The p-Block Elements Test

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The p-Block Elements Test - 66

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Weekly Quiz Competition

Question 1
1 / -0

Each of the following is true for white and red phosphorus except that they:

A
are both soluble in $$CS_{2}$$

B
can be oxidized by heating in air

C
consist of the same kind of atoms

D
can be converted into one another.

Solution

White phosphorus, yellow phosphorus or simply tetraphosphorus ($$P_4$$) exists as molecules made up of four atoms in a tetrahedral structure. As phosphorous molecule is non polar,so it dissolves in non polar solvent like carbon disulfide, $$CS_2$$.
Hence option A is correct.
Question 2
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$$2$$ moles of $$PCl_{5}$$ were heated in a closed vessel of $$2$$ litre capacity. At equilibrium, $$40\%$$ of $$PCl_{5}$$ is dissociated into $$PCl_{3}$$ and $$Cl_{2}$$. The value of equilibrium constant is:

A
$$0.266$$

B
$$0.53$$

C
$$2.66$$

D
$$5.3$$

Solution

$$PCl_5(g)\leftrightharpoons PCl_3(g)+Cl_2(g)$$

2 mol 0 0

-2 $$\alpha$$mol 2 $$\alpha$$mol 2 $$\alpha$$mol

2 (1-$$\alpha$$)mol 2 $$\alpha$$mol 2 $$\alpha$$mol

When degree of dissociation, $$\alpha=40$$%=$$0.4$$

At equilibrium , the molar concentration of the component of mixture

$$[PCl_5(g)]=\cfrac {2(1-\alpha)}{V}=\cfrac {1(1-0.4)}{2}=0.6$$ mol/L

$$[PCl_3(g)]=\cfrac {2\alpha}{V}=\cfrac {2\times 0.4}{2}=0.4$$ mol/L

$$[Cl_2(g)]=\cfrac {2\alpha}{V}=\cfrac {2\times 0.4}{2}=0.4$$ mol/L

So, $$Eq^m$$ constant, $$K_c=\cfrac {[PCl_3(g)]\times [Cl_2(g)]}{[PCl_5(g)]}$$

$$=\cfrac {0.4\times 0.4}{0.6}$$

$$=0.27$$ mol/L

Hence, the correct option is $$A$$.
Question 3
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In Haber's process $$50.0\ g$$ of $$N_{2}(g)$$ and $$10.0\ g$$ of $$H_{2}(g)$$ are mixed to produced $$NH_{3}(g)$$. What are the number of moles of $$NH_{3}(g)$$ formed?

A
$$3.33$$

B
$$2.36$$

C
$$2.01$$

D
$$5.36$$

Solution

$$N_2 + 3H_2 \rightarrow2NH3$$
Here.
28 g $$N_2$$ reacts with 6 g of $$H_2$$
So,
50 g of $$N_2$$ react with $$\dfrac{6}{28}\times50 g -H_2 = 10.7 g$$ but we have only 10g so $$H_2$$ is limiting reagent..

6 g $$H_2$$ gives 34g $$NH_3$$
10g $$H_2$$ will give $$\dfrac{34}{6}\times10=56.67g -NH_3$$
So number of moles of $$NH_3$$ formed is,
$$=\dfrac{Given-mass}{Molar-mass}=\dfrac{56.67}{17}=3.33-moles$$
Question 4
1 / -0

Chlorine has two naturally occuring isotopes $$_{ 17 }^{ 37 }{ Cl }$$ and $$_{ 17 }^{ 35 }{ Cl }$$ and average atomic mass of chlorine is $$35.5$$. The percentage of $$_{ 17 }^{ 37 }{ Cl }$$ is:

A
25%

B
75%

C
40%

D
60%

Solution

Let ,

The percentage of $$Cl^{37} $$ be $$x $$ and The percentage of $$Cl^{37} $$ is $$ y = (100-x) $$

Average atomic mass = $$ \frac{Atomic-mass \times abundance}{100} $$

$$ 35.5 = \frac{37x + 35(100-x)}{100} $$

$$ 37x + 35(100-x) = 3550 $$

$$ 2x = 50 $$

$$ \therefore x = 25 % $$

$$ y = 100 - x = 75 % $$

The percentage of $$Cl^{37} $$ = 25%

The percentage of $$Cl^{35} $$ = 75%

Hence , option A is correct .
Question 5
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Which among the following statements is incorrect?

A
$$XeF_{4}$$ and $$SbF_{5}$$ combine to form salt.

B
$$He$$ and $$Ne$$ do not form clathrates.

C
$$He$$ has highest boiling point in its group.

D
$$He$$ diffuses through rubber and polyvinyl chloride.

Solution

As we move from top to bottom in g group the boiling point of the elements increases as the atomic mass increases.So $$He$$ is the first element in the group it should have less boiling point than the other elements.
So the given statement $$He$$ has highest boiling point is the wrong.
Hence option C is correct.
Question 6
1 / -0

Potassium permanganate $$\rightarrow$$ potassium manganate + $$A$$ + oxygen.
What is $$A$$ in the given reaction?

A
Manganese

B
Manganese dioxide

C
Manganese trioxide

D
Manganese tetraoxide

Solution

The compound $$A$$ in the given reaction is Manganese dioxide. It is obtained by heating of potassium permanganate to form oxygen.
$$2KMnO_4 \rightarrow K_2MnO_4 + MnO_2 + O_2$$
Question 7
1 / -0

Which of the following is the polar molecules ?

A
$$XeF{l_4}$$

B
$$S{F_4}\;$$

C
$$Si{F_4}$$

D
none

Solution
Question 8
1 / -0

Phosphours sulphide $${ P }_{ 4 }{ S }_{ 3 }$$, a well known chemical used in match industry has many $$P-S$$ bonds ?

A
4

B
2

C
6

D
8

Solution
Question 9
1 / -0

Select reaction in which at least one compound is formed as product, in which Xe is in +6 oxidation state?

A
$$XeF_6+CsF \xrightarrow[thermal \, decomposition]{\Delta} product$$

B
$$XeF_6 \overset{Hydrolysis} {\rightarrow} \quad ? \xrightarrow[disproportion]{OH^-} product$$

C
$$Xe+F_2\overset{873\, K, 7 bar}{\rightarrow} \quad ? \quad \overset{SF_6}{\rightarrow}product$$

D
$$Xe_{excess}+F_2 \quad \overset{400^0C,1 bar}{\rightarrow}\xrightarrow[of water]{Excess} product$$
Question 10
1 / -0

The atomic number of $$N$$ is 7. The atomic number of 3rd member of nitrogen family will be:

A
23

B
15

C
33

D
43

Solution

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The p-Block Elements Test - 66

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The p-Block Elements Test - 66

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Question 1

are both soluble in $$CS_{2}$$

can be oxidized by heating in air

consist of the same kind of atoms

can be converted into one another.

Solution

Question 2

$$0.266$$

$$0.53$$

$$2.66$$

$$5.3$$

Solution

Question 3

$$3.33$$

$$2.36$$

$$2.01$$

$$5.36$$

Solution

Question 4

25%

75%

40%

60%

Solution

Question 5

$$XeF_{4}$$ and $$SbF_{5}$$ combine to form salt.

$$He$$ and $$Ne$$ do not form clathrates.

$$He$$ has highest boiling point in its group.

$$He$$ diffuses through rubber and polyvinyl chloride.

Solution

Question 6

Manganese

Manganese dioxide

Manganese trioxide

Manganese tetraoxide

Solution

Question 7

$$XeF{l_4}$$

$$S{F_4}\;$$

$$Si{F_4}$$

none

Solution

Question 8

4

2

6

8

Solution

Question 9

$$XeF_6+CsF \xrightarrow[thermal \, decomposition]{\Delta} product$$

$$XeF_6 \overset{Hydrolysis} {\rightarrow} \quad ? \xrightarrow[disproportion]{OH^-} product$$

$$Xe+F_2\overset{873\, K, 7 bar}{\rightarrow} \quad ? \quad \overset{SF_6}{\rightarrow}product$$

$$Xe_{excess}+F_2 \quad \overset{400^0C,1 bar}{\rightarrow}\xrightarrow[of water]{Excess} product$$

Question 10

23

15

33

43

Solution

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