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# NEET Chemistry Solved Question Paper 2016

#### Multiple Choice Questions

11.

Which of the following hydrocarbons is the most reactive towards addition of H2SO4?

• Ethene

• Propylene

• 3-methyl but-1-ene

• 1-butene

C.

3-methyl but-1-ene

As $\mathrm{\pi }$-bond is more activated, the addition of H2SO4 is carried out lastly. In the given compound, 3-methyl but-1-ene undergoes rapid addition, because $\mathrm{\pi }$-bond is activated by hyperconjugation and after addition the most stable 3$°$-carbocation is formed.

12.

Consider the following structures

Choose the correct statement regarding the above structures.

• Dipole moment varies as II > III > I

• II is more stable than I

• I is the most reactive among three

• All of the above

D.

All of the above

Hence, dipole moment decreases as II > III > I

Stability can be represented as II > III > I

Reactivity can be represented as III > II > I.

# 13.100 mL of a solution contains 2 g of acetic acid and 3g of sodium acetate providing Ka =1.8 x 10-5 , then choose the correct option.This solution is basic in nature This solution is acidic in nature This solution is amphoteric in nature This solution is neutral in nature

B.

This solution is acidic in nature

pH formula is,

pH = -

pH = -

pH = 4.7851

14.

Match the particle with its characteristic

 Column I Column II A. $\mathrm{\alpha }$-particle p. Slow moving B. Isobar q. High penetration power C. $\mathrm{\gamma }$- ray r. Same atomic mass D. $\mathrm{\beta }$- particle s. consists of electron

• A - p; B - r; C - q; D - s

• A - p; B - q; C - r; D - s

• A - r; B - s; C - p; D - q

• A - s; B - r; C - p; D - q

A.

A - p; B - r; C - q; D - s

 Column I Column II A. $\mathrm{\alpha }$- particle p. Slow moving B. Isobar r. Same atomic mass C. $\mathrm{\gamma }$- ray q. High penetration power D. $\mathrm{\beta }$- particle s. Consists of electron

15.

At radioactive equilibrium, the ratio between two atoms of radioactive elements A and B is 3.1 x 109 : 1. If the half-life period of A is 2 x 1010 yrs, then the half-life of B is

• 9.54 yrs

• 2.14 yrs

• 3.29 yrs

• 6.45 yrs

D.

6.45 yrs

Ratio of A : B = 3.1 x 109 : 1

Half life of A = 2 x 1010 yrs.

Half life of B is =

16.

11.2 L of gas at STP Weight 14g. The gas would be

• H2

• CO

• B2H6

• All of these

D.

All of these

Molecular weigth of gas =  = 28

The above formula denotes the formula of mole concept. The gas would be CO, N2, B2H6.

17.

When 2-methyl butyl bromide is treated with sodium ethoxide in ethanol, what will be the major product?

• 2-methyl but-2-ene

• 3-methyl but-1-ene

• 2-methyl but-1-ene

• 2-methyl sodium-butoxide

A.

2-methyl but-2-ene

When 2-methyl butyl bromide is treated with sodium ethoxide in ethanol, two alkenes are possible. The more substituted alkene, i.e. 2-methyl-2-butene is the major product.

18.

Highest energy will be absorbed to eject out the electron in the configuration.

• 1s22s22p1

• 1s22s22p3

• 1s22s22p2

• 1s22s22p4

B.

1s22s22p3

Highest electric will be absorbed to eject out the electron in the configuration of 1s22s22p3

19.

For the formation of Cr2O3 and Al2O3, values of ${∆}_{\mathrm{f}}\mathrm{G}°$ are 540 kJ mol-1 and -827 kJ mol-1 respectively. What will be the possibility for reaction of Cr2O3 by Al?

• Reduction of Cr2O3 by Al will take place

• Oxidation of Cr2O3 by Al will take place

• Neither oxidation nor reduction will take place

• Reaction is not feasible

A.

Reduction of Cr2O3 by Al will take place

On substrating equation (i) from (ii)

2Al + Cr2O3 $\to$Al2O3 + 2Cr

$∆$rG = [-827 - (-540)] kJ mol-1

= -287 kJ mol-1

$∆{\mathrm{G}}^{°}$  value is negative for the reduction by Al. Therefore, the reaction is feasible.

20.

Both Mg and Fe metal can reduce copper from a solution having Cu2+ ion, according to equilibria.

Mg(s) + Cu2+ $⇌$ Mg2+ + Cu(s); K1 = 6 x 1090

Fe(s) + Cu2+ $⇌$ Fe2+ + Cu(s); K2 = 3 x 1026

Choose the correct option regarding above equilibrium

• Mg removes more Cu2+ from solution

• Fe removes more Cu2+ from solution

• Both will equally remove Cu2+ from solution

• Both metals cannot remove Cu2+ from solution

A.

Mg removes more Cu2+ from solution

Since, K1 > K2, the product in the first reaction is much more favoured than in the second one. Mg thus removes more Cu2+ from solution than Fe does.