JEE Main 2020 Chemistry Paper With Solutions Jan 9 Shift 1

Question 9. B has a smaller first ionization enthalpy than Be. Consider the following statements:

(i) It is easier to remove 2p electron than 2s electron

(ii) 2p electron of B is more shielded from the nucleus by the inner core of electrons than the 2s electron of Be

(iii) 2s electron has more penetration power than 2p electron

Atomic radius of B is more than Be (Atomic number B=5, Be=4)

The correct statements are:

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

Be (4): 1s²2s²

B (5): 1s²2s²2p¹

The electron in 2p¹ can easily be extracted.

The penetrating power is of the order: s > p > d > f The shielding power order: s > p > d > f

As we move along the period, the size decreases, as Z_eff increases. Hence the radius of B is smaller than the radius of Be.

Answer: (a)

a) > 1400⁰C
b) < 1400⁰C
c) > 1200⁰C
d) < 1200⁰C

Solution: In Ellingham’s diagram, the line of the element that lies below can reduce the oxide of the element which lies above it. Therefore, for A to reduce BO₂ , the temperature when the line for element A is below that of BO₂, according to the graph when T > 1400 ℃.

For T > 1400 ℃ , ΔG_r < 0 for A + BO₂-> B + AO₂

Answer: (a)

It was found that the 𝐸𝑎 is decreased by 30 kJ/mol in the presence of catalyst. If the rate remains unchanged, the activation energy for catalysed reaction is (Assume pre exponenetial factor is same)

a) 75 kJ/mol
b) 135 kJ/mol
c) 105 kJ/mol
d) 198 kJ/mol

K = Ae^(-Ea/RT)

K_catalyst = K_{without catalyst}

E_ac = E_a – 30

-(E_a-30)/T_500k = -E_a/T_700k

On solving E_a = 105 kJ mol^-1

Answer: (c)

Hence, major product formed is that of option c.

Answer: (c)

Out of the four options given, only aniline and phenol show strong +R effects, but as we know, aniline is a Lewis base and can react with a Lewis acid that is added during the reaction. Hence, Phenol gives the highest yield in Friedel-Craft’s reaction.

Answer: (b)

Answer: (a)

a) C > B > A
b) B > A > C
c) A > B > C
d) C > A > B

Heat of combustion ∝ 1/ stability

The trans-isomer is more stable than the cis-isomer. More the number of trans forms in a structure, higher the stability.

Answer (c)

a) A > B > D > E > C
b) B > A > D > C > E
c) A > B > E > D > C
d) C > E > D > B > A

As we know weaker the conjugate base, stronger the acid.

The order of stability of conjugate base:

Hence, the order of basicity or acidic strength is:

A > B > D> E> C

Answer (a)

(i) A and D both form blue-violet colour with ninhydrin.

(ii) Lassaigne extract of C gives positive AgNO₃ test and negative Fe₄[Fe(CN)₆]₃ test.

(iii) Lassaigne extract of B and D gives positive sodium nitroprusside test.

Based on these observations which option is correct?

a) A – Alitame, B – Saccharin, C – Aspartame, D – Sucralose
b) A –Saccharin, B – Alimate, C – Sucralose, D – Aspartame
c) A – Aspartame, B – Alitame, C – Saccharin , D – Sucralose
d) A – Aspartame, B – Saccharin, C – Sucralose, D – Alitame

It has a free amine group and hence reacts with ninhydrin to give a purple colour known as

Ruhemann’s purple.

It has Sulphur, therefore, it will give a positive test with sodium nitroprusside.

It has chlorine and hence it forms a precipitate with AgNO3 in the Lassaigne’s extract of the sugar.

It has a free amine group and hence reacts with ninhydrin to give purple colour known as Ruhemann’s purple. Also, it has Sulphur, therefore, it will give positive test with sodium nitroprusside.

Answer: (d)

is a methyl ketone, which gives positive Iodoform test.

Answer: (b)