Text Book Back Questions and Answers - Chapter 2 - p-Block Elements – I - 12th Chemistry Guide Samacheer Kalvi Solutions

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p-Block Elements - I
TextBook Evalution
I. Choose the correct answer:
An aqueous solution of borax is
(a) neutral
(b) acidic
(c) basic
(d) amphoteric
Answer:
(c) basic.
Hint: $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+7 \mathrm{H}_2 \mathrm{O} \rightleftharpoons \underset{\text { Strong base }}{2 \mathrm{NaOH}}+\underset{\text { Weak acid }}{4 \mathrm{H}_3 \mathrm{BO}_3}$
Question 2.
Boric acid is an acid because its molecule
(a) contains replaceable $\mathrm{H}^{+}$ion
(b) gives up a proton
(c) combines with proton to form water molecule
(d) accepts $\mathrm{OH}^{-}$from water, releasing proton.
Answer:
(d) accepts $\mathrm{OH}^{-}$from water, releasing proton
Hint: $\mathrm{B}(\mathrm{OH})_3+\mathrm{H}_2 \mathrm{O} \rightleftharpoons\left[\mathrm{B}(\mathrm{OH})_4\right]^{-}+\mathrm{H}^{+}$
Question 3.
Which among the following is not a borane?
(a) $\mathrm{B}_2 \mathrm{H}_6$

(b) $\mathrm{B}_3 \mathrm{H}_6$
(C) $\mathrm{B}_4 \mathrm{H}_{10}$
(d) none of these
Answer:
(a) $\mathrm{B}_2 \mathrm{H}_6$
Hint:
- Nido borane $-\mathrm{B}_{\mathrm{n}} \mathrm{H}_{4+\mathrm{n}}$
- aracno borane $-\mathrm{B}_{\mathrm{n}} \mathrm{H}_{6+\mathrm{n}}, \mathrm{B}_3 \mathrm{H}_6$ is not a borane
Question 4.
Which of the following metals has the largest abundance in the earth's crust?
(a) Aluminium
(b) calcium
(c) Magnesium
(d) Sodium
Answer:
(a) Aluminium
Question 5.
In diborane, the number of electrons that accounts for banana bonds is
(a) $\operatorname{six}$
(b) two
(c) four
(d) three
Answer:
(c) four
Hint: There are two $3 \mathrm{c}-2 \mathrm{e}^{-}$bonds i.e., the bonding in the bridges account for 4 electrons.
Question 6.
The element that does not show catenation among the following p-block elements is
(a) Carbon
(b) silicon
(c) Lead
(d) germanium
Answer:
(c) Lead

Question 7.
Carbon atoms in fullerene with formula $\mathrm{C}_{60}$ have
(a) $\mathrm{sp}^3$ hybridised
(b) sp hybridised
(c) $\mathrm{sp}^2$ hybridised
(d) partially $\mathrm{sp}^2$ and partially $\mathrm{sp}^3$ hybridised
Answer:
(c) $\mathrm{sp}^2$ hybridised
Question 8.
Oxidation state of carbon in its hydrides
(a) +4
(b) -4
(c) +3
(d) +2
Answer:
(a) +4
Hint: $\mathrm{CH}_4{ }^{+}$in which the oxidation state of carbon is 4 .
Question 9.
The basic structural unit of silicates is
(a) $\left(\mathrm{SiO}_3\right)^{2-}$
(b) $\left(\mathrm{SiO}_4\right)^{2-}$
(c) $(\mathrm{SiO})^{-}$
(d) $\left(\mathrm{SiO}_4\right)^{4-}$
Answer:
(d) $\left(\mathrm{SiO}_4\right)^{4-}$
Question 10.
The repeating unit in silicone is

Answer:

Question 11.
Which of these is not a monomer for a high molecular mass silicone polymer?
(a) $\mathrm{Me}_3 \mathrm{SiCl}$
(b) $\mathrm{PhSiCl}_3$
(c) $\mathrm{MeSiCl}_3$
(d) $\mathrm{Me}_3 \mathrm{SiCl}_3$
Answer:
(a) $\mathrm{Me}_3 \mathrm{SiCl}$
Question 12.
Which of the following is not $\mathrm{sp}^2$ hybridised?
(a) Graphite
(b) graphene
(c) Fullerene
(d) dry ice
Answer:
(a) dry ice
Hint: dry ice - solid $\mathrm{CO}_2$ in which carbon is in sp hybridized state
Question 13.
The geometry at which carbon atom in diamond are bonded to each other is
(a) Tetrahedral
(b) hexagonal
(c) Octahedral
(d) none of these
Answer:
(a) Tetrahedral
Question 14.
Which of the following statements is not correct?
(a) Beryl is a cyclic silicate
(b) $\mathrm{Mg}_2 \mathrm{SiO}_4$ is an orthosilicate
(c) $\mathrm{SiO}_4^{4-}$ is the basic structural unit of silicates
(d) Feldspar is not aluminosilicate
Answer:
(d) Feldspar is a three dimensional silicate

Question 15.
$\mathrm{AlF}_3$ is soluble in $\mathrm{HF}$ only in the presence of $\mathrm{KF}$. It is due to the formation of
(a) $\mathrm{K}_3\left[\mathrm{AlF}_3 \mathrm{H}_3\right]$
(b) $\mathrm{K}_3\left[\mathrm{AlF}_6\right]$
(C) $\mathrm{AlH}_3$
(d) $\mathrm{K}\left[\mathrm{AlF}_3 \mathrm{H}\right]$
Answer:
(6) $\mathrm{K}_3\left[\mathrm{AlF}_6\right]$
Hint: $\mathrm{AlF}_3+3 \mathrm{KF} \rightarrow \mathrm{K}_3\left[\mathrm{AlF}_6\right]$
Question 16.
Match items in column - I with the items of column - II ans assign the correct code

Answer:

Answer:
(a) $\mathrm{A}-2, \mathrm{~B}-1, \mathrm{C}-4, \mathrm{D}-3$
Question 17.
Duralumin is an alloy of
(a) $\mathrm{Cu}, \mathrm{Mn}$
(b) $\mathrm{Cu}, \mathrm{Al}, \mathrm{Mg}$
(c) $\mathrm{Al}, \mathrm{Mn}$
(d) $\mathrm{Al}, \mathrm{Cu}, \mathrm{Mn}, \mathrm{Mg}$
Answer:
(d) $\mathrm{Al}, \mathrm{Cu}, \mathrm{Mn}, \mathrm{Mg}$
Hint: $\mathrm{Al}-95 \%, \mathrm{Cu}-4 \%, \mathrm{Mn}-0.5 \%, \mathrm{Mg}-1.1 \%$
Question 18.
Thermodynamically the most stable form of carbon is
(a) Diamond
(b) graphite
(c) Fullerene
(d) none of these
Answer:
(b) graphite
Question 19.
The compound that is used in nuclear reactors as protective shields and control rods is
(a) Metal borides
(b) metal oxides
(c) Metal carbonates
(d) metal carbide
Answer:
(a) Metal borides
Question 20.
The stability of +1 oxidation state increases in the sequence
(a) $\mathrm{Al}<\mathrm{Ga}<\mathrm{In}<\mathrm{Tl}$
(b) $\mathrm{Tl}<\mathrm{In}<\mathrm{Ga}<\mathrm{A} 1$
(c) In $<\mathrm{Tl}<\mathrm{Ga}<\mathrm{A}$
(d) $\mathrm{Ga}<\mathrm{In}<\mathrm{Al}<\mathrm{Tl}$
Answer:
(a) $\mathrm{Al}<\mathrm{Ga}<\mathrm{In}<\mathrm{Tl}$
II. Answer the following questions:

Question 1.
Write a short note on anamolous properties of the first element of p-block.
Answer:
In p-block elements the first member of each group differs from the other elements of the corresponding group. The following factors are responsible for this anomalous behaviour.
1. Small size of the first member.
2. High ionisation enthalpy and high electronegativity.
3. Absence of d-orbitals in their valance shell.
The first member of the group-13, boron is a metalloid while others are reactive metals. Moreover, boron shows diagonal relationship with silicon of group -14. The oxides of boron and silicon are similar in their acidic nature.
Question 2.
Describe briefly allotropism in p-block elements with specific reference to carbon.
Answer:
Some elements exist in more than one crystalline or molecular forms in the same physical state. This phenomenon is called allotropism. Most common allotropes of carbon are,
1. Graphite
2. Diamond
3. Fullerenes
4. Carbon nanotubes
5. Graphene.
1. Graphite:
- It is the most stable allotropic form of carbon at normal temperature and pressure.
- It is soft and conducts electricity.
- It is composed of flat two dimensional sheets of carbon atoms.
- Each sheet is a hexagonal net of $\mathrm{sp}^2$ hybridised carbon atoms with a $\mathrm{C}-\mathrm{C}$ bond length of $1.41 \mathrm{~A}$.
- Structure of graphite,

2. Diamond:
- It is very hard.
- The carbon atoms in diamond are $\mathrm{sp}^1$ hybridised, with a $\mathrm{C}-\mathrm{C}$ bond length of $1.54 \mathrm{~A}$.
- In the diamond, carbon atoms are arranged in tetrahedral manner.
- Structure of Diamond,

3. Fullerenes:
- It is newly synthesised allotropes of carbon.
- The $\mathrm{C}_{60}$ molecules have a soccer ball like structure and is called buckminster fullerene or buckyballs.
- It has a fused ring structure consists of 20 six membered rings and 12 five membered rings.
- Each carbon atom is $\mathrm{sp}^2$ hybridised.
- The $\mathrm{C}-\mathrm{C}$ bond distance is $1.44 \mathrm{~A}$ and $\mathrm{C}=\mathrm{C}$ distance is $1.38 \mathrm{~A}$.
- Structure of fullerene,

4. Carbon nanotubes:
1. It is recently discovered allotropes, have graphite like tubes with fullerene ends.
2. These nanotubes are stronger than steel and conduct electricity.
3. Structure of Carbon nanotubes.

5. Graphene:
- It has a single planar sheet of $\mathrm{sp}^2$ hybridised carbon atoms that are densely packed in a honeycomb crystals lattice.
- Structure of Graphene,

Question 3.
Boron does not react directly with hydrogen. Suggest one method to prepare diborane from $\mathrm{BF}_3$.
Answer:
Boron does not react directly with hydrogen. However it forms a variety of hydrides called boranes. Treatment of gaseous boron trifluoride with sodium hydride around $450 \mathrm{~K}$ gives diborane.

Question 4.
Give the uses of Borax.
Answer:
Uses of borax:
- Borax is used for the identification of coloured metal ions.
- In the manufacture optical and borosilicate glass, enamels and glazes for pottery.
- It is also used as a flux in metallurgy and also acts as a good preservative.
Question 5.
What is catenation? describe briefly the catenation property of carbon.
Answer:
Catenation:
It is the phenomenon of an atom to form a strong covalent bond with the atoms of itself. Carbon shares the property of catenation to maximum extent because it is small in size and can form pn-pn multiple bonds to itself. The following conditions are necessary for catenation.
1. The valency of element is greater than or equal to two.
2. Element should have an ability to bond with itself.
3. The self bond must be as strong as its bond with other elements.
4. Kinetic inertness of catenated compound towards other molecules. Carbon possesses all the above properties and forms a wide range of compounds with itself.
Question 6.
Write a note on Fisher tropsch synthesis.
Answer:
The reaction of carbon monoxide with hydrogen at a pressure of less than 50 atm using metal catalysts at 500-700 $\mathrm{K}$ yields saturated and unsaturated hydrocarbons.
$
n \mathrm{CO}+(2 n+1) \mathrm{H}_2 \rightarrow \mathrm{C}_n \mathrm{H}_{(2 n+2)}+2 \mathrm{H}_2 \mathrm{O} ; \quad n \mathrm{CO}+2 n \mathrm{H}_2 \rightarrow \mathrm{C}_n \mathrm{H}_{2 n}+n \mathrm{H}_2 \mathrm{O}
$

Question 7.
Give the structure of $\mathrm{CO}$ and $\mathrm{CO}_2$.
Answer:
Structure of $\mathrm{CO}$ :

Structure of $\mathrm{CO}_2$ :

Question 8.
Give the uses of silicones.
Answer:
Uses of silicones:
1. Silicones are used for low temperature lubrication and in vacuum pumps, high temperature oil baths etc.
2. They are used for making water proofing clothes.
3. They are used as insulting material in electrical motor and other appliances
4. They are mixed with paints and enamels to make them resistant towards high temperature, sunlight, dampness and chemicals.
Question 9.
$\mathrm{AlCl}_3$ behaves like a lewis acid. Substantiate this statement.
Answer:
In $\mathrm{AlCl}_3, \mathrm{Al}$ in electron deficient it needs two electrons to complete octet so it act as lewis acid. $\mathrm{AlCl}_3$ usually exist as a dimer to achieve octet by bridged $\mathrm{Cl}$ atom electron deficient compounds are lewis acids.
Question 10 .
Describe the structure of diborane.
Answer:
In diborane two $\mathrm{BH}_2$ units are linked by two bridged hydrogens. Therefore, it has eight $\mathrm{B}-\mathrm{H}$ bonds. However, diborane has only 12 valance electrons and are not sufficient to form normal covalent bonds. The four terminal B-H bonds are normal covalent bonds (two centre - two electron bond or $2 \mathrm{c}-2 \mathrm{e}$ bond). The remaining four electrons have to used for the bridged bonds, i.e. two three centred B-H-B bonds utilise two electrons each.

Hence, these bonds are three centre - two electron bonds. The bridging hydrogen atoms are in a plane as shown in the figure. In dibome, the boron is $\mathrm{sp}^3$ hybridised. Three of the four $\mathrm{sp}^3$ hybridised orbitals contains single electron and the fourth orbital is empty.

Two of the half filled hybridised orbitals of each boron overlap with the two hydrogens to form four terminal 2c-2e bonds, leaving one empty and one half filled hybridised orbitals on each boron. The Three centre - two electron bonds, B-H-B bond formation involves overlapping the half filled hybridised orbital of one boron, the empty hybridised orbital of the other boron and the half filled 1s orbital of hydrogen.

Question 11.
Write a short note on hydroboration.
Answer:
Diborane adds on to alkenes and alkynes in ether solvent at room temperature. This reaction is called as hydroboration and is highly used in synthetic organic chemistry especially for anti-Markovnikov
addition.
$
\mathrm{B}_2 \mathrm{H}_6+3 \mathrm{RCH}=\mathrm{CHR} \rightarrow \mathrm{B}\left(\mathrm{CH}_2-\mathrm{CH}_2 \mathrm{R}\right)_3+6 \mathrm{H}_2
$
Question 12.
Give one example for each of the following:
1. icosogens
2. tetragen
3. prictogen
4. chalcogen
Answer:
1. Icosogens:
- Boron
- Aluminium
- Gallium
2. Tetragen:
- Carbon
- Silicon
- Germanium
3. Prictogen:
- Oxygen
- Sulfur
- Selenium
4. Chalcogen:
- Fluorine
- Chlorine
- Bromine

Question 13.
Write a note on metallic nature of p-block elements.
Answer:
1. The tendency of an element to form a cation by loosing electrons is known as electro-positive or metallic character.
2. This character depends on the ionisation energy.
3. Generally on descending a group the ionisation energy decreases and hence the metallic character increases.

In p-block, the elements present in lower left part are metals while the elements in the upper right part are non metals. Elements of group 13 have metallic character except the first element boron which is a metalloid, having properties intermediate between the metal and nonmetals. The atomic radius of boron is very small and it has relatively high nuclear charge and these properties are responsible for its nonmetallic character.

In the subsequent groups the non-metallic character increases. In group 14 elements, carbon is a nonmetal while silicon and germanium are metalloids. In group 15 , nitrogen and phosphorus are non metals and arsenic \& antimony are metalloids. In group 16, oxygen, sulphur and selenium are non metals and tellurium is a metalloid. All the elements of group 17 and 18 are non metals.
Question 14
Complete the following reactions:
(a) $\mathrm{B}(\mathrm{OH})_3+\mathrm{NH}_3 \rightarrow$
(b) $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4+\mathrm{H}_2 \mathrm{O} \rightarrow$
(c) $\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \rightarrow$
(d) $\mathrm{B}_2 \mathrm{H}_6+\mathrm{CH}_3 \mathrm{OH} \rightarrow$
(e) $\mathrm{BF}_3+9 \mathrm{H}_2 \mathrm{O} \rightarrow$
(f) $\mathrm{HCOOH}+\mathrm{H}_2 \mathrm{SO}_4 \rightarrow$
(g) $\mathrm{SiCl}_4+\mathrm{NH}_3 \rightarrow$
(h) $\mathrm{SiCl}_4+\mathrm{C}_2 \mathrm{H}_5 \mathrm{OH} \rightarrow$
(i) $\mathrm{B}+\mathrm{NaOH} \rightarrow$
(j) $\mathrm{H}_2 \mathrm{~B}_4 \mathrm{O}_7 \underline{\text { Red }}$ hot

Answer:

(a) $\mathrm{B}(\mathrm{OH})_3+\mathrm{NH}_3 \stackrel{\Delta}{\longrightarrow}$ $\underset{\text { (Boron nitride) }}{\mathrm{BN}}+3 \mathrm{H}_2 \mathrm{O}$

(b) $\mathrm{Na}_2 \mathrm{~B}_4 \mathrm{O}_7+\mathrm{H}_2 \mathrm{SO}_4+5 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { (Boric acid) }}{4 \mathrm{H}_3 \mathrm{BO}_3}+2 \mathrm{Na}_2 \mathrm{SO}_4$

(c) $\mathrm{B}_2 \mathrm{H}_6+2 \mathrm{NaOH}+2 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { (Sodiummeteborate) }}{4 \mathrm{NaBO}_2}+6 \mathrm{H}_2$

(d) $\mathrm{B}_2 \mathrm{H}_6+6 \mathrm{CH}_3 \mathrm{OH} \longrightarrow \underset{\text { (Trimethyl borate) }}{2 \mathrm{~B}\left(\mathrm{OCH}_3\right)_3+6 \mathrm{H}_2}$

(e) $4 \mathrm{BF}_3+9 \mathrm{H}_2 \mathrm{O} \longrightarrow \underset{\text { (Boric acid) }}{4 \mathrm{H}_3 \mathrm{BO}_3}+3 \mathrm{H}^{+}+3\left[\mathrm{BF}_4\right]^{-}$

(f) $\mathrm{HCOOH}+\mathrm{H}_2 \mathrm{SO}_4 \longrightarrow \underset{\text { (Carbon monoxide })}{\mathrm{CO}}+\mathrm{H}_2 \mathrm{O}+\mathrm{H}_2 \mathrm{SO}_4$

(g) $\mathrm{SiCl}_4+\mathrm{NH}_3 \underset{\text { ether }}{\stackrel{330 \mathrm{~K}}{\longrightarrow}} \mathrm{Cl}_3 \mathrm{Si}-\mathrm{NH}-\mathrm{SiCl}_3$

(h) 

(i) $2 \mathrm{~B}+6 \mathrm{NaOH} \longrightarrow \underset{\text { (Sodium borate) }}{2 \mathrm{Na}_3 \mathrm{BO}_3+\mathrm{H}_2 \mathrm{O}}$

(j) $\mathrm{H}_2 \mathrm{~B}_4 \mathrm{O}_7 \stackrel{\text { Red hot }}{\longrightarrow} \underset{\text { (Boric anhydride) }}{2 \mathrm{~B}_2 \mathrm{O}_3}+\mathrm{H}_2 \mathrm{O}$

Question 15 .
How will you identify borate radical?
Answer:
When boric acid or borate salt is heated with ethyl alcohol in presence of concentrated $\mathrm{H}_2 \mathrm{SO}_4$, an ester triethyl borate is formed. The Vapour of this ester burns with a green edged flame and this reaction is
used to identify the presence of borate.

Question 16.
Write a note on zeolites.
Answer:
Zeolites:
1. Zeolites are three dimensional crystalline solids containing aluminium, silicon and oxygen in their regular three dimensional framework.
2. They are hydrated sodium alumino silicates with general formula, $\mathrm{Na}_2 \mathrm{O} \cdot\left(\mathrm{Al}_2 \mathrm{O}_3\right) \cdot \mathrm{x}\left(\mathrm{SiO}_2\right) y\left(\mathrm{H}_2 \mathrm{O}\right)$ $(\mathrm{x}=2$ to $10 ; \mathrm{y}=2$ to 6$)$
3. Zeolites have porous structure in which the monovalent sodium ions and water molecules are loosely held.
4. The Si and $\mathrm{Al}$ atoms are tetrahederally coordinated with each other through shared oxygen atoms.
5. Zeolites structure looks like a honeycomb consisting of a network of interconnected tunnels and cages.
6. Zeolite crystal to act as a molecular sieve. They helps to remove permanent hardness of water.
Question 17.
How will you convert boric acid to boron nitride?
Answer:
Fusion of urea with boric acid $\mathrm{B}(\mathrm{OH})_3$, in an atmosphere of ammonia at $800-1200 \mathrm{~K}$ gives
$\underset{\text { (Boric acid) }}{\mathrm{B}(\mathrm{OH})_3}+\mathrm{NH}_3 \stackrel{\Delta}{\longrightarrow} \underset{\text { (Boron nitride) }}{\mathrm{BN}}+3 \mathrm{H}_2 \mathrm{O}$

Question 18 .
A hydride of 2nd period alkali metal
(A) on reaction with compound of Boron
(B) to give a reducing agent
(C) identify A, B and C.
Answer:
1. A hydride of $2^{\text {nd }}$ period alkali metal (A) is lithium hydride ( $\left.\mathrm{LiH}\right)$.
2. Lithium hydride (A) reacts with diborane (B) to give lithiumborohydride (C) which is act as reducing agent.

(A) Lithium hydride $-\mathrm{LiH}$
(B) Diborane $-\mathrm{B}_2 \mathrm{H}_6$
(C) Lithium borohydride $-\mathrm{LiBH}_4$
Question 19.
A double salt which contains fourth period alkali metal
(A) on heating at $500 \mathrm{~K}$ gives
(B). Aqueous solution of (B) gives white precipitate with $\mathrm{BaCl}_2$ and gives a red colour compound with alizarin. Identify $A$ and $B$.
Answer:
1. A double salt which contains fourth period alkali metal (A) is potash alum $\mathrm{K}_2 \mathrm{SO}_4 \cdot \mathrm{Al}_2\left(\mathrm{SO}_4\right)_3 \cdot 24 \mathrm{H}_2 \mathrm{O}$
2. On heating potash alum (A) $500 \mathrm{k}$ give anhydrous potash alum (or) burnt alum (B).

3. Aqueous solution of burnt alum, has sulphates ion, potassium ion and aluminium ion. Sulphate ion reacts with $\mathrm{BaCl}_2$ to form white precipitate of Barium Sulphate
$
\left(\mathrm{SO}_4\right)_2+\mathrm{BaCl}_2 \rightarrow \mathrm{BaSO}_4+2 \mathrm{Cl}^{-}
$
Aluminium ion reacts with alizarin solution to give a red colour compound.
Question 20 .
$\mathrm{CO}$ is a reducing agent. Justify with an example.
Answer:
Both thermodynamic and kinetic factors make carbon monoxide $(\mathrm{CO})$ a better reducing agent. When $\mathrm{CO}$ is used to reduce a metal oxide, it gets oxidized to $\mathrm{CO}_2$ Thermodynamically, $\mathrm{CO}_2$ is much more stable than CO. For example,
$
\mathrm{CO}+\mathrm{Fe}_2 \mathrm{O}_3 \rightarrow 2 \mathrm{Fe}+3 \mathrm{CO}_2
$
Question 1.
Why group 18 elements are called inert gases? Write the general electronic configuration of group 18 elements.
Answer:
The elements of group-18 have completely filled s and p orbitals, hence they are more stable and have least reactivity. Therefore group-18 elements are called inert gases. $\mathrm{ns}_2 \mathrm{np}_6$ is the general electronic configuration of group elements.