Chemistry Test 255

Result

Chemistry Test 255

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Weekly Quiz Competition

Question 1
4 / -1

Identify the pair of elements with the highest and lowest electronegativity respectively

A
K and Rb

B
I and F

C
F and Fr

D
Fr and Li

Solution

Fluorine (F) is the element with highest electronegativity, lying at the extreme right in the second period.

Francium (Fr) is the most electropositive or least electronegative element, lying at the bottom left of the periodic table.
Question 2
4 / -1

Which of the following facts regarding boron and silicon is not true?

A
Boron and silicon are semiconductors

B
Boron and silicon form halides which are not hydrolysed

C
Boron and silicon react with magnesium to form magnesium boride and magnesium silicide which are decomposed by acids to give volatile borane and silane, respectively

D
Both boron and silicon react with alkalis to form borates and silicates containing BO₄ and SiO₄ tetrahedral units, respectively

Solution

The halides of B and Al are hydrolysed.
Question 3
4 / -1

Electrolysis of dilute aqueous NaCl solution was carried out by passing 10 mA current. The time required to liberate 0.01 mol of H₂ gas at the cathode is (1 F = 96500C mol⁻¹)

A
9.65 × 10⁴ s

B
19.3 × 10⁴ s

C
28.95 × 10⁴ s

D
38.6 × 10⁴ s

Solution

2H₂O + 2e⁻→ H₂+ 2OH⁻

For 0.01 mole H2 0.02 mole of electrons are consumed

charge required = 0.02 × 96500 C = i × t
Question 4
4 / -1

The rate of reaction becomes double when its temperature is raised from 300 K to 330 K. The activation energy of the reaction is (Given: log 2 = 0.30)

A
18.96 kJ mol⁻¹

B
23.96 kJ mol⁻¹

C
28.96 kJ mol⁻¹

D
33.96 kJ mol⁻¹

Solution
Question 5
4 / -1

An oxide of nitrogen is reddish brown and paramagnetic at room temperature but it decolourises and also loses its paramagnetism on freezing it. The oxide at room temperature is

A
Pure NO₂

B
Pure N₂O₄

C
equilibrium mixture of N₂O₄ and NO₂

D
N₂O₅

Solution

2Pb(NO₃)₂→ 2PbO + 4NO₂+ O₂

i.e. can be prepared by heating heavy metal nitrate of Pb(NO₃)₂. It is a red brown gas but on cooling below 273 K converts to colourless liquid and retains as dimer. In NO₂ molecules 17 valence e−are there (12 that of O & 5 that of N) i.e. one e−unpaired results to paramagnetic nature. So this is an odd e⁻molecule on being dimer with even e⁻becomes stable & loses paramagnetic characteristics.
Question 6
4 / -1

Ethanal reacts with HCN and the addition product so obtained is hydrolyzed to form a new compound. This compound shows

A
optical isomerism

B
geometrical isomerism

C
tautomerism

D
metamerism

Solution

The asterisked carbon is 'chiral'.
Question 7
4 / -1

The correct combination is:

A
[NiCl₄]^2- - square-planar; [Ni(CN)₄]^2- - paramagnetic

B
[Ni(CN)₄]^2- - tetrahedral ; [Ni(CO)₄]^2- - paramagnetic

C
[NiCl₄]^2- - paramagnetic; [Ni(CO)₄] - tetrahedral

D
[NiCl₄]^2- - diamagnetic; [Ni(CO)₄] - square-planar

Solution
Question 8
4 / -1

On treatment of the following compound with a strong acid, the most susceptible site for bond cleavage is

A
C1 - O2

B
O5 - C6

C
O2 - C3

D
C4 - O5

Solution

The most susceptible site for bond cleavage is O2 - C3 as this cleavage would result in a 2^o carbocation which is more stable.
Question 9
4 / -1

Haemoglobin and gold sol are examples of:

A
Negatively charged sols

B
Positively charged sols

C
Negatively and positively charged sols, respectively

D
Positively and negatively charged sols, respectively

Solution

Haemoglobin and gold sol both are colloids and always carry an electric charge. Haemoglobin is a positively charged sol, because in hemoglobin, Fe²⁺ ion is the central metal ion of the octahedral complex. All metal sols like Au-sol etc. are negatively charged sols.
Question 10
4 / -1

The major product of the following reaction is

A

B

C

D

Solution

This is an example of E₂ elimination.

Conjugated diene is formed with phenyl ring in conjugation which makes it very stable.