The p-Block Elements Test

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The p-Block Elements Test - 61

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Weekly Quiz Competition

Question 1
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The silicate anion in the mineral kinoite is a chain of three $$SiO_4$$ tetrahedral that share corners with adjacent tetrahedral. The mineral also contains $$Ca^{2+}$$ ions, $$Cu^{2+}$$ ions, and water molecules in a $$1:1:1$$ ratio. The mineral is :

A
$$Ca_2Cu_2Si_3O_{10}\cdot 2H_2O$$

B
$$CaCuSi_3O_{10}\cdot H_2O$$

C
$$Ca_2Cu_2Si_3O_8\cdot 2H_2O$$

D
$$Ca_2Cu_2Si_3O_6\cdot 2H_2O$$

Solution

Three $$SiO_4$$ tetrahedral that share corners with adjacent tetrahedral creates $$Si_3O_{10}^{-4}$$ anion.
So to form a neutral compound total poistive charge required is 4; also ratio of cations is 1:1 so compound is $$Ca_2Cu_2Si_3O_{10}.2H_2O$$.
Question 2
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Borax is converted into crystalline boron by the following steps:

$${Borax \overset{X}{\rightarrow} H_3BO_3\overset{\Delta }{\rightarrow} B_2O_3\xrightarrow[\Delta ]{Y} B}$$
Then $$X$$ and $$Y$$ are respectively:

A
$$HCl, Mg$$

B
$$HCl, C$$

C
$$C, Al$$

D
$$HCl , Al$$

Solution

In first step,Boric acid is obtained by treating a hot concentrated solution of borax with hydrochloric acid or sulphuric acid. The resulting solution on concentration and cooling gives crystals of boric acid. here acid used is HCl or $$H_2SO_4$$.

Hence, the correct option is $$A$$.
Question 3
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One type of silicon adhesive uses a polymer that has two $$OH$$ groups left on the ends of the chain:
$$HO-Si(CH_3)_2[O-Si(CH_3)_2]_n-O-Si(CH_3)_2-OH$$
It has been found that this kind of polymer reacts easily with an organic-silicon molecule that contains acetate $$(CH_3COO)$$ group. Which of these three molecules $$CH_3COOSi(CH_3)_3$$, $$(CH_3COO)_2Si(CH_3)_2$$, $$(CH_3COO)_3SiCH_3$$ would you choose to add to the hydroxy polymer in order to make a strong adhesive?

A
$$CH_3COOSi(CH_3)_3$$,

B
$$(CH_3COO)_2Si(CH_3)_2$$

C
$$(CH_3COO)_3SiCH_3$$

D
None of the above

Solution

$${ \left( C{ H }_{ 3 }COO \right) }_{ 3 }SiC{ H }_{ 3 }-$$triacetoxy (methyl) silane.
This should give the cross-linking necessary for strength of adhesives.
Question 4
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Select correct statements:

A
Hydrides of $$B$$ and $$Si$$ are volatile and catch fire on exposure to air.

B
Oxides of $$B$$ and $$Si(B_2O_3$$ and $$SiO_2)$$ are acidic in nature.

C
Borates and silicates have tetrahedral $$BO_4$$ and $$SiO_4$$ structural units.

D
All the above are correct statements.

Solution

Hydrides of silicon and boron are less stable because in these, hydrogen withdraw electrons from M-H bond and due to this they are volatile and catch fire on exposure to air.
Oxides of $$B$$ and Si $$(B_2O_3$$ and $$SiO_2)$$ are electron deficient due to high electronegativity difference and shows acidic nature.
Borates and silicates have tetrahedral $$BO_4$$ and $$SiO_4$$ structural units.
Question 5
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Which among the following statements is false regarding glass?

A
It absorbs ultra-violet light

B
It consists of silicates of sodium and alkali

C
$$HF$$ causes marking on glass

D
All the halogen acids $$(HF, HCl, HBr$$ and $$HI)$$ are stored in glass bottles

Solution

(A) The glass is opaque due to scattering of light by small crystals therein. However, it absorbs ultra-violet light.
Thus, the option $$A$$ is correct.
(B) It consists of silicates of sodium and alkali.
The ordinary glass may be represented as $$Na_2O.CaO.6SiO_2$$.
Thus, the option $$B$$ is correct.
(C) $$HF$$ causes marking on glass as glass dissolves in HF with the liberation of silicon tetrafluoride gas.
Thus, the option $$C$$ is correct.
(D) Except, $$HF$$ all the halogen acids ($$HCl, HBr$$ and $$HI$$) are stored in glass bottles.
Hence, the option $$D$$ is incorrect.
Question 6
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Select correct statement about stability of cations.

A
$$Ge^{4+} > Sn^{4+} > Pb^{4+}$$

B
$$Ge^{2+} < Sn^{2+} < Pb^{2+}$$

C
$$Pb^{2+} > Pb^{4+}, Sn^{4+} > Sn^{2+}$$

D
All the above are correct statements.

Solution

Common oxidation state of IVA group elements are +2, +4.
+2 state becomes gradually more stable down the group & +4 state stability decreases from carbon to Lead due to inert pair effect.
So order is
$$Ge^{4+} > Sn^{4+} > Pb^{4+}$$
$$Ge^{2+} < Sn^{2+} < Pb^{2+}$$
And for lower element Pb II oxidation state is more stable than IV oxidatio state.
$$Pb^{2+} > Pb^{4+}, Sn^{4+} > Sn^{2+}$$
Question 7
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Silicone's are :

A
synthetic polymers containing repeated $$R_2SiO$$ units.

B
silicates with common $$SiO_4$$ unit.

C
ketones with silyl group $$(SiH_3)$$ similar to alkyl $$(SiH_3)_2CO$$

D
Zircon (neso silicates)

Solution

These are organosilicon polymers containing $$R_2SiO$$ repeating units and empirical formula analogous to ketone $$(R_2CO)$$
$$-\underset{|}{\overset{|}{S}}i-O-\underset{|}{\overset{|}{S}}i-linkage$$
Organic polymers containing silicon in them are called as silicones.
These are formed by hydrolysis of alkyl chlorosilanes and their polymerisation. The alkyl or aryl substituted chlorosilanes are prepared by the reaction of Grignard reagent and silicon tetrachlorides.
Question 8
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Select correct statement(s) :

A
$$Si$$ uses all of its valence electrons in an $$sp^3$$-hybridization and crystallizes in an fcc structure similar to diamond.

B
Diamond is an electrical insulator, graphite is an electrical conductor but silicon is a semiconductor.

C
Most common silica is quartz.

D
All of the above are correct statements.

Solution

In diamond, 4 C-C covalent bond are present but in graphite only 3 C-C bond are present and 4th electron is free so it is conductor but diamond is insulator due to absence of free ions.

Silica or silicon dioxide occurs in nature in the free state as sand, quartz and flint.

Hence,option D is correct.
Question 9
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Directions For Questions

Silicone's are synthetic polymers containing repeated $$R_2SiO$$ units. Since, the empirical formula is that of a ketone $$(R_2CO)$$, the name silicone has been given to these materials. Silicone's can be made into oils, rubbery elastomers and resins. They find a variety of applications because of their chemical interness, water repelling nature, heat-resistance and good electrical insulating property.
Commercial silicon polymers are usually methyl derivatives and to a lesser extent phenyl derivatives and are synthesized by the hydrolysis of

...view full instructions

If we mix $$SiMe_3Cl$$ with $$SiMe_2Cl_2$$, we get silicone's of the type:

A

B

C
both (a) and (b)

D
none of the above

Solution

When $$Me_3SiCl$$ is mixed with $$Me_2SiCl_2$$ silicones of the type $$Me_3Si-O-Si(Me_2)-O-Si(Me_2)-OSiMe_3$$ are obtained.
Each of the terminal silicon atom carries three methyl groups and each of the middle silicon atom carries two methyl groups.
Question 10
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Aqueous solution of borax reacts with 2 mol of acids. This is because of?

A
Formation of 2 mol of $$B(OH)_{3}$$ only.

B
Formation of 2 mol of $$\left [B(OH)_{4} \right ]^{-}$$ only.

C
Formation of 1 mol each of $$B(OH)_{3}$$ and $$\left [B(OH)_{4} \right ]^{-}$$ only.

D
Formation of 2 mol each of $$\left [B(OH)_{4} \right ]^{-}$$ and $$B(OH)_{3}$$, of which $$\left [B(OH)_{4} \right ]^{-}$$ reacts with acid.

Solution