NEET Chemistry Solved Question Paper 2016

Entropy is a measure of randomness of the system, so entropy of substance in different physical states is in the order Gas > Liquid > Solid

$\mathrm{\Delta }$S=S_product -S_reactant

_{In option (c), solid is converted into gaseous state, so $\mathrm{\Delta }$S will be positive.}

O^2- has the highest value of ionic radius and this can be explained on the basis of $\frac{\mathrm{z}}{\mathrm{e}}$ (nuclear charge / number of electrons) ratio. when $\frac{\mathrm{z}}{\mathrm{e}}$ ratio increases the size decreases and when $\frac{\mathrm{z}}{\mathrm{e}}$ ratio decreases , the size increases.

$\frac{\mathrm{z}}{\mathrm{e}} \propto \frac{1}{\mathrm{r}}$

$\stackrel{-}{\mathrm{v}}$ =(wave number)

=R_HZ^{2$\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$}

${\mathrm{\nu }}_1$(He⁺ , z=2)=R_H(2²)$\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$

v₂(${\mathrm{Be}}_3^{+}$ , z=4)=R_H(4²)$\left[\frac{1}{{\mathrm{n}}_1^2}-\frac{1}{{\mathrm{n}}_2^2}\right]$

$\frac{{\mathrm{v}}_2}{{\mathrm{v}}_1}=4$

$\therefore$ v₂=4v₁=4x

K_c = $\frac{\left[\mathrm{A}\right] \left[\mathrm{B}\right]}{\left[\mathrm{AB}\right]}$

Now , if [A] = 2 $\times$ [A] , B should be $\frac{1}{2}$ [B].

So , that K_c remains constant.

Number of milliequivalents of NaOH = 0.1 x 10 = 1

Number of milliequivalents of H₂SO₄
= 10 x 0.05 = 0.5

Milliequivalent of H₂SO₄ will neutralise NaOH so that (1-0.5) = 0.5 milliequivalent of NaOH is left unneutralised.

Therefore, pH of the resulting solution will be greater than seven i.e., pH > 7,

The given values are:

${\mathrm{P}}_{\mathrm{A}}^{\circ }=25\mathrm{mmHg} , \mathrm{Ps}=22\mathrm{mmHg} , \mathrm{m} =?$

Using Raoult's law

$\frac{{\mathrm{P}}_{\mathrm{A}}^{\circ }-{\mathrm{P}}_{\mathrm{S}}}{{\mathrm{P}}_{\mathrm{A}}^{\circ }}={\mathrm{X}}_{\mathrm{B}}$

Substituting the values:

$$\begin{gathered} {\mathrm{X}}_{\mathrm{B}}=\frac{25-20}{25}=0.2 \\ \therefore {\mathrm{X}}_{\mathrm{B}}=1-{\mathrm{X}}_{\mathrm{B}} \end{gathered}$$

=1-0.2=0.8

Number of moles of oxygen in the cylinder

=$\frac{\mathrm{mass} \mathrm{in} \mathrm{gram}}{\mathrm{molecular} \mathrm{mass} \mathrm{in} \mathrm{gram}}$

=$\frac{2.5 \times {10}^4}{32}$ = 781.25

$\therefore$ Number of moles of N₂ = 3 x 781.25

=2343.75

Mass of nitrogen in cylinder 

=6.5625 x 10⁴ g

Hence , the total mass of gas cylinder 

= 2.5 x 10⁴ + 6.56259 x 10⁴

= 9.0625 x 10⁴ g

B₂ : Total electrons = 10

Configuration : $\mathrm{\sigma }1{\mathrm{s}}^2 , \stackrel{*}{\mathrm{\sigma }} 1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2 , \stackrel{*}{\mathrm{\sigma }} 2{\mathrm{s}}^2 , \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^{1 }\approx \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^1$

If Hund's rule is violated, then

$\mathrm{\sigma }1{\mathrm{s}}^2 , \stackrel{*}{\mathrm{\sigma }} 1{\mathrm{s}}^2 , \mathrm{\sigma }2{\mathrm{s}}^2 , \stackrel{*}{\mathrm{\sigma }} 2{\mathrm{s}}^2 , \mathrm{\pi }2{\mathrm{p}}_{\mathrm{x}}^{2 }\approx \mathrm{\pi }2{\mathrm{p}}_{\mathrm{y}}^0$

So, diamagnetic bond order = $\frac{6-4}{2}$ = 1

IF₅ has 5 bond pair and 1 lone pair

The geometry is square pyramidal with sp³d hybridisation

The order of energy of orbitals can be calculated from (n + 1) rule. The lower the value of (n +l) For an orbital, lower is its energy. If two orbitals have same (n+J) value, the orbital with lower value of n has lower energy.

(a) n+l = 4  (b) n+l = 3

(c) n+l = 4   (d) n+l = 4

$\therefore$ The correct set which represents the highest energy level is n = 3 , l = 2 , m = -2 and s = + $\frac{1}{2}$