1. B - Group

2. C - Group

1) B - Group

It is interesting to note that the first member of each group differs in many respects from the other members.

These differences are quite striking in Groups 13-16.

The cumulative effects of small size, high electro-negativity and non-availability of d-orbitals for the first member are responsible for these differences.

a) Inert Pair Effect

The p-block elements display two oxidation states.

This is in sharp contrast to the s-block elements that display only one oxidation state, the group number.

- The higher oxidation state is equal to the group number minus 10 (i.e. number of s and p electrons in the valence shell) and the lower one is two units less than the group number (i.e. number of p-electron in the valence shell).

- The lower oxidation state becomes more stable on descending the group. This is referred to as the Inert Pair Effect

- The higher oxidation state is displayed only when both the ns and np electrons are involved in bond-formation.

- On the other hand, the lower oxidation state is observed when only the np electron(s) participate in bond formation.

- On moving down the group the ns electrons tend to remain inert and do not participate in bond formation.

- This reluctance of the outermost s orbital electron pair to participate in bond formation is called Inert Pair Effect

Physical properties of Group 13 Elements

1.1) Boranes:

Structure and Bonding of Boranes

Rather than exhibiting the simple chain and ring configurations of carbon compounds, the boron atoms in the more complex boranes are located at the corners of polyhedrons, which can be considered either as deltahedrons (polyhedrons with triangular faces) or deltahedral fragments.

Developing an understanding of these boron clusters has done much to help chemists rationalize the chemistry of other inorganic, organometallic, and transition-metal cluster compounds.

1.1.1) The Boron Hydrides

- The elements form tri-hydrides (MH3), the stability decreases on moving down the group.

- They are electron deficient compounds.

- Boron forms a series of volatile hydrides called boranes (by analogy with alkanes and silanes).

They fall into two series:

Bn Hn+4 : B2H6, B5H9, B6H10, B8H12, B10H14

Bn Hn+6: B4H10, B5H11, B6H12, B9H15

They are named by indicating the number of boron atoms.

If two or more boranes have the same number of B atoms, then the H atoms are also specified,

Example : B5H9 and B5H11 are named pentaborane - 9 and pentaborane - 11 respectively

Diborane is the simplest and most extensively studied hydride.

It is an important reagent in synthetic organic chemistry.

It may be prepared by various ways:

a) Boron tri-fluoride etherate

b) Sodium borohydride

2NaBH4 + H2SO4 / 2H3PO4 → B2H6 + 2H2 + Na2SO4 / 2Na2HPO4

- It is a colourless gas, which burns in air and is readily hydrolyzed.

B2H6 + 3O2 → B2O3 + 3H2O

B2H6 + 6H2O → 2H3BO3 + 6H2

Diborane undergoes addition reaction with alkenes and alkynes in ether at room temperature to form organo-boranes.

6RCH = CH2 + B2H6 → 2B(CH2CH2R)3

The terminal B- H bond distances are the same as in non-electron deficient compounds. These are normal two centre - two – electron bonds (2c-2e)

Electron deficiency is thus associated with the bridge bonds.

The four bridge bonds involve only four electrons – a pair of electrons is involved in binding three atoms – B, H and B. These bonds are called three-centre-two-electron-bonds (3c-2e)

- Each B atom is sp3 hybridized giving four sp3 hybrid orbitals.

- B has three valence electrons so three orbitals are filled singly.

- Two of the sp3 hybrid orbitals on each B overlap with the 1s orbitals of H forming four 2c – 2e bonds.

- Then one singly filled sp 3 hybrid orbital on one B atom, and one vacant sp3 hybrid orbital on another B atom overlap with a singly filled 1s orbital on one H atom to form a bonding orbital shaped like a banana embracing all three nuclei Another 3c – 2e bond is formed similarly

STYX codes generally employ to find the structure of Boranes as shown below :

p = number of 2c-2e B-H bonds = B atoms = normal terminal H atoms (i. e., one per B atom).

q = number of additional H atoms beyond those in p.

c = net charge = number of protons added or subtracted.

s = number of 3c-2e B-H-B bonds.

t = number of 3c-2e B-B-B bonds.

y = number of 2c-2e B-B bonds.

x = number of additional 2c-2e B-H bonds beyond those in p.

Bonding in closo-boranes:

1.2) Boron Halides

B-F bond is much shorter than the single bond covalent radii predict.

Dative π delocalization of the fluorine lone pairs to the vacant p orbital of boron.

Effect is much reduced for Cl, Br, and I.

Dative π bonds occur for oxygen and nitrogen.

The electron deficiency on boron increases and the relative acid strength follows the order:

BF3 < BCl3 < BBr3 < BI3

a) Boron Subhalides

- B2X4 are planar in the solid state, only B2F4 remains planar in the gas phase.

- The formation of an "extended π system" is more important than steric effects for the small F.

1.3) Boron-nitrogen compounds

The B-N bond is isoelectronic with the C-C bond and parallels between boron-nitrogen compounds with organic compounds are known.

One of the best known of these pairs is benzene and the isoelectronic borazine also known as inorganic benzene, B3N3H6

The compounds are structurally similar and also exhibit similar physical properties.

However, the chemical properties are widely different.

The nature of the л-bond in borazine and benzene differ.

In benzene the л – bond is formed by sideways overlap of the 2p orbitals of carbon atoms and the two atoms involved in л bond formation do not differ in electro negativity.

In borazine the nitrogen atom donates an unshared electron pair to a vacant p orbital on B, thereby forming a л bond

The polarity of the B-N bond is less than what is expected on basis of electro negativity difference between boron and nitrogen.

This is because during σ - bond formation the electron density shifts towards the more electronegative atom, nitrogen while the reverse happens during л bond formation.

The molecule however is polar and undergoes addition reactions.

This is in sharp contrast to benzene, which does not undergo addition reaction.

Another interesting compound is boron nitride, BN.

Like carbon it exists in two forms – a diamond – like form and another form like graphite, which comprises of six-membered rings fused together.

Similarity in structures of Boron Nitride and Graphite

1.4) Organometallic Compounds

Tri-alkyl and tri-aryl derivatives of boron, aluminium, gallium and indium are known.

The tri-alkyl derivatives of aluminium are important and are obtained by treating aluminium chloride with the appropriate Grignard reagent.

AlCl3 + 3RMgI → AlR3

These are electron deficient compounds and exists as bridged dimers.

The terminal Al-C bond lengths are shorter than the bridge Al – C bond lengths.

The bridge bonds are three – centre – two – electron bonds.

Aluminium tri-ethyl is referred to as Ziegler catalyst and is used to carry out polymerization of ethene to give polythene

Polymerization is quicker if TiCl4 (Natta Catalyst) is used along with Ziegler catalyst and the reaction does not require high pressure.

Vast quantities of polythene (over 15 million tons annually) are made by Ziegler- Natta catalysis.

Long chain alcohols can be obtained from ethane and aluminium tri-alkyl.

These are converted to sulphonates and used in biodegradable detergents.

Aluminium alkyls catalyze the dimerization of propene to isoprene.

1.5) Complex Formation

The Group 13 elements form complexes more readily than the s-block elements.

Because of lack of d orbitals, boron exhibits a coordination number of 4 and forms tetrahedral complexes like NaBH4, HBF4, BF3.NH3 etc.

The lower members form octahedral complexes also : Example [AlCl6]2- [GaCl6]2- etc.

Apart from complexes, aluminium forms double sulphates (alums) of the general formula M2SO4.Al2(SO4).24H2O where M is a monovalent metal or ammonium

Example : K2SO4.Al2(SO4).24H2O is potash alum.

Aluminium can be replaced by cations of comparable size and charges like iron, chromium etc.

1.6) Anomalous behaviour of Boron

The chemistry of boron closely resembles that of silicon (diagonal relationship).

In fact the similarity between boron and silicon is more than that of boron and the lower members of Group 13.

This is exemplified by the fact that boron is a non-metal while Al, Ga, In and Tl are metals.

Since boron is a non- metal its oxide B2O3 (like SiO2) is acidic, whilst Al2O3 and Ga2O3 are amphoteric and In2O3 basic.

Boron like silicon forms hydrides which are volatile liquids; in contrast (AlH3)n is a polymeric solid.

The halides of B and Si undergo ready hydrolysis while aluminium halides undergo partial hydrolysis.

The oxoanions of boron and silicon – borates and silicates polymerize by sharing oxygen atoms to give chains and rings.

No such compounds of aluminium are known.


Problem : P4N4Cl6(NHC6H5)2

A2X2 Spin system two type of P – centre 2 triplet 31P NMR

Problem : B2H6

Two types of protons

4 – terminal and 2 – Bridging

11B = I = 3/2

Intensities → 4 : 2

The two B – center are equivalent and they give one signal which split in triplet by the two equivalent terminal proton.

= 2 × 2 × 1/2 +1 = 3 by two equivalent bridging proton (JB – Ht > JB – H6) → 11B NMR of B2H6 gives triplet of triplets

(Considering the coupling with 11B): I = 3/2

= 2 × 3/2 + 1 = 4 → quartet

First order effect →

11B → 3 × 3 = 9 (Triplet of triplet)

Hterminal → 4 × 3 = 12 (quartet of triplet)

Hbridging → 2 × 2 × 3/2 + 1 = 7(septet)

→ 7 × 5 (Hterminal) = 35(septet of quintet)

First order effect :

1) quintet of triplet for terminal proton

2) septet of quintet for the bridging proton

11B NMR spectrup give triplet of triplet

Problem: NF3 → pyramidal → C3V as in PF3

Because of the rapid nuclear quadruple relaxation of 14N(I = 1) the single 19F – NMR peak is not split by 14N at very low temperature(-250 0C).

It gives a single unsplite peak for 19F but with the increase of temperature the NMR peak start to broaden at about 20 0C it gives a sharp triplet at higher temperature i.e. 19F-NMR peak is split into a triplet peak by 14N (I = 1).

Problem: W2O2F9- : (intensits 8 : 1)

Problem: B3H8-

11B = I = 3/2

→ 2nI + 1m= 2 × 3 × 3/2 + 1 = 7

8 equivalent proton gives single PMR signal which is split into 7 fine components by three 11B (I = 3/2) nuclei

Problem: B10H14:

No. of signal 4, intensity ratio 4 : 2 : 2 :2

Problem: [FCl2C – CCl2F] – 1,2 difluorotetrachloroethane

At room temperature a) only one 10F NMR signal at very low temperature (about – 120oC) two 19F NMR signal

Quadrapole nuclei N and Cl do not couple with the methyl proton (11B = I = 3/2)

7 – components 2 × 2 × 3/2 + 1 = 7