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NEET Chemistry Solved Question Paper 2017

Multiple Choice Questions

A gas is allowed to expand in a well-insulated container against a constant external pressure of 2.5 atm from an initial volume of 2.50 L to a final volume of 4.50 L. The change in internal energy ΔU of the gas in joules will be

1136.5 J
-500 J
-505 J
+505 J

-505 J

ΔU = q + w
For adiabatic process, q = 0
∴ ΔU = w
= – P·ΔV
= –2.5 atm × (4.5 – 2.5) L
= –2.5 × 2 L-atm
= –5 × 101.3 J
= –506.5 J
= –505 J

2584 Views

With respect to the conformers of ethane, which of the following statements is true?

Bond angle remains same but bond length changes
Bond angle changes but bond length remains same
Both bond angle and bond length change
Both bond angles and bond length remains same

Both bond angles and bond length remains same

There is no change in bond angles and bond lengths in the conformations of ethane. There is only change in dihedral angle.

7424 Views

For a given reaction, ΔH = 35.5 kJ mol^–1 and ΔS = 83.6 JK^–1 mol^–1. The reaction is spontaneous at (Assume that ΔH and ΔS do not vary with temperature)

T < 425 K
T > 425 K
All temperatures
T > 298 K

T > 425 K

∵ ΔG = ΔH – TΔS
For a reaction to be spontaneous, ΔG = –ve
i.e., ΔH < TΔS

therefore, T> ΔH/ΔS
= 35.5 x 103J/83.6JK^-1
i.e T>425 K

2477 Views

20-litre container at 400 K contains CO₂(g) at pressure 0.4 atm and an excess of SrO (neglect the volume of solid SrO). The volume of the containers is now decreased by moving the movable piston fitted in the container. The maximum volume of the container, when pressure of CO₂ attains its maximum value, will be

Given that: SrCO₃ ⇌ SrO(s) + CO₂(g) kp = 1.6 atm)

5 litre
10 litre
8 litre
3 litre

5 litre

Max. the pressure of CO2= Pressure of CO2 at equilibrium
For reaction, SrCO₃ ⇌ SrO(s) + CO₂(g)

K_p= P_CO2 = 1.6 atm = maximum pressure of CO₂
Volume Container at this stage,
V= nRT/P .... (i)

Since container is sealed and reaction was not earlier at equilibrium
therefore, n = constant
n= Pv/RT .... (ii)
from equation (i) and (ii) we get
V = (0.4 x 20)/RT = 5 Litre

3397 Views

HgCl₂ and I₂ both when dissolved in water containing I^– ions the pair of species formed is

HgI₂ , I₃^-
Hgl₂, I^-
HgI₄^3-, I₃^-
Hg₂I₂, I^-

HgI₄^3-, I₃^-

In a solution containing HgCl₂, I₂ and I–, both HgCl₂ and I₂
compete for I^–.
Since formation constant of [HgI₄]^2– is 1.9 × 10³⁰
which is very large as compared with I₃^– (K_f= 700)
therefore, I^– will preferentially combine with HgCl₂
HgCl₂ + 2l^- → Hgl₂↓ + 2Cl^-
Hgl₂ + 2l^- → [Hgl₄]2^-

2970 Views

Match the interhalogen compounds of column I with the geometry in column II and assign the correct code

Column I	Column II
a) XX'	(i) T-shape
b) XX₃'	(ii) Pentagonal bipyramidal
c) XX'₅	(iii) Linear
d) XX'₇	(iv) Square-pyramidal
	(v) Tetrahedral

(a) (b) (c) (d)
(iii) (iv) (i) (ii)
(a) (b) (c) (d)
(iii) (i) (iv (ii)
(a) (b) (c) (d)
(iv) (v) (iii) (ii)
(a) (b) (c) (d)
(iv) (iii) (ii) (i)

(a)	(b)	(c)	(d)
(iii)	(i)	(iv	(ii)

1176 Views

The correct statement regarding electrophile is

Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from a nucleophile
Electrophile is a negatively charged species and can form a bond by accepting a pair of electrons from another electrophile
Electrophiles are generally neutral species and can form a bond by accepting a pair of electrons from a nucleophile
Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile

Electrophile can be either neutral or positively charged species and can form a bond by accepting a pair of electrons from a nucleophile

1176 Views

Which one is the wrong statement?

de-Broglie's wavelength is given by λ = h/mv,
where m = mass of the particle, v = group velocity of the particle
The uncertainty principle is $increment straight E space straight x space increment straight t space greater or equal than space fraction numerator straight h over denominator 4 straight pi end fraction$
Half-filled and fully filled orbitals have greater stability due to greater exchange energy, greater symmetry and more balanced arrangement
The energy of 2s orbital is less than the energy of 2p orbital in case of Hydrogen like atoms