Chemistry Test

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Chemistry Test - 18

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Weekly Quiz Competition

Question 1
1 / -0

A solution of sulphur dioxide in water reacts with H₂S precipitating sulphur. Here sulphur dioxide acts as

A
an oxidising agent

B
a reducing agent

C
an acid

D
a catalyst

Solution

First - let us write down the reaction:
\(S O_{2}+2 H_{2} S \longrightarrow 3 S+2 H_{2} O\)
\(\ln H_{2} S,\) sulphur has an oxidation number of \(-2 . \ln S O_{2}\) Sulphur has an oxidation number of +4
On the right-hand side, we only have elemental sulphur whose oxidation number is 0. The sulphur in \(S O_{2}\) changes its oxidation number from +4 to
0
This means that the sulphur is gaining electrons. We know that when a
reactant gains electrons, it gets reduced. Hence, \(S O_{2}\) is getting reduced.
Also, the reactant that gets reduced is the oxidising agent. Hence, here,
\(S O_{2}\) acts as an oxidising agent.
Question 2
1 / -0

Which pair of elements has same chemical properties?

A
13, 22

B
3, 11

C
4, 24

D
2, 4

Solution

The pair which belongs to some group i.e., in which both the elements have same outer electronic configuration has same chemical properties.
\[
\begin{array}{l}
\therefore 3 \Rightarrow 1 s^{2}, 2 s^{1} \\
11 \Rightarrow 1 s^{2}, 2 s^{1} 2 p^{2}, 3 s^{1}
\end{array}
\]
Question 3
1 / -0

The solubility of II group hydroxides increases with molecular weight. This is because:

A
lattice energy is same but hydration energy decreases down the group.

B
hydration energy is same but lattice energy decreases down the group

C
both lattice and hydration energy increase down the group

D
decrease in lattice energy down the group is faster than decrease in hydration energy.

Solution

On moving down the group II, the atomic and ionic size increases . The size of Be²⁺ is smallest and the size of Ba²⁺ is highest. As the size increases, the decrease in the lattice energy is much more than the decrease in the hydration energy. Due to this, the solubility increases with increase in the molecular weight on moving down the group. Hence, barium hydroxide is more soluble than beryllium hydroxide.
Question 4
1 / -0

In a molecule of phosphorus (V) oxide, there are:

A
4P−P,10P−O, and 4P=O bonds

B
12P−O and 4P=O bonds

C
2P−O and 4P=P bonds

D
6P−P, 12P−O and 4P=P bonds

Solution

As can be seen from the image, it contains \(12 P-O\) bonds and \(4 P=O\) bonds.
Question 5
1 / -0

The wave length of first member of Balmer series of a hydrogen atom is nearly (The value of Rydberg constant is 1.08 × 10 7 m⁻¹)

A
4400A°

B
5500A°

C
6600A°

D
7700A°

Solution

\(n_{1}=2 \quad n_{2}=3\)
\(\frac{1}{\lambda}=R\left[\frac{1}{4}-\frac{1}{9}\right]\)
\(\lambda=\frac{36}{5 R}=\frac{36}{5 \times 1.08 \times 10^{7}} m\)
\(=\frac{36}{5 \times 1.08} \times 10^{-7} m\)
\(\approx 6600 A^{0}\)
Question 6
1 / -0

Which one of the following is the correct quadratic form of the Ostwald's dilution law equation?

A
α²C+αK−K=0

B
α²C−αK−K=0

C
α²C−αK+K=0

D
α²C+αK+K=0

Solution

Ostwald's dilution law equation is \(K=\frac{\alpha^{2} C}{(1-\alpha)}\)
\(K(1-\alpha)=\alpha^{2} C\)
\(\alpha^{2} C-K(1-\alpha)=0\)
\(\alpha^{2} C-K+\alpha K=0\)
or \(\alpha^{2} C+\alpha K-K=0\)
Question 7
1 / -0

The nature of the chemical bond in MgO is:

A
electrovalent

B
covalent

C
dative

D
polar covalent

Solution

MgMg is a metal and completes its octet by the loss of e⁻ . Oxygen is a non-metal and completes its octet by the gain of e⁻ . So, the formation of MgO involves transfer of e⁻from metal to non-metal. Hence, the bonding is electrovalent or ionic in nature.
Question 8
1 / -0

Tautomerism is exhibited by :

A
CH₃CNO

B
(CH₃)₂NH

C
R₃CO₂H

D
RCH₂NO₂

Solution

Nitro compound exhibit tautomerism RCH₂NO₂
Question 9
1 / -0

H₂O₂ is used as

A
An oxidant only

B
A reductant only

C
An acid only

D
An oxidant, a reductant and an acid

Solution

Solution As an oxidant
\[
2 F e^{2+}(a q)+H_{2} O_{2}+2 H^{+}(a q) \longrightarrow 2 F e^{3+}(a q)+2 H_{2} O(l)
\]
As as reductant (these reactions are apt to produce \(O_{2}\) in lab):
\[
\begin{array}{l}
\mathrm{NaOCl}+\mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow \mathrm{O}_{2}+\mathrm{NaCl}+\mathrm{H}_{2} \mathrm{O} \\
2 \mathrm{KMnO}_{4}+3 \mathrm{H}_{2} \mathrm{O}_{2} \longrightarrow 2 \mathrm{MnO}_{2}+2 \mathrm{KOH}+2 \mathrm{H}_{2} \mathrm{O}+3 \mathrm{O}_{2}
\end{array}
\]
Hydrogen peroxide is a weak acid, forming hydroperoxide or peroxide
salts with many metals.
Question 10
1 / -0

The bond dissociation energy of H - H, C - C, and C - H bonds respectively are 104.2, 83.1 and 98.8 kcal/ mole. The electronegativity of C is?

A
2.53

B
2.51

C
2.50

D
2.52

Solution

\(\triangle=E_{C-H}-\sqrt{E_{H-H} \times E_{C-C}}\)
\(=98.8-[(104.2) \times(83.1)]^{\frac{1}{2}}=5.75 \mathrm{kcal}\)
\(\left(X_{C-H}\right)\) E.N of C-H bond \(=0.218 \sqrt{\triangle}=0.218(5.75)^{\frac{1}{2}}\)
\(=0.43\)
\(X_{C}=0.43+X_{H}=0.43+2.1=2.53\)

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Chemistry Test - 18

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Chemistry Test - 18

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Question 1

an oxidising agent

a reducing agent

an acid

a catalyst

Solution

Question 2

13, 22

3, 11

4, 24

2, 4

Solution

Question 3

lattice energy is same but hydration energy decreases down the group.

hydration energy is same but lattice energy decreases down the group

both lattice and hydration energy increase down the group

decrease in lattice energy down the group is faster than decrease in hydration energy.

Solution

Question 4

4P−P,10P−O, and 4P=O bonds

12P−O and 4P=O bonds

2P−O and 4P=P bonds

6P−P, 12P−O and 4P=P bonds

Solution

Question 5

4400A°

5500A°

6600A°

7700A°

Solution

Question 6

α2C+αK−K=0

α2C−αK−K=0

α2C−αK+K=0

α2C+αK+K=0

Solution

Question 7

electrovalent

covalent

dative

polar covalent

Solution

Question 8

CH3CNO

(CH3)2NH

R3CO2H

RCH2NO2

Solution

Question 9

An oxidant only

A reductant only

An acid only

An oxidant, a reductant and an acid

Solution

Question 10

2.53

2.51

2.50

2.52

Solution

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α²C+αK−K=0

α²C−αK−K=0

α²C−αK+K=0

α²C+αK+K=0

CH₃CNO

(CH₃)₂NH

R₃CO₂H

RCH₂NO₂