Chemistry Test

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Chemistry Test - 25

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Weekly Quiz Competition

Question 1
1 / -0

If the nitrogen atom had electronic configuration 1s⁷, it would have energy lower than that of the normal ground state configuration 1s² 2s²2p³, because the electrons would be closer to the nucleus, yet 1s⁷ is not observed because it violates

A
Heisenberg uncertainty principle

B
Aufbau rule

C
Pauli exclusion principle

D
Bohr postulate of stationary orbits

Solution

1s⁷violate Pauli exclusion principle, according to which an orbital cannot have more than two electrons.
Hence, the correct option is (C)
Question 2
1 / -0

\(\mathrm{Ph}-\mathrm{CH}_{2}-\mathrm{C} \equiv \mathrm{CH} \underset{\mathrm{Y}}{\stackrel{\mathrm{X}}{\rightleftarrows}} \mathrm{Ph}-\mathrm{C} \equiv \mathrm{C}-\mathrm{CH}_{3}\)
The reagents \(X\) and \(Y\) respectively are

A
Lindlar catalyst, NaNH₂

B
NaNH₂and alc. KOH

C
Pt catalyst, Wilkinson's catalyst

D
Alc. KOH and NaNH₂

Solution

Internal alkynes are more stable than terminal alkynes (because of hyperconjugation), isomerization is favored thermodynamically. The reaction is carried out with moderately strong bases, at high temperatures (> 100 °C ), which are not able to completely deprotonate terminal alkynes.

The deciding step is the tautomerization of the acetylide anion to the propargyl anion which is stabilized by mesomerism.

On heating with alc. KOH in inert solvent, the triple bond of 1-alkyne is shifted towards the centre to form an isomeric 2-alkyne.

The triple bond migrates from the terminal position into the C-C chain.

Isomerization in the opposite direction leading to the formation of a terminal alkyne can be accomplished with strong bases, e.g. sodium amide at 150 °C, which are able to completely deprotonate terminal alkynes.

The reaction proceeds in the opposite direction because the most stable anion (acetylide) is formed under the strong basic conditions and not the more stable hydrocarbon (internal alkyne) which is formed under less basic conditions.

Hence, the correct option is (D)
Question 3
1 / -0

The first orbital of \(\mathrm{H}\) is represented by \(: \psi=\frac{1}{\sqrt{\pi}}\left(\frac{1}{a_{0}}\right)^{3 / 2} \mathrm{e}^{-\mathrm{r} / a_{0}},\) where \(\mathrm{a}_{0}\) is Bohr's radius. The probability of finding the electron at a distance \(r\), from the nucleus in the region \(d V\) is :

A
∫ψ²4πr²dv

B
∫ψdv

C
ψ²dr

D
ψ²4πr²dr

Solution

P( r ) = ψ²4πr²dr

4πr²dr = dv = vol of thin spherical shell around nucleus at distance ( r )
Hence, the correct option is (D)
Question 4
1 / -0

Following is the graph between log t_1/2 and log a (a = initial concentration) for a given reaction at 27^oC. Hence, order is :

A
0

B
1

C
2

D
3

Solution

\(\mathrm{A} \rightarrow\) Products
From graph \(\rightarrow\)
\(\log t_{\frac{1}{2}} \propto \tan 45 \log a\)
\(\log t_{\frac{1}{2}} \propto \log a\)
\(t_{\frac{1}{2}} \propto a\) (Initial Condition)
for zero order reaction
\(\therefore t_{\frac{1}{2}} \propto(a)^{1-n}\)
Hence, the correct option is (A)
Question 5
1 / -0

Amongst the following, the most basic compound is

A
Aniline

B
Acetanilide

C
p-nitroaniline

D
Benzylamine

Solution

Lone pair is not involved in resonance, most basic. In all other cases, lone-pair of nitrogen is involved in resonance, less basic.
Hence, the correct option is (D)
Question 6
1 / -0

\(\mathrm{ }_{11}^{23}{Na}\) is the more stable isotope of Na. Find out the process by which \({ }_{11}^{24} \mathrm{Na}\) can undergo radioactive decay.

A
β^- - emission

B
α - emission

C
β⁺ - emission

D
K - electron capture

Solution

In stable isotope of Na,there are 11 protons and 12 neutrons.In the given radioactive isotope of sodium (Na²⁴) there are 13 neutrons, one neutron more than that required for stability. A neutron rich isotope always decay by β -emission as

₀n¹ → -₁β⁰ +₁H¹
Hence, the correct option is (A)
Question 7
1 / -0

The reagent with which both acetaldehyde and acetone react easily is

A
Tollen’s reagent

B
Schiff’s reagent

C
Grignard reagent

D
Fehling reagent

Solution

Grignard’s reagent reacts with both aldehydes and ketones while other three reagents reacts only with aldehydes, not with ketones.
Hence, the correct option is (C)
Question 8
1 / -0

Two organic compounds A and B both containing C and H yield on analysis, the same percentage composition by weight, C=(12/13)×100; H=(1/13)×100. Which among the following will be the correct formula of the compounds 'A' and 'B' ? ('A' decolourises bromine water but 'B' does not)

A
A = C₂H₂, B = C₆H₆

B
A = C₆H_6, B = C₂H₂

C
A = C₂H₄, B = C₂H₆

D
A = C₂H₂, B = C₃H₈

Solution

Thus empirical formula = CH

The two which satisfy this empirical formula are C₂H₂ and C₆H₆.

As 'A' decolourise bromine water thus 'A' should be C₂H₂ and 'B' does not decolourise bromine water thus 'B' should be C₆H₆.
Hence, the correct option is (A)
Question 9
1 / -0

When the same amount of zinc is treated separately with excess of sulphuric acid and excess of sodium hydroxide, the ratio of volumes of hydrogen evolved is

A
1:1

B
1:2

C
2:1

D
9:4

Solution

\(Zn + H _{2} SO _{4} \longrightarrow ZnSO _{4}+ H _{2} \uparrow\)
\(Zn +2 NaOH \longrightarrow Na _{2} ZnO _{2}+ H _{2} \uparrow\)

Thus, one mole of Zn gives one mol of H2 gas.

Hence, the correct option is (A)
Question 10
1 / -0

20% of N₂O₄molecules are dissociated in a sample of gas at 27^∘C and 760 torr. Mixture has the density at equilibrium equal to:

A
1.48 g/L

B
1.84 g/L

C
2.25 g/L

D
3.12 g/L

Solution

\(\mathrm{N}_{2} \mathrm{O}_{4}(\mathrm{g}) \rightleftharpoons 2 \mathrm{NO}_{2}(\mathrm{g})\)
\(1-\alpha\)
Total Moles \(1+\alpha\) at equilibrium observed Molar Mass at equilibrium
\(\mathrm{M}_{\mathrm{eq}}=\frac{\mathrm{M}}{1+\alpha}=\frac{92}{1-2}=76 \cdot 67 \mathrm{g} / \mathrm{mole}\)
\(\rho_{\operatorname{mix}}=\frac{P M_{e q}}{R T}=\frac{1 \times 76 \cdot 67}{0 \cdot 082 \times 300}\)
\(=3 \cdot 12 \mathrm{g} / \mathrm{L}\)
Hence, the correct option is (D)

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Chemistry Test - 25

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Chemistry Test - 25

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Question 1

Heisenberg uncertainty principle

Aufbau rule

Pauli exclusion principle

Bohr postulate of stationary orbits

Solution

Question 2

Lindlar catalyst, NaNH2

NaNH2 and alc. KOH

Pt catalyst, Wilkinson's catalyst

Alc. KOH and NaNH2

Solution

Question 3

∫ψ24πr2dv

∫ψdv

ψ2dr

ψ24πr2dr

Solution

Question 4

0

1

2

3

Solution

Question 5

Aniline

Acetanilide

p-nitroaniline

Benzylamine

Solution

Question 6

β- - emission

α - emission

β+ - emission

K - electron capture

Solution

Question 7

Tollen’s reagent

Schiff’s reagent

Grignard reagent

Fehling reagent

Solution

Question 8

A = C2H2, B = C6H6

A = C6H6, B = C2H2

A = C2H4, B = C2H6

A = C2H2, B = C3H8

Solution

Question 9

1:1

1:2

2:1

9:4

Solution

Question 10

1.48 g/L

1.84 g/L

2.25 g/L

3.12 g/L

Solution

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Lindlar catalyst, NaNH₂

NaNH₂and alc. KOH

Alc. KOH and NaNH₂

∫ψ²4πr²dv

ψ²dr

ψ²4πr²dr

β^- - emission

β⁺ - emission

A = C₂H₂, B = C₆H₆

A = C₆H_6, B = C₂H₂

A = C₂H₄, B = C₂H₆

A = C₂H₂, B = C₃H₈