \mathrm{KE} \propto \mathrm{T} \end{gathered}$$

Here, the temperature of gas remains constant hence kinetic energy of molecules remains the same.

For n = 6 electron are filled as

6s $\to$ 4f $\to$ 5d $\to$ 6p

This is because electron first enters in that orbitals for which (n + 1) is lower. In case, if (n + 1) is same, it goes first in that orbital for which n is lower.

According to Bohr's concept, an electron always move in the orbit with angular momentum (mvr) equal to $\frac{\mathrm{nh}}{2\mathrm{\pi }}$

$$\begin{gathered} \therefore \mathrm{mvr} = \frac{\mathrm{nh}}{2\mathrm{\pi }} \\ \mathrm{or} \mathrm{r} = \frac{\mathrm{n}}{2\mathrm{\pi }}\times \left(\frac{\mathrm{h}}{\mathrm{mv}}\right) \\ \mathrm{or} \mathrm{r}= \frac{\mathrm{n\lambda }}{2\mathrm{\pi }} (\because \mathrm{\lambda } = \frac{\mathrm{h}}{\mathrm{mv}}) \\ \mathrm{For} \mathrm{fourth} \mathrm{orbit} (\mathrm{n} = 4) \\ \mathrm{r} = \frac{2\mathrm{\lambda }}{4} \\ \therefore \mathrm{Circumference} = 2\mathrm{\pi r} = 2\mathrm{\pi } \times \frac{2\mathrm{\lambda }}{\mathrm{\pi }} = 4\mathrm{\lambda } \end{gathered}$$

N₂(g) + O₂(g) $\rightleftharpoons$2NO(g); K₁  ...(i)

2NO(g) + O₂(g) $\rightleftharpoons$2NO₂(g); K₂  ...(ii)

N₂(g) + 2O₂(g) $\rightleftharpoons$2NO₂(g); K = K₁ $\times$ K₂

$\therefore$ For NO₂(g)  $\rightleftharpoons$$\frac{1}{2}$N₂(g) + O₂(g); K = ${\left[\frac{1}{{\mathrm{K}}_1{\mathrm{K}}_2}\right]}^{\frac{1}{2}}$

Among the given compounds, SO₂ has unsymmetrical or angular structure. Thus, it has the highest dipole moment.

Single bond contains= 1 $\mathrm{\sigma }$ bond

Double bond contains = 1 $\mathrm{\sigma }$ bond + 1 $\mathrm{\pi }$ bonds

Triple bond contains= 1$\mathrm{\sigma }$ bond + 2 $\mathrm{\pi }$ bonds

The molecule of O₃ is bent with an angle 116.8° and equal O - O distance of 128 pm. The ozone molecule consists of 2 sigma and 1 pi bonds. The angular shape of molecule is represented as

Bond angles depend upon number of lone pair (s) of electrons. Higher the number of lone pairs of electrons, lesser the bond angle.

$2{\mathrm{AB}}_2 \left(\mathrm{g}\right) \rightleftharpoons 2\mathrm{AB} \left(\mathrm{g}\right) + {\mathrm{B}}_2 \left(\mathrm{g}\right)$

Initial moles           1             0           0

At equillibrium    2(1-x)         2x          x

where, x = degree of dissociation

Total moles at equilibrium = 2 - 2x + 2x + x

                                      =  (2 + x)

So, ${\mathrm{p}}_{{\mathrm{AB}}_2} = \frac{2(1 - \mathrm{x})\mathrm{p}}{(2 + \mathrm{x})}$

$$\begin{gathered} {\mathrm{p}}_{\mathrm{AB}} = \frac{2\mathrm{xp}}{(2 + \mathrm{x})} \\ {\mathrm{p}}_{{\mathrm{B}}_2} = \frac{\mathrm{xp}}{(2 + \mathrm{x})} \\ {\mathrm{K}}_{\mathrm{p} }= \frac{({\mathrm{p}}_{\mathrm{AB}}{)}^2 ({\mathrm{p}}_{{\mathrm{B}}_2})}{({\mathrm{p}}_{{\mathrm{AB}}_2}{)}^2} \\ = \frac{{\left({\displaystyle \frac{2\mathrm{xp}}{2 + \mathrm{x}}}\right)}^2 \left({\displaystyle \frac{\mathrm{x}}{2 + \mathrm{x}}}\mathrm{p}\right)}{{\left({\displaystyle \frac{2(1 - \mathrm{x})}{2 + \mathrm{x}}}\mathrm{p}\right)}^2} \\ = \frac{{\mathrm{x}}^3\mathrm{p}}{(2 + \mathrm{x})(1 - \mathrm{x}{)}^2} \\ \because \mathrm{x} \ll 1 \mathrm{and} 2, \mathrm{so} (1 - \mathrm{x}) \approx 1, (2 + \mathrm{x})\approx 2 \\ = \frac{{\mathrm{x}}^3\mathrm{p}}{2} \\ \mathrm{x} = {\left(\frac{2{\mathrm{K}}_{\mathrm{p}}}{\mathrm{p}}\right)}^{\frac{1}{3}} \end{gathered}$$

pOH = pK_b + $\log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]}$

        = $-\log {\mathrm{K}}_{\mathrm{b}} + \log \frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Base}\right]}$

        = $-\log 1.8 \times {10}^{-5} + \log \frac{0.20}{0.30}$

        = 5 - 0.25 + (-0.176)

        = 4.75 - 0.176 = 4.57

$\therefore$ pH = 14 - 4.57 = 9.43

Boron in BF₃ is electron poor or electron deficient species. It has an empty orbital, so it can accept a pair of electrons, therefore, making it a lewis acid.