The p-Block Elements Test

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The p-Block Elements Test - 42

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Weekly Quiz Competition

Question 1
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The relative strength of trichlorides of boron group to accept a pair of electron is given by:

A
$$GaCl_3 > AlCl_3 > BCl_3$$

B
$$AlCl_3 < BCI_3 < GaCl_3$$

C
$$AlCl_3 < GaCl_3 < BCl_3$$

D
$$BCl_3 < AlCl_3 < GaCl_ 3$$

E
$$GaCl_3 < BCl_3 < AlCl_3$$

Solution

Lewis acidic strength is in the order
$$GaX_3 < AlX_3 < BX_3$$
So, acidic strength of trichlorides is as follows:
$$GaCl_3 < AlCl_3 < BCl_3$$
Hence, strength of accepting electron is as follows:
$$GaCl_3 > AlCl_3 > BCl_3$$
Question 2
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The correct Lewis acid order for boron halides is:

A
$$BBr_3 > BCl_3 > BI_3 > BF_3$$

B
$$BI_3 > BF_3 > BBr_3 > BCl_3$$

C
$$BF_3 > BCl_3 > BBr_3 > BI_3$$

D
$$BI_3 > BBr_3 > BCl_3 > BF_3$$

Solution

(i) On the basis of back bonding, smaller be the halide atom, more effective be the back bonding and show less tendency to accept a pair of electrons.

(ii) Size of halides increase down the group, hence the ability of back bonding decreases making the molecule more acidic.

Hence, correct order of Lewis acid for boron-halides is $$Bl_3 > BBr_3 > BCl_3 > BF_3$$.
Question 3
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What is the chief constituent of Pyrex glass ?

A
$$B_2O_3$$

B
$$SiO_2$$

C
$$Al_2O_3$$

D
$$NaO_2$$

Solution

Pyrex glass is obtained by fusing toghether 60 to 80% $$\displaystyle SiO_2$$, 10 to 25 % $$\displaystyle B_2O_3$$ and remaining amount of $$\displaystyle Al_2O_3$$.

$$\displaystyle SiO_2$$ is the chief constituent of Pyrex glass.

Hence, the correct answer is option $$B$$.
Question 4
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The outermost electronic configuration of the most electronegative element is:

A
$${ ns }^{ 2 }{ np }^{ 3 }$$

B
$${ ns }^{ 2 }{ np }^{ 4 }$$

C
$${ ns }^{ 2 }{ np }^{ 5 }$$

D
$${ ns }^{ 2 }{ np }^{ 6 }\quad $$

Solution

$${ ns }^{ 2 }{ np }^{ 5 }$$ configuration represents the most electronegative element as after gaining one electron it becomes more stable (inert gas configuration) [electronegativity is the tendency of attracting electron].
Fulorine (atomic number 9) is the most electronegative element. Its electronic configuration is $$1s^2{ 2s }^{ 2 }{ 2p }^{ 5 }$$. It is a halogen.
Question 5
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An ionic compound is:

A
$$Sn{ Cl }_{ 2 }$$

B
$$C{ Cl }_{ 4 }$$

C
$$Ge{ Cl }_{ 4 }$$

D
$$Si{ Cl }_{ 4 }$$

Solution

An ionic compound is $$Sn{ Cl }_{ 2 }$$.
All tetrahalides of carbon family are covalent except $$ \displaystyle SnF_4$$ which is ionic.
Question 6
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The hybridised state of $${ Al }^{ 3+ }$$ in the complex ion formed when $$Al{ Cl }_{ 3 }$$ is treated with aqueous acid is:

A
$${ sp }^{ 3 }$$

B
$${ dsp }^{ 2 }$$

C
$${ sp }^{ 3 }{ d }^{ 2 }$$

D
$${ sp }^{ 2 }d$$

E
$${ sp }^{ 2 }$$

Solution

$$Al^{ 3+ }$$ aqueous acid $$\rightarrow { \left[ Al{ \left( { H }_{ 2 }O \right) }_{ 6 } \right] }^{ 3+ }$$.

$${ Al }^{ 3+ }$$ ions give a complex ion in which $${ Al }^{ 3+ }$$ ion form six coordinate covalent bonds with six $${ H }_{ 2 }O$$ molecules.

Hence, it shows $${ sp }^{ 3 }{ d }^{ 2 }$$ hybridisation.

(Due to availability of vacant $$3d$$-orbitals.)

$$Al=3{ s }^{ 2 }3{ p }^{ 1 }$$

$${ Al }^{ 3+ }=3{ s }^{ 0 }$$

Four electron pairs denoted by $$6{ H }_{ 2 }O$$ molecules.
Question 7
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Colemanite is

A
$$Na_{2}B_{4}O_{7}\cdot 10H_{2}O$$

B
$$Ca_{2}B_{6}O_{11} \cdot 5H_{2}O$$

C
$$NaBO_{2}$$

D
$$H_{3}BO_{3}$$

Solution

Colemanite, an ore of boron is $$Ca_{2}B_{6}O_{11}\cdot 5H_{2}O$$ is a borate mineral found in evaporite deposits of alkaline lacustrine environments.

Colemanite is a secondary mineral that forms by alteration of borax and ulexite.

Hence, option B is correct.
Question 8
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Boron has an extremely high melting point because of:

A
the strong van der Waals forces between its atoms

B
the strong binding forces in the covalent polymer

C
its ionic crystal structure

D
allotropy

Solution

Boron has very high melting point because of its small atomic size , it forms strong covalent bonds with the neighbouring atoms. Thus boron atom are closely packed in its solid state so a large amount of heat is needed to break the bonds between atoms.
That is why boron has an extremly high melting point.
Question 9
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Elements of 14 group :

A
Exhibit oxidation state of + 4 only

B
Exhibit oxidation states of +2 and + 4

C
form $$M^{2-}$$ and $$M^{4+}$$ ions

D
form $$M^{2+}$$ and $$M^{4+}$$ ions

Solution

Due to inert pair effect, elements of group 14 exhibit oxidation states of $$+ 2$$ and $$+ 4$$.
Question 10
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Addition of HCl to an aqueous solution of $$Pb(NO_3)_2$$ gives a

A
Yellow Precipitate

B
Baum Precipitate

C
White Precipitate

D
Black Precipitate

Solution

$$Pb (NO_2)_2(aq)+ 2HCl (aq)\rightarrow PbCl_2 (s)+ 2 HNO_3 (aq)$$ gives white precipitate

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Re-Attempt Weekly Quiz Competition

The p-Block Elements Test - 42

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The p-Block Elements Test - 42

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Question 1

$$GaCl_3 > AlCl_3 > BCl_3$$

$$AlCl_3 < BCI_3 < GaCl_3$$

$$AlCl_3 < GaCl_3 < BCl_3$$

$$BCl_3 < AlCl_3 < GaCl_ 3$$

$$GaCl_3 < BCl_3 < AlCl_3$$

Solution

Question 2

$$BBr_3 > BCl_3 > BI_3 > BF_3$$

$$BI_3 > BF_3 > BBr_3 > BCl_3$$

$$BF_3 > BCl_3 > BBr_3 > BI_3$$

$$BI_3 > BBr_3 > BCl_3 > BF_3$$

Solution

Question 3

$$B_2O_3$$

$$SiO_2$$

$$Al_2O_3$$

$$NaO_2$$

Solution

Question 4

$${ ns }^{ 2 }{ np }^{ 3 }$$

$${ ns }^{ 2 }{ np }^{ 4 }$$

$${ ns }^{ 2 }{ np }^{ 5 }$$

$${ ns }^{ 2 }{ np }^{ 6 }\quad $$

Solution

Question 5

$$Sn{ Cl }_{ 2 }$$

$$C{ Cl }_{ 4 }$$

$$Ge{ Cl }_{ 4 }$$

$$Si{ Cl }_{ 4 }$$

Solution

Question 6

$${ sp }^{ 3 }$$

$${ dsp }^{ 2 }$$

$${ sp }^{ 3 }{ d }^{ 2 }$$

$${ sp }^{ 2 }d$$

$${ sp }^{ 2 }$$

Solution

Question 7

$$Na_{2}B_{4}O_{7}\cdot 10H_{2}O$$

$$Ca_{2}B_{6}O_{11} \cdot 5H_{2}O$$

$$NaBO_{2}$$

$$H_{3}BO_{3}$$

Solution

Question 8

the strong van der Waals forces between its atoms

the strong binding forces in the covalent polymer

its ionic crystal structure

allotropy

Solution

Question 9

Exhibit oxidation state of + 4 only

Exhibit oxidation states of +2 and + 4

form $$M^{2-}$$ and $$M^{4+}$$ ions

form $$M^{2+}$$ and $$M^{4+}$$ ions

Solution

Question 10

Yellow Precipitate

Baum Precipitate

White Precipitate

Black Precipitate

Solution

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