[DOCUMENT] discusses inorganic chains, rings, cages, and clusters. It provides examples of:
- Silicate chains including different types of silicates based on their structure.
- Borazine and phosphazenes which are cyclic like benzene.
- Cages including white phosphorus, boranes, and carboranes.
- Polymeric sulfur nitride ((SN)x) which is a superconductor.
- Cluster valence electron theory and Wade's rule for classifying boranes and carboranes.
1 of 39
More Related Content
Inorganic chains, rings, cages and clusters
1. 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
3. 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)
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
4. DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
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
5. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
6. 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’.
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
7. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
BORAZINE
8. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
Phosphazenes
9. 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.
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
S4N4
10. 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-
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
Homocyclic Rings
11. 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.
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
CAGES
12. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
Boranes
14. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
PENTABORANE(9) - B5H9
STYX Code is 4120
STYX Code is 4012
TETRABORANE - B4H10
15. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
STYX Code is 3203 STYX Code is 4220
PENTABORANE(11) - B5H11
16. 4 B-H-B bonds
6 B-B-B bond
2 B-B bond
0 BH2 groups
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
STYX Code is 4620
DECABORANE(14) – B10H14
17. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
18. 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).
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
Number of electron
pairs
Type of Borane/
Carborane
n+1 Closo
n+2 Nido
n+3 Arachno
n+4 Hypo
19. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
20. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
21. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
22. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
23. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
24. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
25. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
26. 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.
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
27. 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.
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
28. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
Four B-H-B bonds
One B-B-B bond
Two B-B bonds
Zero BH2 group
29. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
C2B10H12 = B12 H12
2- is icosahedral borane.
BnHn
2- is closo borane
30. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
31. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
32. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
33. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
34. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
35. Q. No 16: All forms of phosphorous upon melting, exist as [CSIR-UGC-NET
Jun2015]
ANSWER: WHITE PHOSPHOROUS
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.
36. 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
DR. A. ANTO AROCKIA RAJ, SXC, PALAYAMKOTTAI.