Ebook chemistry practical application part 2

Non transition Elements 5 Non transition Elements The Structural Features of (a) B4H10 (b) B5H9(c) B5H11 (a) Tetraborane, B4 HI0 In this molecule four B atoms may be regarded as a portion of slightly[.]

Non-transition Elements

5

Non-transition Elements

The Structural Features of : (a) B4 H 10 (b) B 5 H 9 (c) B 5 H 11

octahedron The structure of this molecule contains :

(i) Fourt bridging (3c2e) BHB bonds

(ii) One direct (2c2e) BB bond

(iii) Six terminal (2c2a) BH bonds

The structure can be represented as follows :

The structure and line representation of the bonding in tetraborane, B 4 H 10

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Thus six BH electron pair, one BB electron pair and four BHB bridge bonds together account for twenty-two valence electrons contributed by four boron and ten hydrogen atoms Thus skeletal electrons are 22

located in the base of the square-pyramid are bonded to each other by four (3c _ 2e) BHB bonds while the B-atom located at the apex of the pyramid is bonded to the two basal B-atoms (2c _ 2e) BB bond This

(ii) Four bridging BHB bonds

(iii) Two BB bonds

(iv) One closed BBB bond

Total 24 skeletal electrons are involved in various bonds formation in the molecule

The geometrical structure of B 5 H 9 , and one of its four-bond resonance structures

B-atoms occupy the five comes of a square pyramid This molecule contains :

(i) Eight terminal BH bonds

(ii) Three bridging (3c - 2e) BHB bonds

(iii) Two closed (3c - 2e) BBB bonds

Non-transition Elements

The structure is shown below :

Structure of pentaborane-11, B 5 H 11

Total number of skeletal electrons are 26

The wave functions of molecular orbitals of diborane in terms of orbitals

It contains two type of bonds :

(i) Four terminal (2c - 2e) BH bonds (normal covalent a bonds)

(ii) Two bridging (3c - 2e) BHB bonds

In terms of molecular orbitals, the three centre BHB orbital may be considered to result from the

2e) BHB bonds are

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is empty and other is singly filled Other two singly filled

BH bonds A three centre bond has a banana shape

(banana bond) This is due to the repulsion between

positive charges on the two bridge hydrogen atoms, causing the three centre bonds to be bent away from each other in the middle

(A) Qualitative picture of bonding in diborane (B) A common method of depicting BHB bridges

Wave mechanical picture of diborane : Consider each borane atom to be sp 3 hybridised Two terminal

BH bonds are cr bonds involving a pair of electron each This accounts for eight of the total twelve

electrons available for bonding Each of the bridging BHB linkage then involves a delocalized or three

The diagrammatically possibilities of overlap together with the resulting MOs and their energies are given

in Fig :

Non-transition Elements

Qualitative description of atomic orbitals (left), resulting three-centre molecular orbitals (right), and the approximate energy level diagram (centre) for one BHB bridge in diborane

The preparation, properties and structure of S 4 N 4 :

S 4 N 4 , Tetrasulphur Tetranitride : It is prepared as follows :

The compound is also formed when sulphur reacts with anhydrous liquid ammonia

nitrogen temperatures it is almost colourless, but at room temperature it is orange-yellow, and at 100°C it

is red

ring :

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Structure of S 4 N 4

The structure is an eight membered heterocyclic ring and cradle shaped

Since decomposition of tetrasulphur tetranitride with alkali gives ammonia, and reduction with stannous chloride followed by decomposition with alkali also gives ammonia, there canbe no NN links in molecule

If the group NN were present, direct decomposition would yield ammonia and reduction followed by decomposition would yield hydrazine or its derivatives

SS distance = 2.63Å,

NN distance = 1.47

Band angles are SNN = 110°,

SNS = 98°, NSN = 76°

B 10 C 2 H 12 is isostructural and isoelectronic with what borane ion B x H y 2_ :

borane ion, the replacement of two carbon atoms with boron atoms must be accompanied by addition of two extra electrons

Non-transition Elements

Framework of theB 12 H 12 2_ ion

Carboranes

The carboranes are mixed hydrides of carbon and boron having both carbon and boron atoms in electron deficient skeletal framework

Geometrically carboranes are classified into two types :

(i) Those in which the boron atom framework closes in on itself to form a polyhedron These are closo

(ii) Those which have the open cage structures, derived formally from one or other of several boranes and

containing from one to four carbon atoms in the skeleton These are nido (nest) compounds

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have the same closed polyhedral structures with one hydrogen atom bonded to each carbon and boron

The nido -carboranes formally related to B6 H 10 All have eight pairs of electrons bonding the six cage atoms together Hydrogen bridges are represented by curved lines

An account of phosphazenes with their structural aspects:

Phosphazenes or Phosphonitrilic Compounds : Phosphazenes are a group of compounds represented by

polymeric, in these compounds P atom is in oxidation state (+V) and N is in the

(+ III) state The compounds are formally unsaturated

from 3 to 7 and

Non-transition Elements

upwards The monomer (x = 1) and dimer (x = 2) are not known These compounds were originally called

phosphonitrilic halides, but are now named systematically poly (chlorophosphazenes)

These can be prepared as follows :

By the ammonolysis phosphorus pentachloride

3PC15 + 3NH3 ¾¾® (NPC12)3 + 9HC1

By the reaction of ammonium chloride with phosphorus pentachloride

The chlorine atoms are reactive, and most reactions of chlorophosphazenes involve replacement of Cl by

or grignard reagents

On hydrolysis chlorine atoms of chlorophosphazenes can replaced by OH groups

Structure : X-ray examination reveals that the trimer and tetramer chlorophosphazenes are having the six

and eight membered rings and composed of alternate nitrogen and phosphorus atoms Thus these

compounds may be regarded as the phosphorus-nitrogen analogues of benzene and cyclo-octatetraene

opposite side of the rings All the NP bonds in the ring are also of equal length i.e., 1.59 ± 0.02Å

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The Structure of N 3 P 3 Cl 6 trimer

In the above structure it is evident that two chlorine atoms are attached to each phosphorus atom This is confirmed by two chemical experiments such as :

have been attached to the same phosphorus atom

d-orbital These orbitals combine to give delocalised pp - dp orbitals which extend over the whole ring

structure The presence of negative chlorine atoms on

Non-transition Elements

phosphorus makes the diffuse phosphorus d-orbitals more compact and favours their overlap

As shown below, resonance structures can be drawn analogous to those for benzene indicating aromaticity

in the ring

The resonance in the trimer is justified by the fact that all the NP distances are the same (1-67 Å) which are less than that expected for a PN single bonds (1.78°A) as in sodium phosphoramidate

nitrogen atoms, with two chlorines on each phosphorus atom

The Structure of trimer and tetramer of phosphonitrilic chlorides

Flat structure of (NPCl 2 ) 3 ; Chair and bot form of (NPCl 2 ) 4

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The reaction completed and given the structure of the main product :

[PNC12]3 + excess (CH3)2 NH ® [P3N3{(CH3)2 N}6]

Balanced Reactions

(i) Reaction of potassium superoxide with carbon dioxide

(ii) Hydrolysis of phosphorus sulphide

(iii) Reaction of sodium chlorite with nitrogen trichloride

(iv) Hydrolysis of calcium cyanamide

(v) Reaction of borontrioxide with cobalt oxide

(vi) Hydrolysis of phosgene

(vii) Reaction of carbon tetrachloride with hydrogen fluoride in anhydrous conditions (viii) Reaction of hydrazine with zinc in acidic medium

(ix) Diborane reacts with ammonia

(x) Silica reacts with carbon in an electric furnace

(xi) Reaction of ammonia with disulphur dichloride

(xii) Reaction of bromate and chloride in acid medium

(xiii) Hydrolysis of nitrogen trichloride

(xiv) Reaction of bromate and bromide in acid solution

(xv) Hydrolysis of borazine

(xvi) Hydrolysis of silicon tetrafluoride

(xvii) Hydrolysis of tetrasulphur tetranitride

Non-transition Elements

produces oxygen and removes carbondioxide Both functions are important in life support system

Cobalt metaborate

Blue colour

Phosgene

Feron

(ix) B2H6.2NH3

Si + C ¾¾® SiC

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The reaction steps and condition for the following conversions :

(i) PC1 5 to poly-dichlorophosphazene

(ii) CO to Urea

like tetrachloroethane

The reaction steps may be

unstable

(ii) Conversion of CO to Urea : It takes place in following steps :

Carbonylchloride (phosgene)

Non-transition Elements

(i) 3BC1 3 + 3NH 4 C1 ¾¾®

(ii) A1 2 (CH 3 ) 6 + 2H 2 O ¾¾®

(in) NO + O 3 ¾¾®

(iv) n[(CH3 ) 2 SiO 4 ] + (CH 3 ) 3 SiOSi (CH 3 ) 3

(v) [3Ca 3 (PO 4 ) 2 ] CaF 2 + 7H 2 SO 4 ¾¾®

(vi) PtF 6 + O 2 ¾¾®

(vii) SbF 5 + BrF 3 ¾¾®

(viii) PI 3 + 3H 2 O ¾¾®

Correct balanced equation is

(vi) PtF6 + O2 ¾¾® O2+ [PtF6]_

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The balanced equation for the reactions of the following with water under ambient conditions:

(i) PC1 3 (ii) NC1 3 (iii) BrF 3 (iv) Al 4 C 3 (v) SF 6

Explained :

(i) BF bond is larger in BF 4 _ than in BF 3 molecule

(ii) The order of Lewis acid strength of different halides of boron is

BF 3 < BC1 3 < BBr 3

or, BBr 3 is a stronger Lewi's acid than BF 3

(iii) N(CH 3 ) 3 is pyramidal in shape while N(SiH 3 ) 3 has planar triangular arrangement

or

(SiH 3 ) 3 N is a weaker base than (CH 3 ) 3 N

(iv) Dipole moment of NH 3 molecule is larger than that of NF 3

(v) PF 3 can act as donor molecule while NF 3 show little tendency to act as donor

(vi) C1F 3 exists whereas FC1 3 does not

Non-transition Elements

(vii) Ba(OH) 2 is fairly soluble in water but Mg (OH) 2 is not

(viii) Dipole moment of CH 3 CI is greater than that of CH 3 F

(ix) HClO 4 is an acid and an oxidising agent whereas H 2 C 2 O 4 is an acid and a reducing agent

(ii) The order of Lewis acid strength of different halides of boron is

This order is just reversed than that expected on the basis of electronegativity values of halogens The electron density from the filled orbitals on halogens is transferred to the empty orbital present on the

boron atom This is known as back donation This is explained on the basis of overlapping of the 2p-filled orbital of halogens sidewise with the empty 2p-orbital of boron atom forming pn-pn back bonding

Due to back bonding, the electron deficiency in the boron atom is decreased, consequently, its Lewis character or electron accepting tendency decreases Since back bonding is maximum in case of fluorine

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filled orbitals of Cl and Br atoms with empty 2p-orbitals of B atoms does not take place effectively due to difference in the energy state of the orbitals involved

due to lone pair-bond pair repulsion

moments of NH and NF bonds In the first case N is more electronegative but in second case F is more electronegative as shown below

Thus in NH3, the dipole moments of NH bonds are in the same direction as that of the lone pair but in

is less electronegative than nitrogen and phosphorus has a larger size than nitrogen Due to greater

electronegativity N cannot easily part with the lone pair of electrons

orbital of the metal to the empty orbital of F

show an oxidation state of +3 and combine with more electronegative fluorine Outer electronic

fluorine being the most electronegative element, it shows an oxidation state of _ l only

group (alkaline earth metals) Thus in general the solubility of the alkaline earth metal hydroxides in water increases with increase in atomic number down the group This is due to the fact tot lattice energy decreases down the group due to increase in size of alkaline earth metal cation whereas the hydration energy of the cations remains almost unchanged Thus AH solution becomes more negative as we move

due to greater electronegativity of fluorine atom than chlorine but actually CF bond length is much

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agent Chlorine is more electronegative than carbon and hence can accept electron and get converted to

Recovery of elemental silver from silver resides from photographic processing (AgCl) is achieved by converting it into A, using common ionic compound B The compound A upon heating decomposes to give an intermediate compound C before giving metallic silver as the end product A, B and C by giving equations for the reaction involved, identified :