Inorganic chains, rings, cages and clusters

TRICKS & TIPS FOR CSIR-NET, SET, GATE IN
CHEMICAL SCIENCES
Dr. A. Anto Arockia Raj
Assistant Professor& Coordinator(S-II)
Department of Chemistry
St. Xavier’s College(Autonomous)
Palayamkottai-627002

INORGANIC CHAINS, RINGS, CAGES &
CLUSTERS
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
CHAINS
• Silicates
• polythiazyl compound
RINGS
• Borazine
• Phosphazenes
• Tetrasulphurtetranitride
• Homocyclic inorganic systems
CAGES
• White phosphorous
• Boranes
• Carboranes
CLUSTERS
• Organometallic clusters

SILICATES
Silicates are the minerals containing silicon and oxygen in tetrahedral
SiO4
4- units, which are linked together in several patterns.
TYPES OF SILICATES:
Ortho silicates (Neso silicates)
Pyro silicates (Soro silicates)
Cyclic Silicates (Ring silicates)
Simple Chain Silicates (Pyroxenes)
Double Chain Silicates (Amphiboles)
Sheet Silicates (Phyllo silicates)
Three Dimensional Silicates (tectosilicates)

 Ortho Silicates – SiO4
4-- Phenacite (Be2SiO4),
Willemite (Zn2SiO4),
Zircon (ZrSiO4)
 Pyro Silicates - Si2O7
6- - Thortveitite (Sc2Si2O7)
 Cyclic Silicates - (SiO3)n
2n- - Benitonite (BaTi(SiO3)3)
 Chain Silicates – Si2O6
4- - Spodumene Li Al(SiO3)2,
Wollastonite Ca3(SiO3)3
 Double Chain Silicates - (Si4O11)n
6n- - Asbestos,
Tremolite Ca2Mg5(Si4O11)2(OH)2
 Sheet Silicates - (Si2O5)n
2n- - Talc Mg3Si4O10(OH)2
Mica X2Y4–6Z8O20(OH,F)4
Where X = K, Na, or Ca
Y = Al, Mg, or Fe
Z = Si or Al
Muscovite mica - KAl2(AlSi3O10)(F,OH)2

3D-SILICATES
The general formula of three dimensional (3-D) or tecto or Framework silicates is (SiO2)n .
All the oxygen atoms of SiO4 are shared with other tetrahedra and thus by forming three-dimensional
network.
E.g. SiO2 - Quartz, Tridymite and Crystobalite - These are the crystalline forms of silica.
When SiO4
4- units are replaced by AlO4
5- units, three dimensional aluminosilicates are formed.
E.g. Feldspar, Zeolites, Ultramarine
Feldspar- Rock forming minerals
Zeolites- Ion Exchanger, Molecular sieves
Ultramarine- Blueing agents- Lapis lazuli- S3
-,S2
-
ZSM-5 is a Shape selective Catalyst- conversion of methanol
into gasoline. Toluene into p-xylene etc.
High temperature conversion of NO (diesel, auto) into N2 and O2
By CuI/CuII with Zeolites

POLYMERIC SULPHUR NITRIDE (SN)X
S4N4 + 8 Ag → 4 Ag2S + 2 N2S4N4 → (SN)x
 It is a golden bronze colored lustrous material- Superconductor
 It conducts electricity at very low temperature i.e. below 0.26 K
 The chain can be generated from S2N2 of Square planar geometry
 Free Radical Mechanism involved.
 The bond order of S-N bond is 1.5, Bond length is 164.5 pm
 Unless Octet, one odd electron present in pi* antibonding orbitals
 Conduction band will lie along the direction of (SN)x fibers leads to ‘One Dimensional Metal’.

 Borazine, also known as borazole - B3H6N3 – colourless liquid
The compound is isoelectronic and isostructural with benzene.
Hence it is called as ‘Inorganic Benzene’.
 3 B2H6 + 6 NH3 → 2 B3H6N3 + 12 H2
 3 BCl3 + 3 NH4Cl → Cl3B3H3N3 + 9 HCl
 2 Cl3B3H3N3 + 6 NaBH4 → 2 B3H6N3 + 3 B2H6 + 6 NaCl
This is a misnomer of benzene because their chemical properties are different.
Boron acts as Lewis acid and Nitrogen as Lewis base
BORAZINE

 When ammonium chloride heated phosphorous pentachloride
at 146℃ to give Phosphonitrilic chlorides
NH4Cl + PCl5 → PNCl2
It is analogous to benzene and borazine with the formula
(NPX2)n, where n = 3,4…8.
Aromatic like benzene. N is sp2 & P is sp3 hybridised respectively.
Trimer is cyclic, planar but not all Phosphazenes.
Craig and Paddock model- p- d interaction.
Dewar model- island model
Phosphazenes

Golden colour Solid- M. Pt. 187 °C - Molar mass :184.287 g mol−1
6 S2Cl2 + 16 NH3 → S4N4 + S8 + 12 NH4Cl
4 NH4Cl + 6 S2Cl2 → S4N4 + 16 HCl + S8
Tetra sulfur tetranitride can form many sulfur nitrogen compounds such as S2N2, S8 , (SN)x etc.
2 S4N4 → 4 N2 + S8
S4N4 serves as a Lewis base by binding through nitrogen to strongly Lewis acidic compounds such as SbCl5
and SO3
S4N4 reacts with Vaska’s Complex [Ir(Cl)(CO)(PPh3)2] in an oxidative addition reaction to form a six
coordinate Iridium complex where the S4N4 binds through two sulphur atoms and one nitrogen atom.
S4N4

Polysulfide anions are open chain structuresThe structure of S4
2+,
Se4
2+, Te4
2+ ions are isoelectronic & isostructural with with S2N2
It is square planar in geometry. Huckel sextet of electrons
An important oxocarbon anions of formula [CO]n
2-,4- are as
follows
Squarate ion- C4O4
2-
Croconate ion – C5O5
2- -
first inorganic compound - Gmelin in 1825- same year of
isolation of Bz.
It is aromatic
It is a bacterial metabolic product
Rhodizonate ion- C6O6
2-
Homocyclic Rings

The simplest cage type of molecule is white Phosphorous (P4)
Two crystalline forms - form is BCC &  form is HCP
It is tetrahedron with six P-P bonds. It is quite reactive, reacts with oxygen to form P4O10.
White phosphorous can be converted into red (300℃ in sunlight)
and black phosphorous(under high pressure) respectively.
Red phosphorous has indefinite & Black Phosphorous is layered structure.
CAGES

Diborane is a simplest boron hydride
It is an electron deficient compound
8 BF3 + 6 NaH → B2H6 + 6 NaBF4.
NaBF4 + I2 → B2H6 + 2NaI + H2
KBH4 + 2 H3PO4 → B2H6 + 2KHPO4 + 2H2
4 2C-2e- bonds & 2 3C- 2e- bonds.
BONDING IN DIBORANE
Boranes

STYX CODE

4 B-H-B bonds
0 B-B-B bond
1 B-B bond
2 BH2 groups
4 B-H-B bonds
1 B-B-B bond
2 B-B bond
0 BH2 groups
PENTABORANE(9) - B5H9
STYX Code is 4120
STYX Code is 4012
TETRABORANE - B4H10

3 B-H-B bonds
2 B-B-B bond
0 B-B bond
3 BH2 groups
HEXABORANE(10) – B6H10
4 B-H-B bonds
2 B-B-B bond
2 B-B bond
0 BH2 groups
STYX Code is 3203 STYX Code is 4220
PENTABORANE(11) - B5H11

4 B-H-B bonds
6 B-B-B bond
2 B-B bond
0 BH2 groups
STYX Code is 4620
DECABORANE(14) – B10H14

WADE’S RULE
Wade’s rules are used to rationalize the shape of borane clusters by
calculating the total number of skeletal electron pairs (SEP) available for
cluster bonding. BORANES classified into following structural categories:
1) Closo- BnHn
2- - n vertices of polyhedron are occupied by boron atoms
2) Nido – BnHn+4 – one vertex is missing from closo borane
3) Arachno - BnHn+6 – two vertices are missing from closo borane
4) Hypo - BnHn+8 – three vertices are missing from closo borane

WADE-MINGOS RULE
STEP 1: Total number of Skeletal electrons for BH- 2 e-, CH-3e-, H- 1e-, anionic charge has to be added.
STEP 2: Calculate number of electron pairs i.e. Total Skeletal electrons/2
STEP 3: Number of vertices is given by ‘n’ (Number of B & C atoms). Then compare number of electron pairs with number of
vertices(n). Thus the cluster structure derived as
Example: B5H11
Step 1: TEC = 2(B-H) + 3 (C-H) + 1(H) + anionic charge
 = 2(5) + 3(0) + 1(6) + 0 = 10+ 6 = 16.
Step 2: Number of electron Pairs = TEC/ 2 = 16/2 = 8
Step 3: n = number of B+ C atoms = 5+0 = 5.
Compare step 2 & 3 thus the structure belongs to n+ 3 (5+3=8)
Arachno structure (Two vertices are missing).
Number of electron
pairs
Type of Borane/
Carborane
n+1 Closo
n+2 Nido
n+3 Arachno
n+4 Hypo

CLUSTER VALENCE ELECTRON THEORY
STEP 1: count total number of valence electrons (B-3, C-4, H-1 electrons, add anionic charges)
STEP 2: compare n with
4n+ 2 = closo cluster 14n+ 2 = closo cluster
4n+4 = Nido cluster 14n+4 = Nido cluster
4n+6 = Arachno cluster 14n+6 = Arachno cluster
4n + 8 = Hypo cluster 14n + 8 = Hypo cluster
1) [Pb10]2-
step 1: TVEC = 10 (4) + 2 = 42
Step 2: n= 10, 4n+2 = 42, Hence it is closo.
2) Rh6(CO)16
Step 1: TVEC = 6(9) + 16 (2) - 6 X 10 =26 Step 1: TVEC = 6 (9) + 16 (2) = 54 + 32 = 86
Step 2: n= 6, 4n+2= 26 Hence it is Closo Step 2: Reminder of 86/14 is 2, Hence it is Closo Cluster

Q. No. 1: The Material that exhibits the highest electrical conductivity among the following
sulfur-Nitrogen compound is ----- [CSIR-NET/JRF JUNE DEC 2011]
A) S4N4
B) S7NH
C) S2N2
D) (SN)x
Answer: D) (SN)x

Q. No 2: The number of skeletal electrons present in the compounds C2B3H5, C2B4H6 and B5H9 are
respectively [CSIR NET JUNE 2016]
A) 10,12 & 12
B) 12, 14 & 14
C) 10, 12 & 14
D) 12, 14 & 12
ANSWER: B) 12, 14 & 14
C2B3H5 = B5H7 = 5(B-H) + 2H = 5(2)+ 2(1) = 12
C2B4H6 = B6H8 = 6 (B-H)+ 2(H) = 6(2) + 2(1) = 14
B5H9 = 5 (B-H) + 4(H) = 5(2)+ 4 (1) = 14

Q.NO 3: Among the given boranes and hetero boranes, the example which belongs to ‘Closo’ type is---
[GATE2016]
A) B5H8
-
B) [C2B9H11]2-
C) GeC2B9H11
D) B6H10
ANSWER : C) GeC2B9H11
GeC2B9H11 = (B-H) + (B-H)2 +B9H11
= B12H14 = BnHn+2 is Closo

Q. No 4: Ammonolysis of S2Cl2 in an inert solvent gives--- [GATE 2015]
A) S2N2
B) S2N2Cl2
C) S2N2H4
D) S4N4
ANSWER: D) S4N4
6 S2Cl2 + 16 NH3 → S4N4 + S8 + 12 NH4Cl
4 NH4Cl + 6 S2Cl2 → S4N4 + 16 HCl + S8

Q.No 5: An example of nido-borane from the following is [GATE 2014]
A) B4H10
B) B6H10
C) B6H12
D) B8 H14
ANSWER: B) B6H10
B6H10 = BnHn+4 is nido

Q. No 6: BCl3 and NH4Cl were heated with 140 °C to give compound ‘X’, which when treated with NaBH4
gave another compound ‘Y’. Compound X and Y are---- [GATE 2014]
A) X= Cl3B3H3N3 Y= B3H6N3
B) X= Cl3B3H9N3 Y= B3H6N3
C) X= Cl3B3H3N3 Y= B3H12N3
D) X= Cl6B3N3 Y= B3H6N3
ANSWER: A) X= Cl3B3H3N3 Y= B3H6N3
 3 BCl3 + 3 NH4Cl → Cl3B3H3N3 + 9 HCl
 2 Cl3B3H3N3 + 6 NaBH4 → 2 B3H6N3 + 3 B2H6 + 6 NaCl

Q.NO 7: The cluster having Arachno type structure is --------- [CSIR-NET/JRF JUNE 2012]
A) [Os5(CO)16]
B) [Os3 (CO)12]
C) [Ir4 (CO)12]
D) [Rh6(CO)16]
ANSWER: B) [Os3 (CO)12]
STEP 1: TVE = 3 (8) + 12(2) – 3(10)= 24 + 24 – 30 = 18 TVEC = 3 (8) + 12(2) = 48 , 48/14 remainder=6
STEP 2: n=3, 4n+6 = 18, Hence it is Arachno n=3, 14n+6 arachno cluster.

Q.No 8: Structure of a carborane with formula, C2B4H8 is formally derived from [CSIR-NET/JRF DEC
2012]
A) closo borane
B) nido borane
C) Arachno borane
D) conjucto borane
ANSWER: B) nido borane
C2B4H8 = 2 (B-H) + B4H8 = B6H10= BnHn+4 . Nido structure.

Q. NO 9: For higher boranes 3c-2e- ‘B-B-B’ bond may be a part of their structures. In B5H9 the number
of such electron deficient bond (s) present is/are [CSIR-UGC-NET/ JRF
DEC2013]
A) 4
B) 2
C) 0
D) 1
ANSWER: D) 1
Four B-H-B bonds
One B-B-B bond
Two B-B bonds
Zero BH2 group

Q. No. 10: T he number of isomeric derivatives possible for the neutral closo-carborane C2B10H12
[CSIR-UGC-NET/ JRF DEC2013]
A) Three
B) Two
C) Four
D) Six
ANSWER: A) Three
C2B10H12 = B12 H12
2- is icosahedral borane.
BnHn
2- is closo borane

Q. No 11: According to Polyhedral electron count rule, the structure of Ir6 (CO)16 [ GATE 2013]
A) Closo
B) Nido
C) Arachno
D) Hypo
Answer: A) Closo
Ir- 5d76s2 = nine electrons
2) Ir6(CO)16
Step 1: TVEC = 6(9) + 16 (2) - 6 X 10 =26 Step 1: TVEC = 6 (9) + 16 (2) = 54 + 32 = 86
Step 2: n= 6, 4n+2= 26 Hence it is Closo Step 2: Reminder of 86/14 is 2, Hence it is Closo Cluster

Q. No 12: Heating mixture of ammonium chloride and sodium tetrahydridoborate gives one liquid product
(X), along with other products at ambient conditions. Compound X is ---- and is an example of------
[ GATE 2013]
A) NH4[BH4] a) ionic solid
B) [(NH3)2 BH2][BH4] b) saturated Heterocyclic compound
C) B3N3H6 c) Molecular cage
D) B3N3H12 d) Unsaturated heterocyclic compound
ANSWER: C) B3N3H6 and d) Unsaturated heterocyclic compound
 3 BCl3 + 3 NH4Cl → Cl3B3H3N3 + 9 HCl
 2 Cl3B3H3N3 + 6 NaBH4 → 2 B3H6N3 + 3 B2H6 + 6 NaCl

Q. No 13: The total Valence electron and structure type adopted by the complex[Fe5(CO)15C] respectively
are [CSIR-UGC-NET/JRF JUN 2014]
A) 74 & Nido
B) 60 & Closo
C) 84 & Arachno
D) 62 & Nido
ANSWER: A) 74 & Nido
[Fe5(CO)15C]: TVE = 5(8) + 15 (2) + 4 = 40+ 30+4 = 74; 74/14 remainder is 4
n = 5 if 14n+4 is nido

Q. No 14: Co4(CO)12 adopts the structure---- [CSIR-UGC-NET DEC 2014]
o A) closo
B) nido
C) arachno
D) hypo
ANSWER: B) nido
Co4(CO)12 : TVE = 4(9) + 12 (2) – 4(10) = 36 + 24 – 40 = 20; n=4, 4n+4 is Nido

Q. No 15: A borane (X) reacts with ammonia to give salt of borohydride Y, the B11 NMR
spectrum of Y consist of a triplet and a quintet. The borane X is - [CSIR-UGC-NET DEC 2014]
A) B2 H6
B) B3H9
C) B4H8
D) B5H9
ANSWER: A) B2 H6

Q. No 16: All forms of phosphorous upon melting, exist as [CSIR-UGC-NET
Jun2015]
ANSWER: WHITE PHOSPHOROUS

Q. No 17: Total Number of vertices in metal clusters [Ru6C (CO)17], [Os5 C (CO)15] and [Ru5 C (CO)16] are 6, 5
and 5 respectively. The predicted structures of these complexes respectively are
[CSIR-UGC-NET Jun2015]
oA) closo, nido and nido
B) closo, nido and arachno
C) arachno, closo and nido
D) arachno, nido and closo
ANSWER: B) closo, nido and arachno
[Ru6C (CO)17] TEC= 6(8)+4+ 17(2)= 48+4+34 = 86 ; PEC= 86 – 6x12/2= 86-72/2 = 14/2 = 7 (n+1)= closo
[Os5 C (CO)15] TEC= 5(8)+4+ 15(2)= 40+4+30 = 74 ; PEC= 74 – 5x12= 74-60 = 14/2 = 7 (n+2)= nido
[Ru5 C (CO)16] TEC= 5(8)+4+ 16(2)= 40+4+32 = 76 ; PEC= 76 – 5x12= 76-60 = 16/2 = 8 (n+3)= arachno

QUESTIONS FOR YOU

THANK YOU

Inorganic chains, rings, cages and clusters

More Related Content

Inorganic chains, rings, cages and clusters