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Published on: 20/08/2019
The p-Block Elements
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Questions + Answers key
Take MCQ Chemistry Test

1.
What is inert pair effect?
2.
What are Fullerenes? How are they prepared?
3.
Explain the Aluminium utensils should not be kept in water overnight.
4.
In some of the reactions, thallium resembles aluminium, whereas in others it resembles with group I metals. Support this statement by giving some evidences.
5.
Name the building block of zeolites. Why zeolites have high porosity?
6.
Out of \({ PbCl }_{ 2 },{ SnCl }_{ 2 },{ SnCl }_{ 4 }\) and \({ PbCl }_{ 4 }\) which one is most commonly used as a reducing agent and why?
7.
Rationalise the given statement and give chemical reactions:
Lead(II) chloride reacts with Cl2 to give PbCl 4
8.
[SiF6]2 is known whereas [SiCl6]2-not. Give possible reasons.
9.
Standard electrode potential values, \(\mathrm{E}^{\ominus}\) for AI3/AI is -1.66 V and that of TI3+/TI is +1.26 V predict about the formation of M3+ ion in solution and compare the electropositive character of two metals.
10.
How does electron deficient compound BF3 achieve electronic saturation , i.e fully ooccupied outer electron shells?
11.
Discuss the pattern of variation in the oxidation states of B to TI
12.
Explain the differences in properties of diamond and graphite based upon their structures.
13.
How is ultra pure elemental silicon obtained? write its importance.
14.
Complete the following reactions
\({ SiO }_{ 2 }+C\longrightarrow \)
15.
Silicon shows a diagonal relation with
magnesium
phosphorous
carbon
boron
16.
Which of the following molecules have zero dipole moment?
CS2
CO2
CCl2
CH2Cl2
17.
Which of the following is a purely acidic oxide?
Si02
Sn02
PbO
Mn02
18.
Silicon carbide (SiC) is known as
quartz
tridynite
corundum
carborundum
19.
Which of the following compound is an important catalyst as well as a Lewis acid?
Al2S2
BF3
S4N4
N2H4
1.
( )
In the elements of 4th, 5th and 6th period of the p-block the electrons present in the intervening d and f-orbitals do not shield the s-electrons of the valence shell effectively. As a result, ns-2-electrons remain more tightly held by the nucles and hence, do not participate in bonding, this is called inert pair effect.
2.
Fullerenes are the allotropes of carbon. Its structure is like a soccer ball.
They are prepared by heating graphite in electric arc in presence of inert gases such as helium or argon.
3.
Because aluminium reacts with water and oxygen (dissolved in ) to form a thin layer of toxic aluminium oxide on the surface of utensils.
2Al(s) +O2(g) + H2O(l)\(\rightarrow\)Al2O3(s) + H2(g)
4.
Thallium and aluminium both the elements belong to group 13. Their general electronic configurations for the valence shell is ns2np1 . Aluminium shows only +3 oxidation state. Like Al, thallium also shows +3 oxidation state in some compounds like Tl2O3, TlCl3 , etc. Like aluminium, thallium also forms octahedral ions like [AlF6]3- and [TlF6]3- . Like group-I alkali metals, thallium shows +1 oxidation state due to inert pair effect in some compounds like TlCl, Tl2O etc., like alkali metal hydroxides, TiOH is water soluble and its aqueous solution is strongly alkaline. Tl2SO4 also forms alums like alkali metal sulphates. Tl2CO3 is soluble in water like alkali metal carbonates.
5.
The building block of zeolites is sodalite cage. The high porosity of zeolites is because of the sodalite cage packing in a three dimensional network of channels and cavities.
6.
Sn4+ is more stable than Sn2+ thus Sn2+ have a great tendency to get converted into Sn4+ by losing two electrons hence, from the given compounds SnCl2 is used as reducing agent.
7.
Due to inert pair effect, Pb is more stable in +2 state than in +4 oxidation state. Therefore, lead (II) chloride does not react with Cl2 to give lead (IV) chloride.
8.
The main reasons for the existence of [SiF6]2- and non-existence of [SiCl6]2- are as follows.
(i) Six large chloride ions cannot be accommodated around Si4+ due to limitation of its size.
(ii) Interaction between lone pair of chloride ion and Si4+ is not very strong.
9.
Since Eo values are more negative in case of AI, so it has higher tendency to form AI3+(aq) ions, whereas TI3+ is highly unstable in aqueous solution and acts as a good oxiding agent. It readily reduces to more stable TI+ ion AI3+ being stable, exists in this form and hence, AI is more electropositive as compared to TI.
10.
BF3 achieve it by the following ways
(i) Multiple bonding or \(p\pi -p\pi \) back bonding e.g BF3 in which a lone pair of electron present in 2p-orbits of one of the fluorine atoms may be transferred to the vacant p-orbital on the bottom atom.
(ii) Formation of complexes in which electrons are received from a donor molecule,e.g F3B\(\longleftarrow \)NH3 .Boron compounds, thus behave as Lewis acids.
11.
| Element | B | AI | Ga | In | Tl |
| Oxidation state | +3 | +3 | +3,+1 | +3,+1 | +1 |
Boron and aliminium show an oxidation state of +3 only because they do not exhibit insert pair effect due to the absence of d or f-electrons. Elements from Ga to TI show two oxidation states, i.e +1 and +3.The tendency to show +1 oxidation state increases down the group due to the inability of ns2 electrons of valence shell to participate in bonding which is called inert pair effect. Therefore TI+ is more stable than TI3+
12.
| Diamond | Graphite |
| Diamond is the hardest substance on earth. | Graphite is soft and slippery |
| In diamond carbon is Sp3- hybridized | In Graphite carbon is Sp2- hybridized |
| Since all the electrons in diamond are firmly held in C-C,6 bonds there are no free electrons in diamond crystal Therefore diamond is bad conductor of electricity | Since only three electrons of each carbon are used in making hexagonal rings of graphite, fourth valence electron is free to move thus graphite is a good conductor of electricity |
| Because of high refractive index diamond can reflect and refract the light. | Graphite is a black substance and possess a metallic lustre |
13.
Ultra pure elemental silicon is prepared by the reduction of highly pure silicon tetrachloride(SiCl4) or silicon chloroform(SiHCl3) with dihydrogen
\({ SiCl }_{ 4 }+2{ H }_{ 2 }(g)\longrightarrow Si+4HCl\)
\({ SiHCl }_{ 3 }+{ H }_{ 2 }\longrightarrow Si+3HCl\)
It can also be prepared by the pyrolysis of SiH4
\({ SiH }_{ 4 }\longrightarrow Si+{ 2H }_{ 2 }\)
Ultra pure elemental silicon is used as a semiconductor.
14.
\({ SiO }_{ 2 }+C\longrightarrow SiC+2CO\)
silicon carbide Carbon or carborundum monoxide
15.
(d)
boron
16.
(a)
CS2
17.
(a)
Si02
18.
(d)
carborundum
19.
(d)
N2H4
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