NEET Chemistry Solved Question Paper 2017

AsF₅ has sp³d hybridisation. In sp³d hybridisation, the d_z² orbital is used along with the 's' and three 'p' orbitals to form three equatorial bonds and two equally strong axial bonds for a trigonal bipyramid.

Dissociation constant is the reciprocal of the stability constant $\left(\mathrm{\beta } = \frac{1}{\mathrm{K}}\right)$

Overa;; complex dissociation equilibrium constant,

K = $\frac{1}{{\mathrm{\beta }}_4}$

  = $\frac{1}{2.1 \times {10}^{13}} = 4.76 \times {10}^{-14}$

Among all the given option, option a has maximum energy as the two electrons are in excited state.

The order of increasing energy of the sub-atomic orbitals is s < p < d < f. The energy in excited state is more than that in the ground state.

pH = pK_a + log $\frac{\left[\mathrm{Salt}\right]}{\left[\mathrm{Acid}\right]}$

= pK_a + log $\frac{\left[\mathrm{KCN}\right]}{\left[\mathrm{HCN}\right]}$            ...(i)

Let the volume of KCN solution required be V mL

$\therefore$ [KCN] = $\frac{5 \times \mathrm{V}}{\mathrm{V} + 10}$ and [HCN] = $\frac{10 \times 2}{\mathrm{V} + 10}$

Now from equation (i),

pH = -log (5 $\times$ 10^-10) + log $\left[\frac{5 \times \mathrm{V}}{\mathrm{V} + 10}/ \frac{10 \times 2}{\mathrm{V} + 10}\right]$

9 = -log (5 $\times$ 10^-10) + log $\frac{\mathrm{V}}{4}$

On solving, V = 1.99 $\approx$2mL.

Hydride of boron occurs as B₂H₆ but B₂Cl₆ does not exist because small hydrogen atoms can easily fir in between boron atoms but large atoms such as chlorine do not fit easily.

Acids which contain P-H bonds have strong reducing properties. Hypophosphorus acid (H3PO₂) is a good reducing agent as it contains two P-H bonds. It has one P-OH bond, so it is monobasic.

Since ionisation potential of hydrogen atom is 13.6 eV.

$\therefore$E₁ = -13.6 eV

Now, E_n - E₁ = $\frac{-13.6}{{\mathrm{n}}^2}$-(-13.6) = 12.1

$\frac{-13.6}{{\mathrm{n}}^2}$+13.6 = 12.1

Therefore, n = 3

After absorbing 12.1 eV, the electron of H-atom is excited to 3^rd shell.

Thus, possible transitions are 3 i.e., 3 $\to$ 2, 2 $\to$ 1 and 3 $\to$ 1.

For the given reaction, K = $\frac{[\mathrm{HI}{]}^2}{\left[{\mathrm{H}}_2\right]\left[{\mathrm{I}}_2\right]}$

As 1 mole of H₂ reacts with 1 mole of I₂, even at equilibrium, [H₂] = [I₂]

Hence, K = $\frac{[\mathrm{HI}{]}^2}{[{\mathrm{I}}_2{]}^2}$ or $\sqrt{\mathrm{K} } = \frac{\left[\mathrm{HI}\right]}{\left[{\mathrm{I}}_2\right]} = \sqrt{47.6}$

i.e. [HI] > [I₂]

In the above given figure, option b is incorrect. The slope of the figure is 

Slope = -$\frac{{\mathrm{E}}_{\mathrm{a}}}{\mathrm{R}}$

$\frac{d}{p} = \frac{M}{RT}$

Let density of gas B be d.

$\therefore$ density of gas A = 3d

And let molecular weight of A be M.

$\therefore$ molecular weight of B = 2M

Since, R is a gas constant and T is same for both gases,

p_A = $\frac{{\mathrm{d}}_{\mathrm{A}}\mathrm{RT}}{{\mathrm{M}}_{\mathrm{A}}}$ and p_B = $\frac{{\mathrm{d}}_{\mathrm{B}}\mathrm{RT}}{{\mathrm{M}}_{\mathrm{B}}}$

$\frac{{\mathrm{p}}_{\mathrm{B}}}{{\mathrm{p}}_{\mathrm{A}}} = \frac{{\mathrm{d}}_{\mathrm{B}}}{{\mathrm{d}}_{\mathrm{A}}} \times \frac{{\mathrm{M}}_{\mathrm{A}}}{{\mathrm{M}}_{\mathrm{B}}} = \frac{\mathrm{d}}{3\mathrm{d}} \times \frac{\mathrm{M}}{2\mathrm{M}} = \frac{1}{6}$

Therefore, $\frac{1}{6}$ is the ration of the pressures acting on B and A.