p-block elements (package solutions)

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Solutions of Assignment (Set-2) The p-Block Elements (Solutions)

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Section-A

Q.No. Solutions

1. Answer (3)

Diborane is an electron deficient compound to minimize electronicdeficiency it forms. 2 electron- three center bond.

2. Answer (4)

Borax in acidic medium give Boric acid, which on heating forms B 2O3. B2O3 when reduced with somereactive metal forms boron.

MgOBOBor

Mg

Na32

3. Answer (1)

Al4C3 + H2O CH4 + Al(OH)3

4. Answer (1)

As we go down the group higher oxidation states becomes less stable due to inert pair effect

5. Answer (1)

B(OH)3 + OH – B(OH)4 –

6. Answer (1)

Silicon can show a coordination number of six because of availability of low lying d orbitals forhybridization.

7. Answer (1)

Bond order in Carbon mono-oxide is 3. Therefore C—O bond is shortest in CO.

lengthBond

1.O.B .

8. Answer (2)

Boron compounds behave as Lewis acid because of their electron deficient nature.

Thep

-Block Elements

B

H

H

H

H

B

H

H

Banana Bond

or 3c–2e bond –

11Chapter



The p-Block Elements (Solutions) Solutions of Assignment (Set-2)


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Q.No. Solutions

9. Answer (3)

Graphite forms sheet like structure where each carbon is sp2 hybridized and it has conjugated bonds

which result into delocalization of electrons and hence graphite conducts electricity.

10. Answer (3)From B2H6, only four terminal hydrogen can be substituted by (CH3) group, so we cannot prepareB(CH3)6. While other compounds can be prepared.

11. Answer (2)

Graphite is taken as thermodynamically most stable.

12. Answer (3)

sp3 hybridisation is present in B2H6.

13. Answer (1)

N

H H

H B

H HB

Both are tetrahedral and sp3 hybridised.

14. Answer (3)

CO2 is neither combustible nor a supporter of combustion

15. Answer (1)

(BN)x decompose with steam under high pressure to from ammonia

16. Answer (1)

Na2 [B4O5 (OH)4]. 8H2O.

17. Answer (1)

18. Answer (4)

19. Answer (1)

20. Answer (4)

21. Answer (1)

22. Answer (4)





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Q.No. Solutions

23. Answer (1)

24. Answer (2)

25. Answer (4)

Section-B

Q.No. Solutions

1. Answer (1, 3, 4)

Acidic are B(OH)3, BF3 and CO2.

2. Answer (1, 2, 4)

Boron is present in Borax, colemanite and kernite (Na2B4O7.4H2O).

3. Answer (1, 2, 3, 4)

O in ether, H2O and ethyl alcohol form co-ordinate bond with BF3, NH3 also donate lone pair to BF3.

4. Answer (1, 2, 3)

Due to p – p interaction between boron and F. Bond length is shorter and Bond energy is high.

5. Answer (1, 2, 3, 4)

B2H6 react with Cl2, CO, NH3 and (CH3)3 N.

6. Answer (1, 2, 4)

B(OH)3 + H2O B(OH)4 – + H+

B(OH)3 + NaOH Na+ [B(OH)4] –

If sugar or any cis diol is added, equilibrium will shift in forward direction. B(OH)3 behaves as strongacid .

7. Answer (1, 2, 3)

CO2 and CS2 is linear, so dipole moments is zero. CCl4 have regular shape, so = 0.

8. Answer (3, 4)

CBr 4 and CI4 are solid (vander Waal solid).

9. Answer (1, 2)

6332:3

162 HNBNHHB





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Q.No. Solutions

10. Answer (1, 2, 3)

The carbon dioxide molecule behaves as a nonpolar molecule even though two of its resonating

structures. OC –O –

and –O –CO

are dipolar.

O CO H

O HO C O

carbon is sp hybridized.

11. Answer (1, 2, 3)

CaC2 + H2O Ca(OH)2 + C2H2

Mg2C3 + H2O CH3 –CCH+Mg(OH)2

CaC2containCa+2andC2

–2

12. Answer (1, 3)

Possible oxidation states of Boron family’s elements are 1 and 3

13. Answer (3, 4)

It gives H+ in aqueous solution, proton donar, and electron deficient

14. Answer (1, 2)

It is electron deficient compound.

15. Answer (2, 4)

16. Answer (1, 2, 3)

Small bases (NH3 and 1º amine) are able to break the bond of bridged hydrogen in a unsymmetrical

manner.

17. Answer (2, 4)

Property-wise graphite is softer than diamond. However both electrical and thermal conductivity ofgraphite is higher than that of diamond. Bond order (C–C) in graphite is 1.5 and higher than C–C bondorder in diamond. Hence, answer is (2, 4).

Section-C

Q.No. Solutions

Comprehension-I

1. Answer (1)

OH4OBBN2NaCl2ClNH2OBNa 2324742

2. Answer (2)

A is H3BO3 and B is B(OC2H5)3, sp3 hybridised.





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Q.No. Solutions

3. Answer (1)

B

O

B

O

B

O

B

O

O OHHO

OH

–

OH

–

Comprehension-II

1. Answer (2)

C60 fullerene contains 20 six membered rings and 12 five membered rings. In fullerenes no. of fivemembered rings are constant.

2. Answer (3)

Inter layer distance in graphite is 3.35 Å while covalent bond length is 1.41 Å

3. Answer (3)

Layer of graphite is not very tightly packed they have a distance of 3.35 Å.

Comprehension-III

1. Answer (3)

2. Answer (4)

3. Answer (4)

Section-D

Q.No. Solutions

1. Answer (2)

Aluminium chloride exist as

AlCl

Cl

Cl

Cl

AlCl

Cl

In aqueous medium, due to small size and vacant d orbital Al+3 form [Al(H2O)6]+3.

2. Answer (2)

B(OH)3 accepts OH –, so it behaves as weak Lewis acid.





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Q.No. Solutions

3. Answer (2)

B cannot form [BF6] –3 because it does not have vacant orbitals of appropriate energy to extend

coordination sphere co-ordination number of Boron is 4.

4. Answer (2)

xexcess

362 BNNHHB

In B2H6, two 3C – 2e bonds are present.

5. Answer (4)

Due to absence of d orbital, BCl3 and CCl4 do not undergo hydrolysis.

6. Answer (1)

7. Answer (4)

8. Answer (4)

9. Answer (2)

Boron atom has very high sum of 3 Ionization energies.

10. Answer (3)

Orthoboric acid is an OH – acceptor i.e., Lewis acid.

H3BO3 + H2O [B(OH)4] –

+ H+

11. Answer (3)

Statement (2) is false

As we go down the group stability of higher oxidation state decreases and stability of lower oxidationstate increases due to inert pair effect hence Pb+2 is more stable than Pb+4 state.

Section-E

Q.No. Solutions

1. Answer A(q), B(r), C(s), D(p)

Colemanite = Ca2B6O11. 5H2O

Bauxite = Al2O3. 2H2O

Borax = Na2 B4 O7. 10H2O

Inorganic Benzene = B3N3H6





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Q.No. Solutions

2. Answer A(p, q, r, s), B(q), C(p), D(q, r)

Borax = Two borons are sp2 hybridised while 2 are sp

3 tetrahedral unit

BF3 = sp2 hybridisation

Diborane = sp3 hybridisation

Boric acid = sp2 hybridisation, triangular unit.

3. Answer A(q, r, t), B(p, r, t), C(q, r, t), D(q, s, t)

In diamond and graphite carbon is sp3 and sp

2 hybridised respectively. In silicates, silica is sp3

hybridised. Silicates are polymeric forms having 44SiO .

Section-F

Q.No. Solutions

1. Answer (8)

Na2B4O7 . 10H2O has 8 water of crystallization as Na2[B4O5(OH)4].8H2O

2. Answer (8)

3. Answer (3)

4. Answer (4)

Al

Cl

Cl

Cl

Cl Al

Cl

Cl

5. Answer (3)

Be3 Al2Si6O18

Section-G

Q.No. Solutions

1. Answer (1)Na2SiO3 is known as water glass. Lead salts are poisonous and (CO + N2) is producer gas.

2. Answer (1)

O C O O C O O C O

CO is used in Mond’s process.





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Q.No. Solutions

3. Answer (4)

4. Answer (1)

Section-H

Q.No. Solutions

1. Answer (1)

Due to backbonding BF3 becomes poorer acid.

2. Answer (1, 4)

It is prepared by B2H6 + NH3 (excess).

3. Answer (1, 2)

Factual

4. Answer (1, 2, 3, 4)

Factual

5. Answer (1, 2, 3, 4)

Structure is as

Si

O

OSi

O

On

6. Answer (3)

As two ortho-silicate units share two common oxygen atoms.

7. (a) Due to p-p back bonding, the lone pair of electrons of F is denoted to the B-atom. Thisdelocalisation reduces the deficiency of electrons on B thereby increasing the stability of BF 3

molecule.

F

B

–

F+

F

F

B

–

FF

+

F

BF

F

F

BF

F

Due to absence of lone pair of electrons on H-atom, this compensation does not occur in BH3. In

other words, electronic deficiency of B stays and hence to reduce its electronic deficiency, BH3

dimerises to form B2H6.

(b) Because of double bond character of B-F bonds in BF3, it has a shorter B-F bond length than in

BF4 –.





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Q.No. Solutions

8. OH10OBHSONa)conc(SOHOH10·OBNa 2acidTetraboric74242422

Borax742

acidBoric332742 BOH4OH5OBH

OH6OBBOH2 2oxideBoric

32heatdRe

33

MgO3B2Mg3OB 32

Structure of B2H6

B B

H

H

H

H

H

H

Bridge hydrogen

Terminal H

2raneChlorodibo52

AlCl

)Diborane( 62

HClHBHClHB 3

9. B(OH)3 reacts with NaOH to form sodium metaborate.

B(OH)3 + NaOH Na+BO2 – + 2H2O

The end point during the titration of H3BO3 with NaOH is not sharp since NaBO2 undergoes excessive

hydrolysis to give back H3BO3 and NaOH. However, when certain polyhydroxy compounds such as

catechol, glycerol, mannitol or sugars are added to the titration solution the metaborate ion combineswith polyhydroxy compound to form a complex.

OH

OHCatechol

+ BO2

–2H O2

B

O

O2

O

O

Due to the formation of this complex, BO2 – ion does not undergoes hydrolysis. As a result, boric acid

behaves as a strong monobasic acid and the end point can thus be easily detected.

10.

(a) B

F F

F

Triangular planar(No lone pair)

F

Br F

F

T-shaped(2 lone pair)

(b) Due to the absence of low lying vacant d orbital with carbon, its hydrolysis is difficult. While SiCl4 has vacant d orbital of appropriate energy.

SiCl4 + 4H2O Si(OH)4 + 4HCl

OR

H2SiO3·H2O

Hydrated silicic acid

Due to availability of vacant d -orbitals in valence shell of their central atom, they can easily extendtheir coordination number beyond four but this is not possible in case of carbon due to absence ofvacant d -orbital.





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Q.No. Solutions

11. Because of the small size of boron atom and presence of only six electron in its valence shell inB(OH)3, it coordinates with the oxygen atom of the H2O molecules to form a hydrated species.

(HO) B + O3

H

H

[B(OH) + H4] – +

or

[(HO) B OH3 2]

In this hydrated species B3+ ion because of its small size it has a high polarizing power and hence pullsthe electron of the coordinated oxygen atom towards it. The coordinate O-atom, in turn pulls the O-electron of the O-H bond, there by facilitating the release of a proton. As a result, B(OH)3 acts as a

weak monobasic lewis acid and thus reacts with NaOH solution to form sodium metaborate

B(OH) + NaOH3 Na [B(OH)+

4]

Na BO +2H O

sodium metaborate

+ –

2 2

–

12. (a) The mineral (A) is colemnite, Ca2B6O11· 5H2O

boiled2 6 11 2 3 (aq.) 3 2 4 7 2

(C) (D)(B)

Ca B O +2Na CO 2CaCO +Na B O +2NaBO Soluble

(b) (C) the borax is crystallized

The mother liquor consisting sodium metaborate is treated with CO2

4 NaBO2 + CO2 Na2B4O7 + Na2CO3

(c) OH10OBNaOH10·OBNa 2 Anhydrous

742heated

2742

strongly2 4 7 2 2 3heated

(D) (E)

Na B O 2NaBO + B O

(d) heat2 3 2 2

Cobaltmeta-borate(E)

Bluecoloured(F)

CoO+B O Co(BO )

13. Na B O 10H O Na B O2 4 2 2 4 77

2NaBO + B O Cu(BO )2 2 3 2 2

CuO

A B C D

14.B O2 3

HCl

BCl + H3 2O

NH3

A B C

B H2 6

BNNH3

D E

15.

B

HO OH

OH HO — CH2

HO — CH2

+ B

HO

HO O — CH2

O — CH2

–

+ H+

(–H O)2

p-block elements (package solutions)

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