NEET Chemistry Solved Question Paper 2014

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

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

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.

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.

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.

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

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

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

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

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

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