The p-Block Elements Test

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The p-Block Elements Test - 10

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Weekly Quiz Competition

Question 1
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The compound that will react most readily with gaseous bromine has the formula is:

A
$$C_4H_{10}$$

B
$$C_2H_4$$

C
$$C_3H_6$$

D
$$C_2H_2$$

Solution

The compound that will react most readily with gaseous bromine has the formula $$C_3H_6$$.

Unsymmetrical alkenes generally are more reactive than the symmetrical alkenes, alkynes and alkanes. That is why, propene is more reactive than the other given compounds.

Hence, option C is correct.
Question 2
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The correct order of atomic radii in group $$13$$ elements is _________.

A
$$B < Ga < Al < Tl < In$$

B
$$B < Al < In < Ga < Tl$$

C
$$B < Ga < Al < In < Tl$$

D
$$B < Al < Ga < In < Tl$$

Solution

The correct order of atomic radii in group $$13$$ elements is:

$$B (\text { 85 pm }) < Ga (\text {135 pm }) < Al (\text {143 pm }) < In (\text { 167 pm }) < Tl (\text {170 pm })$$.

On moving down the group, the atomic radius increases from boron to thallium. The atomic radius of Ga is less than that of Al. Ga has 10 d electrons which do not screen the nuclear charge in an effective manner and hence electrons in the outermost shell of Ga experience more force of attraction towards the nucleus than Al. This results in a decrease in atomic radius.

Hence, option C is correct.
Question 3
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Which of these is not a monomer for a high molecular mass silicone polymer?

A
$$MeSiCl_{3}$$

B
$$Me_{2}SiCl_{2}$$

C
$$Me_{3}SiCl$$

D
$$PhSiCl_{3}$$
Question 4
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The element which, forms oxides in all oxidation states varying from +1 to +5 is:

A
$$N$$

B
$$P$$

C
$$As$$

D
$$Sb$$

Solution

Nitrogen atom can form oxides in all oxidation states varying from +1 to +5.

$$N_2O$$ Nitrogen oxidation state is +1
$$NO$$ Nitrogen oxidation state is +2
$$N_2O_3$$ Nitrogen oxidation state is +3
$$NO_2$$ Nitrogen oxidation state is +4
$$N_2O_5$$ Nitrogen oxidation state is +5.

So answer is A.
Question 5
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Among the following substituted silanes, the one which will give rise to cross-linked silicon polymer on hydrolysis is:

A
$$R_3SiCl$$

B
$$R_4Si$$

C
$$RSiCl_3$$

D
$$R_2SiCl_2$$

Solution

Among the following substituted silanes, the one which will give rise to cross-linked silicone polymer on hydrolysis is trichloroalkylsilane $$RSiCl_3$$.

On hydrolysis, $$RSiCl_3$$ will give $$RSi(OH)_3$$ which will polymerize to cross-linked silicone polymer.

$$ \displaystyle RSiCl_3 \xrightarrow [-3HCl]{3H_2O} RSi(OH)_3 \xrightarrow {\text {polymerisation}}$$ cross-linked silicone polymer.
Question 6
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Which of the element shows $$+4$$ oxidation state?

A
$$Sn$$

B
$$Ra$$

C
$$Fr$$

D
$$Sc$$

Solution

Only tin is capable of exhibiting stable +4 oxidation state while others cannot. $$Sn^{+4}$$ state is stable and forms $$SnCl_4$$ compound.
Question 7
1 / -0

In diborane, the two H-B-H angles are nearly :

A
$$60^o, 120^o$$

B
$$90^o, 120^o$$

C
$$95^o,150^o$$

D
$$120^o, 180^o$$

Solution

In diborane, the bonds between boron and the terminal hydrogen atoms are conventional 2-center, 2-electron covalent bonds. The two $$H-B-H$$ angles are nearly $$90^o and \ 120^o$$.
Question 8
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Which of the following shows bond in silicone?

A
$$Si-C-Si-C-Si$$

B
$$Si-Si-Si-Si$$

C
$$-Si-O-Si-O-Si$$

D
$$Si-C-Si-O-Si$$

Solution

Silicones are organic silicon polymers containing $$Si-O-Si$$ linkages.

Hence, option $$C$$ is correct.
Question 9
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Which one of the following pentafluorides cannot be formed?

A
$$PF_5$$

B
$$AsF_5$$

C
$$SbF_5$$

D
$$BiF_5$$

Solution

Bismuth due to inert pair effect exists with only +3 oxidation state. +5 oxidation state is uncommon for Bi. So, bismuth does not form pentahalide.

Hence, option $$D$$ is correct.
Question 10
1 / -0

Which of the following is the most basic oxide?

A
$${ Sb }_{ 2 }{ O }_{ 3 }$$

B
$${ Bi }_{ 2 }{ O }_{ 3 }$$

C
$$Se{ O }_{ 2 }$$

D
$${ Al }_{ 2 }{ O }_{ 3 }$$

Solution

$${ Bi }_{ 2 }{ O }_{ 3 }$$ is the most basic oxide.
More the oxidation state of the central metal atom, more is its acidity. Hence, $$Se{ O }_{ 2 }$$ (O.S of Se $$=+4$$ ) is acidic. Further, for a given oxidation state, the basic character of the oxides increases with increasing size of the central atom.

Thus, $${ Al }_{ 2 }{ O }_{ 3 }$$ and $${ Sb }_{ 2 }{ O }_{ 3 }$$ are amphoteric and $${ Bi }_{ 2 }{ O }_{ 3 }$$ is basic.

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The p-Block Elements Test - 10

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The p-Block Elements Test - 10

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Question 1

$$C_4H_{10}$$

$$C_2H_4$$

$$C_3H_6$$

$$C_2H_2$$

Solution

Question 2

$$B < Ga < Al < Tl < In$$

$$B < Al < In < Ga < Tl$$

$$B < Ga < Al < In < Tl$$

$$B < Al < Ga < In < Tl$$

Solution

Question 3

$$MeSiCl_{3}$$

$$Me_{2}SiCl_{2}$$

$$Me_{3}SiCl$$

$$PhSiCl_{3}$$

Question 4

$$N$$

$$P$$

$$As$$

$$Sb$$

Solution

Question 5

$$R_3SiCl$$

$$R_4Si$$

$$RSiCl_3$$

$$R_2SiCl_2$$

Solution

Question 6

$$Sn$$

$$Ra$$

$$Fr$$

$$Sc$$

Solution

Question 7

$$60^o, 120^o$$

$$90^o, 120^o$$

$$95^o,150^o$$

$$120^o, 180^o$$

Solution

Question 8

$$Si-C-Si-C-Si$$

$$Si-Si-Si-Si$$

$$-Si-O-Si-O-Si$$

$$Si-C-Si-O-Si$$

Solution

Question 9

$$PF_5$$

$$AsF_5$$

$$SbF_5$$

$$BiF_5$$

Solution

Question 10

$${ Sb }_{ 2 }{ O }_{ 3 }$$

$${ Bi }_{ 2 }{ O }_{ 3 }$$

$$Se{ O }_{ 2 }$$

$${ Al }_{ 2 }{ O }_{ 3 }$$

Solution

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