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NEET Chemistry Solved Question Paper 2017

Multiple Choice Questions

1.

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

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

3. n =3, l= 2

4. 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

A.

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)

2.

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.

 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

3.

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

• H2CO3

• SiF4

• BF4

• HClO3

A.

H2CO3

SiF4 and HClO3 have sp3 - hybridisation.

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

Therefore, the Correct answer is H2CO3

4.

Which of the following is least soluble in water?

• C2H6

• CH3OH

• CH3NH2

• C6H5OH

A.

C2H6

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

5.

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

B.

pH > 7

nHCl = MHCl x VHCl and nNaOH = MNaOH x VNaOH

where n = number of moles

M = molarity, V = volume

∴ nHCl  = 0.5 x 750 = 375 milimoles

nNaOH = 2.0 x 250 = 500 milimoles

∵ nNaOH > nHCl

The solution is basic and has pH > 7

6.

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

A.

 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

7.

The most stable carbonium ion among the following is

• ${\mathrm{CH}}_{3}\stackrel{+}{{\mathrm{CH}}_{2}}$

• ${\mathrm{C}}_{6}{\mathrm{H}}_{5}\stackrel{+}{\mathrm{C}}{\mathrm{H}}_{2}$

C.

Due to more possibility for delocalisation in

8.

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

• KO2 < RbO2 < CsO2

• CsO2 < RbO2<KO2

• RbO2

• KO2 < CsO2 <RbO2

A.

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

9.

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

• Vm > 22.4 L

• Vm < 22.4 L

• Vm = 22.4

• Vm = 44.8 L

B.

Vm < 22.4 L

10.

The energy released when 6 moles of octane is burnt in air will be [Given, ΔHf 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

D.

-35.5 kJ

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

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