NEET Chemistry Solved Question Paper 2017

Reason:

(i) More be the sum of n + l → more be the energy.

(ii) For the

same sum, the more value of n, have higher energy.

Because, n + l for →

(a) 4 + 1 = 5
(b) 4 + 0 = 4
(c) 3+ 2 = 5
(d) 3+1 = 4

therefore, correct order is (iv)<(ii) < (iii) < (i)

The correct match is

SiF₄ and HClO₃ have sp³ - hybridisation.

BF₃ and H₂CO₃ have sp² - hybridisation but BF₃ is non-polar.

Therefore, the Correct answer is H₂CO₃ 

C₂H₆ is non-polar compound and other compounds are polar in nature. Hence, it is least soluble in water.

n_HCl = M_HCl x V_HCl and n_NaOH = M_NaOH x V_NaOH

where n = number of moles

M = molarity, V = volume

∴ n_HCl  = 0.5 x 750 = 375 milimoles

n_NaOH = 2.0 x 250 = 500 milimoles

∵ n_NaOH > n_HCl

The solution is basic and has pH > 7

Principal quantum number = n
Azimuthal quantum number = l = 0 to (n-1)
Magnetic quantum number = m = -l to +1
Spin quantum number = s = +1/2 or -1/2
Now,

(i) In first option the given values are,

n=3; l=2; m=-3; s=-1/2

So according to Azimuthal quantum number

l = 0 to (n-1) = 0 to 2 = here it is 2 which is permissible

m = -l to +l = -2 to +2 = here it is 3 which is  not permissible

s = +1/2 or -1/2 = here it is -1/2 whch is permissible

(ii) In the second option given values are,

n = 4; l = 0, m = 0; s= 1/2

So according to Azimuthal quantum number

l = 0 to (n-1) = 0 to 3 = here it is 0 which is permissible

m =-l to +l = -3  to +3 = here it is 0 which is permissible

s = +1/2 or -1/2 = here it is 1/2 whch is permissible

(iii) In the third option given values are,

n =5; l = 3; m= 0; s= -1/2

So according to Azimuthal quantum number

l = 0 to (n-1) = 0 to 4 = here it is 3 which is permissible

m = -l to +l = -4 to +4 = here it is 0 which is permissible

s = +1/2 or -1/2 = here it is -1/2 whch is permissible

(iv) it the forth option given values are,

n=3; l =2; m=-2; s = 1/2

So according to Azimuthal quantum number

l = 0 to (n-1) = 0 to 2 = here it is 2 which is permissible

m = -l to +l = -2 to +2 = here it is 2 which is permissible

s = +1/2 or -1/2 = here it is -1/2 whch is permissible

Due to more possibility for delocalisation in 

$$\begin{gathered} {\mathrm{C}}_6{\mathrm{H}}_5 - \stackrel{+}{\mathrm{C}} -{\mathrm{C}}_6{\mathrm{H}}_5 (\mathrm{of} \mathrm{\pi } -\mathrm{electrons}), \\ {\mathrm{C}}_6{\mathrm{H}}_5 \mathrm{\_}\stackrel{+}{\mathrm{C}} -{\mathrm{C}}_6{\mathrm{H}}_5 \mathrm{is} \mathrm{the} \mathrm{most} \mathrm{stable} \mathrm{ion}. \end{gathered}$$

With a progressive increase in the size of alkali metal ions, the stability of superoxides increases because the size of superoxide ion is large and large cation can be stabilised more by large anion. 

Therefore, the order of stability is

KO₂ <RbO₂ < CsO₂

$$\begin{gathered} \mathrm{compressibility} \left(\mathrm{Z}\right) = \frac{\mathrm{PV}}{\mathrm{nRT}} \\ \mathrm{Z} \mathrm{less} \mathrm{than} \mathrm{unity}, \mathrm{therefore} \\ \frac{\mathrm{PV}}{\mathrm{nRT}} <1 \\ \mathrm{or} \mathrm{PV}<\mathrm{nRT} \\ [\mathrm{At} \mathrm{STP}, \mathrm{P} =1 \mathrm{atm}, \mathrm{T} = 273\mathrm{K}, \mathrm{R} =0.0821 \mathrm{L} \mathrm{atm} {\mathrm{K}}^{-1}] \\ \mathrm{i}.\mathrm{e} \\ \mathrm{or} 1 \mathrm{atm} \mathrm{x} \mathrm{V} < 1 \mathrm{x} 0.0821 \mathrm{x} 273 \\ \mathrm{or} \mathrm{V} <\hspace{0.17em}22.4 \mathrm{litres}. \end{gathered}$$

$$\begin{gathered} \mathrm{C} +{\mathrm{O}}_2 \stackrel{ }{\to } {\mathrm{CO}}_2; ∆\mathrm{G} = - 490 \mathrm{J} ...\left(\mathrm{i}\right) \\ {\mathrm{H}}_2 + \frac{1}{2}{\mathrm{O}}_2 \stackrel{ }{\to } {\mathrm{H}}_2\mathrm{O} ; ∆\mathrm{H} =- 240 \mathrm{J} ...\left(\mathrm{ii}\right) \\ 8\mathrm{C} + 9{\mathrm{H}}_2 \stackrel{ }{\to } {\mathrm{C}}_8{\mathrm{H}}_{18}; ∆\mathrm{H} = +160 \mathrm{J} .... \left(\mathrm{iii}\right) \end{gathered}$$

On applying, 8 x Eq. (i)  + 9x Eq. (ii) -Eq. (iii), we get

$$\begin{gathered} {\mathrm{C}}_8{\mathrm{H}}_{18} + \frac{25}{2}{\mathrm{O}}_2 \stackrel{ }{\to } 8{\mathrm{CO}}_2 + 9{\mathrm{H}}_2\mathrm{O} \\ ∆{\mathrm{H}}_{\mathrm{R}}^0 = [8 \mathrm{x} (-490\left)\right] + [9 \mathrm{x} (-240\left)\right] + 160 \\ = -5920 \mathrm{J} {\mathrm{mol}}^{-1} \end{gathered}$$

Hence, energy exchange when 6 moles of octane is burnt in air = - 5920 x 6 = -35520 J = -35.5 kJ