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JEE Chemistry Solved Question Paper 2016

Multiple Choice Questions

11.

Consider the following sets of quantum numbers. Which of the following setting is not permissible arrangement of electrons in an atom?

n = 4; l = 0; m = 0; s = $- \frac{1}{2}$
n = 5; l = 3; m = 0; s = $+ \frac{1}{2}$
n = 3; l = 2; m = -2; s = $- \frac{1}{2}$
n = 3; l = 2; m = -3; s = $+ \frac{1}{2}$

12.

The increasing order of bond order of O₂, O $_{2}^{+}, O_{2}^{-} and O_{2}^{2 -}$ is

$O_{2}^{+} < O_{2} < O_{2}^{-} < O_{2}^{2 -}$
$O_{2}^{2 -} < O_{2}^{-} < O_{2}^{+} < O_{2}$
$O_{2} < O_{2}^{+} < O_{2}^{-} < O_{2}^{2 -}$
$O_{2}^{2 -} < O_{2}^{-} < O_{2} < O_{2}^{+}$

13.
HCl gas is covalent and NaCl is an ionic compound. This is because

sodium is highly electropositrve

hydrogen is a non-metal

HCl is a gas

electronegativity difference between H and Cl is less than 2.1

electronegativity difference between H and Cl is less than 2.1

HCl (g) is covalent whereas NaCl is an ionic solid due to small difference in their electronegativities. In HCl, the difference in electronegativity of H and Cl-atoms is less than 2.10, whereas in NaCl, the difference in electronegativities of Na and Cl-atoms is more than 2.10.

Electronegativity of (Cl) = 3.00 (approx. value)
Electronegativity of (Na) = 0.88 (approx. value)
Electronegativity difference = 2.12

14.

Main axis of diatomic molecule is z. The orbitals p_x and p_y overlap to form

$π$ -molecular orbital
σ-molecular orbital
δ-molecular orbital
No bond is formed

15.

An organic compound contains C = 40%, H = 13.33% and N = 46.67%. Its empirical formula is

C₂H₂N
C₃H₇N
CH₄N
CHN

16.

IUPAC name of the compound,

1-bromobut-2-ene
2-bromo-2-butene
bromobutene
1-bromobut-3-ene

17.

Benzene carbaldehyde is reacted with conc. NaOH solution to give the products A and B. The product A can be used as a food preservative and the product B is an aromatic hydroxy compound, where -OH group is linked to sp³-hybridised C-atom next to benzene ring. The products A and B respectively are

sodium benzoate and phenol
sodium benzoate and phenyl methanol
sodium benzoate and cresol
sodium benzoate and picric acid

18.

An organic compound A on reduction gives compound B, which on reaction with trichloromethane and caustic potash forms C. The compound C on catalytic reduction gives N-methyl benzenamine, the compound A is,