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NEET Chemistry Solved Question Paper 2017

Multiple Choice Questions

When 750 mL of 0.5 M HCl is mixed with 250 mL of 2 M NaOH solution, the value of pH will be

pH = 7
pH > 7
pH < 7
pH = 0

pH > 7

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

Which of the following is the correct order of stability for the given superoxides?

KO₂ < RbO₂ < CsO₂
CsO₂ < RbO₂<KO₂
RbO₂
KO₂ < CsO₂ <RbO₂

KO₂ < RbO₂ < CsO₂

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

The most stable carbonium ion among the following is

$C_{6} H_{5} - {CH}_{2} \overset{+}{C} H_{2}$
${CH}_{3} \overset{+}{{CH}_{2}}$
$C_{6} H_{5} - \overset{+}{C} - C_{6} H_{5}$
$C_{6} H_{5} \overset{+}{C} H_{2}$

$C_{6} H_{5} - \overset{+}{C} - C_{6} H_{5}$

Due to more possibility for delocalisation in

$C_{6} H_{5} - \overset{+}{C} - C_{6} H_{5} (of π - electrons), C_{6} H_{5}_\overset{+}{C} - C_{6} H_{5} is the most stable ion .$

Among the following compounds, which compound is polar as well as exhibit sp²-hybridisation by the central atom.

H₂CO₃
SiF₄
BF₄
HClO₃

H₂CO₃

SiF₄ and HClO₃ have sp³ - hybridisation.

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

Therefore, the Correct answer is H₂CO₃

Match the type of series given in Column I with the wavelength range given in Column II and choose the correct option.

Column I		Column II
A	Lyman	1.	Ultraviolet
B.	Paschen	2.	Near-infrared
C.	Balmer	3.	Far Infrared
D.	Pfund	4.	Visible

A B C D

1 2 4 3
A B C D

4 3 1 2
A B C D

3 1 2 4
A B C D

4 3 2 1

A	B	C	D
1	2	4	3

The correct match is

A → Lyman ultraviolet → (1) Ultraviolet

B → Paschen → (2) → Near Infrared

C → Balmer → (4) → Visible

D → Pfund → (3) → Far Infrared

Among the following set of quantum numbers, the impossible set is

n l m s

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

4 0 0 1/2
n l m s

5 3 0 -1/2
n l m s

3 2 -2 1/2

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

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

The energy released when 6 moles of octane is burnt in air will be [Given, ΔH_f for CO₂ (g). H₂O(g) and C₈H₁₈ (l), respectively are -490, -240 and +160J/mol]

-37.4 kJ
-20 kJ
-6.2 kJ
-35.5 kJ

-35.5 kJ

$C + O_{2} \overset{}{\to} {CO}_{2}; ∆ G = - 490 J . . . (i) H_{2} + \frac{1}{2} O_{2} \overset{}{\to} H_{2} O; ∆ H = - 240 J . . . (ii) 8 C + 9 H_{2} \overset{}{\to} C_{8} H_{18}; ∆ H = + 160 J . . . . (iii)$

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

$C_{8} H_{18} + \frac{25}{2} O_{2} \overset{}{\to} 8 {CO}_{2} + 9 H_{2} O ∆ H_{R}^{0} = [8 x (- 490)] + [9 x (- 240)] + 160 = - 5920 J {mol}^{- 1}$

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

The electrons identified by quantum numbers n and l, are as follows

n =4, l=1
n = 4, l = 0
n =3, l= 2
n = 3, l = 1

If we arrange them in order of increasing energy, i.e., from lowest to highest, the correct order is

IV<II<III<I
II<IV<I<III
I < III < II < IV
III < I < IV < II

IV<II<III<I

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 compressibility of a gas is less than unity at STP, therefore,

V_m > 22.4 L
V_m < 22.4 L
V_m = 22.4
V_m = 44.8 L

V_m < 22.4 L

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