csirugcjrfnetchemistry

Prepared by V. Aditya vardhan adichemadi(at)gmail.com WARANGAL

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This part (Series-2 from Q.no 51 - 60) is updated on 22nd Nov, 2009 and going to beupdated frequently. For recent updates and other parts, join the following group

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SOLVEDCSIR UGC JRF NET

CHEMICAL SCIENCESPAPER 1 (PART-B)

SERIES-2NOTE: Related and additional questions appeared in previous GATE exams are alsosolved.

51 52 53 54 55 56 57 58 59 60

51) CO bond order is lowest in1. uncoordinated CO 2. CO bonded to one metal3. CO bridging two metals 4. CO bridging three metals

Explanation:* Carbon monoxide is a -donor as well as -acceptor. It not only forms -bond and alsoforms a -bond with the metal as a result of back donation of electrons from metal. This bond is formed due to overlapping of filled d-orbital of metal with empty * -LUMO ofcarbon monoxide. Due to this -back donation, the bond order of carbon monoxide de-creases.* The bond order decreases with the number of metal centres with which carbon monoxideforms bonds. Hence the bond order in CO bridging three metals is lowest.

Additional information:Molecular Orbital Energy Level Diagram of Carbon monoxide (CO)* A rather advanced version of MOELD for CO is shown below.


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CO O2s

2px2pz 2py

C

Ener

gy

2s

2pz2px 2py

nb

*nb

* *

*

HOMO

LUMO

very much increase in energy

very much decrease in energy

* Because of some interaction between 2s of carbon and 2pz of oxygen, the energy of one

-orbital is decreased very much and at the same time, the energy of one of * is alsoincreased very much.* Also look at non bonding nature of one and one * molecular orbitals. This is because,these orbitals are not present between (along the internuclear axis) carbon and oxygen atoms.

* Using HOMO( * -non bonding) , carbon monoxide forms a bond with empty d or p-orbital of metal atom and thus acts as -donor..

σ - bond formation between Metal and CO

M C O..M C O..+filled σ-orbital (HOMO)

lobe of an empty p or d- orbital

* CO can also accepts electrons into it’s LUMO( * - antibonding) during back donation ofelectrons of one of filled d-orbital of metal atom.


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- Back bond formation between Metal and CO

*empty -orbital (LUMO)

filled d- orbital

C O+M

..

C OM

..

* Because of -back bond formation, the C-O bond order decreases due to increase in thenumber of electrons in the antibonding orbitals of carbon monoxide. (Note: M-C bond orderincreases). This is reflected in decrease of C-O bond stretching frequency in IR spectrum.* For example:

Uncoordinated or "free" CO: 2143 cm-1

Terminal M-CO: 2125 to 1850 cm-1

Doubly bridging ( -2): 1850 to 1750 cm-1

Triply bridging ( -3): 1675 to 1600 cm-1

* Bridging carbonyls tend to have weaker and broader IR bands.

* Following are some illustrations:

1) Ni(CO)4 -1CO

Terminal carbonylν =2057 cm

2) Ir Ir

C

O

OC CO

2-1 -1 -1

CO

µ bridging carbonyl= 2002 cm (s), 1959 cm (w), 1776 cm (m)

Additional questions:51.1) ‘A’ and ‘B’ with the formula, [M(CO)(NO)Cl4] are two geometrical isomers. Which

among them is cis if the stretching frequency of C-O bond in ‘A’ is greater than that in‘B’?

Ans:- Two -acceptor ligands in a trans configuration will compete for the same d orbitals.Placing a strong acceptor trans to CO will therefore lessen the availability of electrons forback-bonding and so the CO stretching frequency will be higher than otherwise.

As NO is a strong field -acceptor ligand, it competes with CO and hence the stretchingfrequency of C-O in trans isomer will be higher than in case of cis isomer.

Therefore ‘A’ is a trans isomer.

Note:Based on CO IR stretching frequencies, the following ligands can be ranked from best -acceptor to worst:

NO+ > CO > PF3 > RN C > PCl3 > P(OR)3 > PR3 > RC N > NH3

For example, if NH3 occupies the position trans to the CO, there will be no competition forthe d-electrons of metal. Hence the stretching frequecy of CO is not affected.

51.2) Arrange the values of CO stretching frequencis of 1) Ni(CO)4 2) Ni(CO)3(PMe3) 3)


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Ni(CO)2(PMe3)2 in their increasing order.Ans:- The PMe3 ligand can donate electron pair strongly to the metal and thus by encouraging the

-back donation from metal to carbon (of carbonyl group). This inturn weakens the C-Obond and thus by decreasing the CO stretching frequency.

P M

Me

Me

Me C O

greaterthe donation

greater is the pi back bonding

and weaker the CO bond

Hence the increasing order of CO stretching frequency is: Ni(CO)3(PMe3)2 < Ni(CO)2(PMe3) < Ni(CO)4

51.3) Which among Ni(CO)3(P(OMe)3) and Ni(CO)2(PMe3)2 the CO stretching frequency ishigher.

Ans:- Phosphine ligand can form bond by donating its lone pair as well as involve in -backbonding interaction between filled metal d-orbital and empty d-orbital of phosphorus.

+M

..

P

R

R

R: M

..P

R

R

R:

Phosphine ligand forming both sigma and pi bonds with Metal

But -back bonding is not very important unless R groups are electron withdrawing.

* In case of PMe3 ligand, the Me groups are electron donating and hence enhance the donat-ing tendency of phosphorus. This strengthen the M-C bond and weakens the C-O bond. (asexplained in the previous problem).

* But in case of P(OMe)3 ligand, the -OMe group is electron withdrawing. Hence the -donating ability of phosphorus decreases. This encourages the -back donation from Metalto Phosphorus and weakens the M-C bond but strengthens the C-O bond.

* Hence the CO stretching frequency in Ni(CO)3(P(OMe)3) is stronger than in case ofNi(CO)2(PMe3)2.

51.4) Arrange [Ti(CO)6]2-, [V(CO)6]

-, [Cr(CO)6] and CO in their increasing order of CO

stretching frequencies.Ans:- Uncoordinated or "free" CO: 2143 cm-1

Terminal M-CO: 2125 to 1850 cm-1

And moreover in case of [Ti(CO)6]2- and [V(CO)6]

- the negative charge enhances theability of metal to back donate electrons to CO. This leads to more weaking of CO bond.


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Therefore the correct order of CO stretchin frequencies is: CO > [Cr(CO)6]

> [V(CO)6]- > [Ti(CO)6]

2-

51.5) Solid Co2(CO)8 shows infrared CO stretching bands at 1857, 1886, 2001, 2031, 2044,2059, 2071 and 2112 cm-1. When Co2(CO)8 is dissolved in hexan, the carbonyl bands at1857 and 1886 cm-1 disappear. These changes in the infrared spectrum in hexane aredue to:

a) Loss of terminal CO.b) Structural changes of Co2(CO)8 involving conversion of terminal CO to bridging

CO.c) Dissociation of Co2(CO)8 to Co(CO)4.d) Structural changes of Co2(CO)8 involving conversion of bridging CO to terminal

CO.Ans:- Very good question:

The bridging CO strecthing frequencies (1857 and 1886 cm-1) are disappeared in hexane.Hence we can propose following structural change.

3 3 42 42(OC) Co Co(CO) (OC) C( -CO) o-Co(CO)

The bridging CO’s are converted to terminal type.Hence option ‘d’ is correct.

Practice questions1) The CO stretching frequency of which of the following is higher?

1) Mo(CO)3(PF3)3 2) Mo(CO)3(pyridine)33) Mo(CO)6 4) Mo(CO)3(PPh3)3

2) The HOMO in CO is1) -bonding 2) -antibonding 3) -bonding 4) σ -antibonding

3) The M-C stretching frequencies of (i) [V(CO)6]-, (ii) [Cr(CO)6] & (iii) [Mn(CO)6]

+ followthe trend:

a) i > ii > iii b) ii > i > iii c) ii > iii > i d) iii > ii > i

4) Which of the following complexes will have the highest CO stretching frequency in theIR? Why?

A) [Mn(CO)3{P(OPh)3}3]+

B) W(CO)3(PEt3)3C) [(PF3)Ag(CO)]+

5) Which of the following complexes will have the lowest CO stretching frequency in theIR? Why?

A) Ni(CO)(PMe3)3B) Fe(CO)4(PPh3)C) [Re(CO)2{P(OMe)3}4]

+

6) The correct order of infrared stretching frequencies, CO of the following:P) Mn(CO)6

+, Q) CO, R) H3BCO and S) [V(CO)6]-

1) P > R > S > Q 2) S > P > R > Q3) Q > S > P > R 4) R > Q > P > S


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Note: In borane carbonyl, the CO bond stretching frequency is 2165 cm-1. This is due toprominant bonding from HOMO (which is non bonding but slightly antibonding) of carbonto LUMO on boron (the CO bond order increases). However, the -bonding is also possiblefrom B to C. But it does not affect the bond order of CO.

The values of COMn(CO)6

+ 2090 cm-1

[V(CO)6]- 1860 cm-1

CO 2143 cm-1

7) Which one of the following IR frequencies is the closest to that of a triply bridged COgroup?

a) 1700 cm-1 b) 1810cm-1 c) 1920cm-1 d) 2140 cm-1

52) The most unstable species among the following is1. Ti(C2H5)42. Ti(CH2Ph)43. Pb(CH3)44. Pb(C2H5)4

Explanation:* Ti(C2H5)4

is kinetically most unstable due to possibility of -elimination of hydride.

Ti

CH2

CH2

H

CH2

CH2

Ti

Hbeta hydride elimination

The reaction is simply the transfer of a hydride (hydrogen atom) from the -position on aligand to the metal center. Usually alkyl metal complexes containing a -hydrogen decom-pose by this process. Hence metal complexes containing ethyl ligands are kinetically lessstable.

An important prerequisite for -hydride elimination is the presence of an open coordina-tion site on the metal complex. In above case the no. of VE on titanium is 8(electron defi-cient) instead of 18 and there are available empty d-orbitals.

-hydride elimination is the reverse of olefin insertion reaction.* Ti(CH2Ph)4 is kinetically stable because it has no -hydrogen.* Pb(CH3)4 and Pb(C2H5)4

are thermally stable. Most probably because of octet configurationin the valence shell of Pb and non availability of d-orbitals. Also Pb-C bond is relativelystronger.

Additional information:Stability of alkyl metal complexes* Stability of alkyl metal complexes is influenced by many factors (both kinetic and thermody-namic), which are summarized below. (Most of them are also applicable to other type ofcomplexes)

1) The M-C bond in case of transition metal complexes is less stable when compared to


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that of main group metals like Sn, Pb etc.,2) M-C cleavage leads to the formation of more stable molecules like H2O and stable

organic molecules.3) Among the transition metals, 5d and 4d series form more stable metal alkyl complexes

than those of 3d series3d < 4d < 5d

4) In a given series, stability increases from left to right i.e., complexes of late transitionmetals(last five in a series) are more stable than those of early transtion metals(first five in aseries).

5) Metals in high coordination number are more stable than those in low coordinationnumbers. Probably due to steric factors in high coordination sphere.

6) Metals with 18 VE (valence electrons) are more stable.18 VE > 17 VE > 16 VE > and so on

7) Population of antibonding orbitals leads to bond breaking and hence any factor thatincreases the energy gap ( ) will improve the stability. The following factors increase the oand hence the stability

a) Strong field ligands increase the stability of complexes.b) Metals with higher oxidation states have more and hence more stable.c) 4d and 5d metals are more stable as is more.d) octahedral tetrahedralΔ >Δ . Hence octahedral complexes tend to be more stable.

8) d0 configurations are more stable as there are no electrons to occupy antibondingorbitals.

9) In octahedral complexes, metals with substitution inert configurations (e.g. d3 and lowspin d6) are more stable (more appropriately we have to say coordinatively inert). But d7

configurations are coordinatively labile and undergo substitution easily. (Why? Ans: In d3

configuration t2g is half-filled. In low spin d6 configuration, when coordinated with strong fieldligands, all the six electrons occupy the low lying t2g and has more CFSE. And in case of d7,atleast one electron will be in eg which lowers the CFSE)

e.g. Co(III) with low spin d6 configuration is coordinatively inert, whereas Co(II) withd7 configuration is coordinatively labile.

Cr(III) with d3 configuration is more inert.Question: What about high spin d6 configurations? Are they labile or inert?

10) Complexes formed by Os, Re and Pt are more stable than those of other metals11) Absence of low energy pathways like -eliminations stabilize the complex (this is a

kinetic factor and already explained). Based on this, stability confered by the ligands is in thefollowing order.

1-norbornyl > benzyl > trimethylsilyl > neopentyl > Ph ~ Me >> Et (1oR) > 2o, 3o RIn case of norbornyl, the -hydrogen not only has a difficult time approaching the metal

center, but that the olefin that would be generated would be highly strained (and violateBredt's rule - which states that a double bond cannot be placed at the bridgehead of a bridgedring system, unless the rings are large enough).

Mbeta elimination is not possible due to ring strain in the formed olefin.

12) Chelating ligands confer more stability to the complex.13) The ligands which can have -interaction with metal stabilise the M-C bond. e.g. aryl,


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acetylide etc., There is a possibility of -interaction between d-orbital of metal and *orbital of carbon. This will strenghen the bond.

14) The metal carbon bond is polar ( M C ). The organic ligand which delocalize thenegative charge on carbon can improve the stability of M-C bond. Hence complexes withperfluoro alkyl ligands (e.g. CF3) are more stable.

15) Complexes with bulky ligands are generally more stable.

Irving-Williams stability series: For a given ligand the stability of highspin octahedralcomplexes of dipositive metal ions increases in the following order:

Ba2+ < Sr2+ < Ca2+ < Mg2+ < Mn2+ < Fe2+ < Co2+ < Ni2+ < Cu2+ > Zn2+

Cu2+ ions are more stable due to Jahn-Teller distortion.

Additional questions52.1) Explain why high-spin Fe3+ complexes of siderophores are kinetically labile whereas

analogous model complexes containing Cr3+ are kinetically inert.Ans:- High spin Fe3+ with d5 configuration has t2g

3eg2 arrangement and has low CFSE value. Hence

low stability and more reactivity (labile).Whereas Cr3+ with d3 configuration has t2g

3 arrangement (exactly half filled) and has highCFSE value. Hence high stability and kinetically inert.

52.2) Explain why Cr(CO)6 is stabler than V(CO)6.Ans:- In Cr(CO)6 , VE = 36 (inert gas configuration); hence more stable.

In V(CO)6 , VE = 35: hence less stable.

52.3) Arrange the follolowing complexes in order of their increasing energy.P - [Mn(H2O)6]

2+ Q - [V(H2O)6]2+ R - [Ni(H2O)6]

2+ S - [Ti(H2O)6]2+

Ans:- Hydration energy is mainly proportional to CFSE.

Ion Configuration CFSE Mn2+ t2g

3eg2 -0.4 O

V2+ t2g3eg

0 -1.2 O Ni2+ t2g

6eg2 -1.6 O

Ti2+ t2g2eg

0 -0.8 O

Hence the increasing order of hydration energy is P < S < Q < R .52.4) Ru(C2H5)Cl(PPh3)2 is stable only under a pressure of ethene because:

1) It is a 16-electron complex.2) It forms an 18 electron adduct with ethene.3) One of the decomposition product is ethene.4) It prevents β -elimination of ethene.

Ans:- It is a 14e complex and can undergo β -hydride elimination easily. But under excess pres-sure of ethene, backward reaction (olefin insertion) is favored. (Le Chatelier’s principle)


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RuCl CH2

Ph3P PPh3

CH2H

RuCl H

Ph3P PPh3

+ CH2 CH2

52.5) Statement: For the reaction, - +n nL MH L M + H , the important factors are the

strength of M-H bond and the nature of the ligand, L.Reason: The key here is the stability of the complex ion, LnM

-.Assertion: Weak π bonding will stabilize LnM

- and so will disfavour the forwardreaction.

a) Both Reason and Assetion are correct.b) Both Reason and Assertion are wrong.c) Reason is correct but Assertion is wrong.d) Reason is wrong but Assertion is correct.

Ans:- TO BE SOLVED

53) In which of the following species quadrupole bonding is involved?1. Mo2(NMe2)62. Mn2(CO)103. Fe2(CO)94. Re2Cl8

2-

Explanation:

ReClCl

ClCl

ReClClClCl

* Quadruple bond is formed between two metal atoms and posseses “one -bond, two -bonds and one -bond ”.* -bond is special kind of bond formed due to overlapping of four lobes of d-orbitals in aface-to-face manner.* Re2Cl8

2- (octachlorodirhenate(III) anion)is the first compound reported to posses quadruple

bond.* Eclipsed confirmation and short Re-Re bond length leads to proposal of this kind ofbonding. Presence of -bond favors eclipsed confirmation rather than expected staggeredconfirmation.


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Re

Cl

Cl

Cl

Cl

Re

Cl

Cl

Cl

ClRe

Cl

ClCl

Cl

Re

Cl

Cl

Cl

Cl

-Bond is formed between d-orbitals in eclipled confirmation.

Zero overlap between d-orbitals in staggered confirmation.

* In the terminology of molecular orbital theory, the bonding is described as 2 4 2 .* The HOMO is molecular orbital and LUMO is * . The energy gap between these orbit-als corresponds to visible wavelengths of light and hence compounds with quadruple bond arebrilliantly colored.* Other examples with this quadruple bonding are

- [Re2(CO)10]- dimolybdenum tetraacetate [Mo2(O2CCH3)4]- Potassium Octachlorodimolybdate(II) K4[Mo2Cl8]- dichromium tetraacetate [Cr2(O2CCH3)4]

Additional information:Metal-Metal Bonding:* M-M bonding may be of following types:

- covalent- dative- symmetry interactions

* In M-M bonding, 2 , , , and xz yz xyzd d d d orbitals are involved (assuming z-axis is along M-

M bond)

2 orbitals are involved in bond formation. and are involved in bond formation. orbitals are involved in bond formation.

z

xz yz

xy

dd dd

Note: 2 2x yd orbitals are involved in M-L bonding.


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2 2-z z

d d

- yz yzd d

- xz xzd d

- xy xyd d

bond

bond

bond

* Following is the MO diagram for the M-M interaction of two square planar metal centers.

Energy2z

d 2zdxzd xzdyzd yzdxyd xyd

*

*

*

M LLL

L

M LLL

L

An M-M bond is formed, as each bonding orbital is filled with an electron pair. The num-ber of bonds reaches 4 (maximum) with d4-d4 systems. But filling of antibonding orbitalscancel out the bonding interactions. Hence there is no M-M bond possible with d8-d8 systems.

Note: Above MO diagram is valid only for metal centres with square planar geometry.


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Electron Count Resulting M-M Bond

d1 - d1 Single bond d2 - d2 Double bond

d3 - d3 Triple bond

d4 - d4 Quadruple bond optimum

d5 - d5 Triple bond

d6 - d6 Double bond (M-L bonding usually dominates) d7 - d7 Single bond

d8 - d8 No bond (symmetry interaction)

Additional questions53.1) A) The number of M-M bonds in Ir4(CO)12 are:

1) four 2) six 3) eight 4) zeroB) Explain the fluxionality of Ir4(CO)12 using any of its substituent.

Ans:- A) Ir4(CO)12 - Tetrairidium dodecacarbonyl has following structure.

Ir Ir

Ir

Ir

OC CO

CO

OC

OC

OC

OC CO

CO

CO

COOC

Symmetry: Td

B) Fluxionality of Ir4(CO)12 can be explained by merry-go-round mechanism, just like inCo4(CO)12.

Illustration: Ir4(CO)11PEt3 exists as two isomers in solution as shown below. This gives riseto fourteen different carbonyl signals.

Ir Ir

Ir

Ir

OC CO

PEt3

OC

OC

OC

OC CO

CO

CO

COOC

Ir Ir

Ir

Ir

OC CO

PEt3

OC

OC

OC

OC CO

CO

COCO

CO

The all terminal structure The bridged structure


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53.2) Considering the quadrupolar nature of M-M bond in [Re2Cl8]2-, the M-M bond

order in [Re2Cl4(PMe2Ph)4]+ and [Re2Cl4(PMe2Ph)4] respectively are:

Ans:- Re belongs to Mn group with d5s2 configuration.

Compound Oxidation state of Re Configuration Bond order

[Re2Cl8]2- +3 d4 4 [Re2Cl4(PMe2Ph)4]+ +1.5 d5.5 2.5 [Re2Cl4(PMe2Ph)4] +2 d5 3

53.3) Using the 18-electron rule, find out the Mo–Mo bond order in [(η 5–Cp)Mo(CO)3]2.Draw the possible isomers for the molecule (Cp = Cyclopentadienyl anion).

Ans:- Mo belongs to Cr group with d5s1 configuration.In the given complex, the oxidation state of Mo is +1. Hence the d-count is 5.Now the number of VE = 5 + 6 + 3(2) = 17eHence there must be a single bond between Mo-Mo. The bond order is 1.

Mo

OC CO

CO

Mo

COOC

CO

Mo

OC CO

CO

Mo

COOC

CO

Mo

OC CO

OC Mo

COOC

CO

Practice questions:1) M-M bond is either not possible or only weak interactions are possible when configura-tions in metals is:

a) d5-d5 b) d2-d2 c) d4-d4 d) d8-d8

2) The molecular orbital configuration of the Re-Re bond in [Re2Cl8]2- is

A) 2 4 1 *1 B) 2 2 2 *2 C) 2 2 2 *2 D) 2 4 2σ π δ

3) Statement: The characteristic spectroscopic feature of the quadruply bonded [Re2Cl8]2- is

a strong royal blue color.Reason: This is due to an absorption band in the visible region due to excitation of an

electron from 2 4 2σ π δ ground state to 2 4 1 1σ π δ * excited state.Assertion: This transition is quantum mechanically allowed.


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a) Both Reason and Assetion are correct.b) Both Reason and Assertion are wrong.c) Reason is correct but Assertion is wrong.d) Reason is wrong but Assertion is correct.

4) The number of M-M bonds present in Ir4(CO)12 are:A) 4 B) 2 C) 8 D) 6

5) An example of the species having a quadrupole bond isa) Mn2(CO)10 b) Cr2O7

2- c) Re2Cl82- d) Hg2(CH3COO)2

54) In which one of the following pairs the species have similar geometry?1. CO2 and SO22. NH3 and BH33. CO3

2– and SO32–

4. SO42– and ClO4

–

Explanation:1. CO2 is linear whereas SO2 is angular with a lone pair2. NH3 is pyramidal with a lone pair whereas BH3 is trigonal planar3. CO3

2– is trigonal planar whereas SO32– is pyramidal with a lone pair

4. Both SO42– and ClO4

– are tetrahedral.

Additional information:VALENCE SHELL ELECTRON PAIR REPULSION (VSEPR) THEORY* Following are the important postulates of VSEPR's theory proposed by Nyholm, Gillespie.1) The shape of a molecule can be determined from the arrangement and repulsions betweenthe electron pairs present in the valence shell of central atom of that molecule.2) There are two types of valence shell electron pairs viz.,

i) Bond pair and ii) Lone pair3) The electron pairs in the valence shell the repel each other and determines the shape of themolecule. The magnitude of the repulsion depends upon the type of electron pair.4) The bond pair is attracted by nuclei the occupies less space and hence it causes less repul-sion. Whereas, the lone pairs are only attracted by one nucleus and hence occupy more space.As a result, the repulsion caused by them is greater.

The order of repulsion between different types of electron pairs is as follows :Lone pair - Lone pair > Lone Pair -Bond pair > Bond pair- Bond pair

5) When the valence shell of central atom contains only bond pairs, the molecule gets sym-metrical structure, whereas the symmetry is distorted when there are lone pairs along withbond pairs.6) The bond angle decreases due to the presence of lone pairs.7) The repulsion increases with increase in the number of bonds between two atoms.

E.g. Triple bond causes more repulsion then double bond which in turn causes morerepulsion than single bonds.8) The repulsion between electron pairs increases with increase electronegativity of centralatom and hence the bond angle increases.9) Shapes of molecules can be predicted from the number of electron pairs in the valence shell ofcentral atom as follows:


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THE GEOMETRY OF MOLECULES CONTAINING ONLY BOND PAIRS IN THE CENTRAL ATOM Number of bond pairs Formula Molecular geometry Examples

2 AB2 Linear A BB BeCl2, BeF2

3 AB3 Trigonal planar

B

B B

A

BF3, BCl3

4 AB4 Tetrahedral A

B

B

BB

CH4, CCl4

5 AB5 Trigonal

bipyramidal AB

BB

B

B

PCl5, PF5

6 AB6 Octahedral AB

B

B

B

B

B

SF6


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GEOMETRY OF MOLECULES CONTAINING CONTAINING ONE OR MORE LONE PAIRS IN CENTRAL ATOM

Total number

of electron

pairs

Number of bond

pairs

Number of lone pairs

Formula Shape of molecule Bond angle Examples

3 2 1 AB2E Angular 120o SO2

3 1 AB3E Trigonal Pyramidal

107o48’ 102o30’

NH3

NF3 4

2 2 AB2E2 Angular

(V- shaped) 104o28’

103o H2O F2O

4 1 AB4E See-Saw SCl4 , SF4 3 2 AB3E2 T-Shape 90o ClF3 5 2 3 AB2E3 Linear 180o

XeF2, I3-

5 1 AB5E Square pyramidal 90o BrF5

6 4 2 AB4E2

Square planar

90o XeF4

Where A = central atom B = atom linked to the central atom E = Lone electron pair

Explanatory examples:1) BeCl2: The valence shell of central atom, beryllium contains only two bond pairs. Hence itis linear in shape with 180o bond angle.

Cl ClBe

180o

Linear molecule

2) BF3: The valence shell of the central atom - boron contains only three bond pairs. Henceit's shape is trigonal planar with 120o bond angle.

120o

B

Cl Cl

Cl

Trigonal planar shape

3) CH4: The valence shell of the central atom - carbon contains only four bond pairs. Hence itis tetrahadral in shape with 109o28' bond angle.

The bond pairs are arranged tetrahedral symmetry so as to minimize repulsions. If the


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bond pairs are arranged in square plane, the angles between them will be only 90o and therepulsions will be more than in case of tetrahedral arrangement. Hence tetrahedral structure ismore favorable than square planar structure.

109o28'

C

H

HH

HTetrahedral structure of methane

4) NH3: There are three bond pairs and one lone pair in the central atom, nitrogen. The bondangle is decreased from 109o28'

to 107o48' due to repulsion caused by lone pair.

107o48'

NH H

H

Trigonal pyramidal structure of ammonia molecule

5) H2O: There are two bond pairs and two lone pairs in the central atom, oxygen. The bondangle is decreased from 109o28'

to 104o28' due to repulsion caused by two lone pairs.

104o28'

OH

H

Angular shape of water molecule

Additional questions53.1) The shape of BrF3 is:

a) Trigonol pyramidal b) Trigonal plannarc) Trigonal bipyramidal d) T-shaped

Ans:- BrF3 is a T-shaped molecule. The central ‘Br’ atom undergoes, sp3d hybridization in 1stexcited state. It is an AX3 type of interhalogen compound. Other e.g. ClF3, BrCl3 etc.,

Br F

F

F:

:

53.2) Discuss the structures of XeF2, XeF4, XeF6, XeO3 and XeO4, XeO2F2 .


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Ans:- Z of Xe = 63.Ground state valence E.C of Xe = 5s2 5p6 5d0

Structure of XeF2E.C in 1st ex.state: 5s2 5p5 5d1

5s 5p

sp3dhybridization

5d

Structure is linear (based on trigonal bipyramidal).

Xe

F

F:

:

:

Logic: Xe is forming two bonds which require two unpaired electrons. Hence sp3dhybridization in 1st excited state.

Structure of XeF4E.C in 2nd ex.state: 5s2 5p4 5d2

5s 5p

sp3d2

hybridization

5d

Its shape is square planar (based on octahedral structure).

:

Xe

F

F

F

F

:

Logic: Xe is forming four bonds which require four unpaired electrons. Hence sp3d2

hybridization in 2nd excited state.

Structure of XeF6E.C in 3rd ex.state: 5s2 5p3 5d3

5s 5p

sp3d3

hybridization

5d


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Shape is distorted octahedral. Xe

F

F

F

F

F

F

:

Logic: Xe is forming six bonds which require six unpaired electrons. Hence sp3d3 hybridizationin 3rd excited state.

Structure of XeO3E.C in 3rd ex.state: 5s2 5p3 5d3

5s 5p

sp3

hybridization

5d

3 bonds

Its shape is trigonal pyramidal (based on tetrahedral structure).

Xe

O

O

O

:

Logic: Xe is forming three bonds and three bonds which require six unpaired electrons.Hence 3rd excited state. But as it forms only 3 bonds , the hybridization is sp3.

Remember, the hybrid orbitals form bonds and the remaining pure orbitals form bonds.

Structure of XeO4E.C in 4th ex.state: 5s1 5p3 5d4

5s 5p

sp3

hybridization

5d

4 bonds

Its shape is tetrahedral: Xe

O

O

O

O

Logic: Xe is forming four bonds and four bonds which require eight unpaired electrons.Hence 4th excited state. But as it forms only 4 bonds , the hybridization is sp3.


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Structure of XeO2F2E.C in 3rd ex.state: 5s2 5p3 5d3

5s 5p

sp3dhybridization

5d

2 bonds

4 bonds

It has see-saw shape (based on trigonal bipyramidal)

Xe

O

O

F

F

:

Note: The less electronegative (comparatively) oxygen atoms occupy the equatorial positions.H.W: What would be the structure of XeOF4?

53.3) The shape of TeCl4 is:Ans:- The structures of SCl4, SeCl4 and TeCl4 are similar. The central atoms belong to VIA group.

Their outer electronic configuration in the ground state is ns2 np4. They form tetrachlorides inthe first excited state: ns2np3nd1. The hybridization is sp3d.

ns np

sp3dhybridization

nd

They have see-saw shapes: e.g. Te

Cl

Cl

Cl

Cl

:

53.4) BrO3- is isostructural with a noble gas species _____ .

Ans:- BrO3- is trigonal pyramidal. It is isostructural with XeO3. In both the cases, the central atom

undergoes sp3 hybridization.


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BrO

OO

-

:

Note: There are 2 d p bonds.53.5) Compare the structure and conductivity of hexagonal boron nitride with graphite.Ans:- Graphite has layered structure. Each layer consists of fused hexagonal rings of carbon

atoms. These layers are stacked on top of each other such that carbons in alternate layers areare arranged over eahc other.

Each carbon undergoes sp2 hybridization. It is an electrical conductor due to presence offree electrons in the delocalized -molecular orbitals.

Boron nitride (BN) may exist as hexagonal boron nitride (h-BN) or cubic boron nitride(c-BN). The structure of h-BN resembles graphite. The layers stack on top of each other suchthat the B and N atoms lie on top of each other; this contrasts with the staggeredarrangement found in graphite.

Boron nitride is an insulator. It is thermally and chemically stable and find its use in hightemperature equipment (ceramics) and nanotechnology.

BN

N

B

BN

BN

N

B

BN

N

BN

BN

BNB

BN

BN

B

NBN N

BN

B

BN

N

B

BN

BN

N

B

BN

N

BN

BN

BN

B

BN

BN

B

NB

N NB

NB

53.6) The Lewis acid character of BF3, BCl3, BBr3 follows the order:a) BF3 < BBr3 < BCl3 b) BCl3 < BBr3 < BF3c) BF3 < BCl3 < BBr3 d) BBr3 < BCl3 < BF3

Ans:- The effectiveness of formation of -back bonding between halogen and boron atom de-creases from B-F to B-Cl to B-Br. Smaller fluorine atom forms -bond effectively withsmaller boron atom. Hence the Lewis acidic nature increases from BF3

to BBr3. (This iscontrary arguments using electronegativity.)

Another reason is the energy required for the pyramidalization of geometry around boronduring the formation of complex by the boron halide is higher for BF3.

BI3 is strongest lewis acid among all due to weak pi-bonding and low pyramidalizationenergy.


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53.7) Which among (CH3)3NBF3, (CH3)3NBCl3 and (CH3)3NBBr3 is less stable. Explain.Ans:- The geometry around the boron atom in above complexes is tetrahedral (mentioned as

pyramidalization in above answer). Transformation from planar geometry to tetrahedralgeometry requires more energy in case of BF3 (as it requires more energy to break pi-backbonding). Hence it will form complexes less readily. As a result, (CH3)3NBF3 is less stable.

N B

CH3

CH3

CH3

X

X

X

N

CH3

CH3

CH3B

X X

X

: +

Note: (CH3)3NBBr3 is more stable.

53.8) The bond angle of Cl2O is1) Smaller than that of F2O 2) Greater than that of H2O3) Smaller than that of H2O 4) Same as that of F2O

Ans:-Factors deciding the bond angle1) Hybdrization of central atom. (e.g. sp3 - 109o28’)

comment: The bond angle is equal to the angle between hybrid orbitals when there are nolone pairs on the central atom.2) The repulsion caused by lone pairs on the central atom.

comment: Lone pairs cause more repulsion than bond pairs and hence the bond angledecreases.3) The electronegativity of central as well as ligand atoms (which are bonded to thecentral atom).

comment: Greater the electronegativity of central atom, lower is the bond angle. Whenelectronegativity of central atom is higher, then the bond pairs will be closer to it. It causesthe bond pairs to move apart.

If the electronegativity of ligand atoms is more, then the bond angle tends to increase.4) The size of central as well as ligand atoms.

comment:Greater the size of central atom lesser is the bond angle. (as the repulsion is minimized)Greater the size of ligand atoms greater is the bond angle due to repulsion.

Explanation: The central atom (oxygen) in Cl2O, F2O and H2O undergoes sp3 hybridization.But there are two lone pairs on it.

In case of H2O, the bond angle is decreased to 104.5o due to repulsion from lone pair.In case of F2O, the bond angle is decreased to 103o. (here, the lone pair repulsion is

predominant factor)In case of Cl2O, the bond angle is increased to 117.6o due to repulsion between bulky Cl

atoms.

Remember, Cl2O is a paramagnetic molecule. But it does not dimerize as the unpaired elec-tron (odd electron) is localized over the whole molecule (especially on O atoms).

53.8) Describe the structural features of oxyacids of phosphorus.


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Ans:- Structural features of oxyacids of phosphorus:1) Phosphorus can show +1 or +3 or +5 oxidation states in oxoacids.2) It is surrounded by oxide or hydroxide or hydrogens in tetrahedral geometry.3) Phosphorus undergoes sp3 hybridization in first excitation state.4) Only the hydrogens of hydroxy groups are acidic.5) The hydrogen atoms directly attached to phosphorus are not acidic but they possess

reducing properties.6) It forms a d -p bond with oxygen.

Formula Name Structure H3PO3 Phosphonic acid

or Phosphorus acid P

O

OH

OHH

Preparation: 3 2 3 3PCl + 3H O H PO + 3HC1 Properties: Dibasic;

Oxidation state of ‘P’ = +3 One P-H bond & hence also a reducing agent; Can form salts containing H2PO3

- or HPO32- ;

Salts are known as phosphonates or phosphites

H3PO2 Phosphinic acid or Hypophosphorus acid P

O

OH

HH

Preparation:

2 2 2 2 4 3 2 4Ba(H PO ) + H SO 2H PO + BaSO Note: Hypophosphites salts are in turn are obtained when Phosphorus is treated with alkalis. e.g.

4 2 2 2 3P + 3NaOH + 3H O 3NaH PO + PH Properties: Monobasic;

Oxidation state of P = +1; Two P-H bonds & hence also a reducing agent; Forms salts containing H2PO2

- Salts are known as phosphinates or hypophosphites.


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H3PO4 Orthophosphoric acid IUPAC name is trihydroxidooxidophosphorus

P

O

OH

OHOH

Preparation:

4 10 2 3 4P O + 6H O 4H PO Properties: Tribasic;

Oxidation state of P = +5; Forms salts containing H2PO4

-, HPO42- and PO4

3-

H4P2O7

Pyrophosphoric acid or Diphosphoric acid P

O

OHOH

OP

O

OHOH

Preparation: 3 4 4 2 7 22H PO H P O H O Properties: Tetrabasic;

Oxidation state of P = +5; Na2H2P2O7 – Sodium dihydrogen pyrophosphate is used in bread making. Na4P2O7 is used in detergents. Ca2P2O7 is used in tooth pastes.

H5P3O10 Triphosphoric acid

P

O

OHOH

OP

O

OOH

P

O

OHOH

Preparation: 3 4 4 2 7 5 3 10 2H PO H P O H P O H O

When phosphoric acid is heated, initially pyrophosphoric acid is formed, which upon further heating gives triphsophoric acid.

Properties: Pentabasic; Oxidation state of P = +5; Na5P3O10 – sodiumtripolyphosphate is used in sulfonate detergents as sequestering agent for Ca2+ and Mg2+ ions. (softens water)


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H4P2O6 Hypophosphoric acid

P

O

OHOH

P

O

OHOH

Preparation: Red P + NaOCl gives Na2H2P2O6 which upon acidification yields hypophosphoric acid.

Properties: Tetrabasic; Oxidation state of P = +4; There is a P-P bond. Acid is thermodynamically unstable and disproportionates to phosphorus and phosphoric acids.

53.9) Discuss the structure of N2O5 in the solid and gas phases.Ans:- In solid phase it exists as a salt [NO2

+][NO3-]

Cation is linear nitronium ion-[NO2+], wheras anion is planar [NO3

-] ion.

In gas phase, as well as when dissolved in non polar solvents like CCl4, it exists as non polarmolecule shown below.

N

O

N

O

O O

O

Do you know?: It is a good nitrating agent. It was used to nitrate benzene to nitrobenzene.But nowadays it is completely replaced by NO2BF4.

53.10) Explain why AlBr3 dimerises to Al2Br6. while BCl3 is monomeric.Ans:- Aluminium cannot form a -back bond with bromine (of same molecule) due to poor

sidewise overlapping of p-orbital of Br and that of Al. But it can form a -bond with abromine atom of another molecule and can accept an electron pair from it. As a result, it candimerize.

Al

BrBr

Br

Al

Br Br

Br

Whereas, boron can form - bond due to back donation from halogen (of same mol-ecule). Hence BCl3 cannot dimerize.

BClCl

Cl53.11) Using the VSEPR theory, predict the number of lone pairs and draw the structures of

PCl5, IF5, SOF4 and XeOF4.Ans:- No. of lone pairs in the central atonm can be calculated by using following logic.

No. of electrons that are not participating the bonding = valence electrons in the centralatom - no. of bonds ( & ) formed by it.

Hence the formula is:


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v-bNo. of lone pairs = 2

wherev = no. of valence electrons in the central atom.b = no. of ( & ) bonds formed by central atom.

The structure and hybridization are decided by no. of lone pairs and no. of atoms around thecentral atom.For e.g. if this number is 5, then the hybridization is sp3d and the shape is based on trigonalbipyramidal structure.

For PCl5v-b 5-5No. of lone pairs = = 02 2

No. of lone pairs + no. of ligand atoms = 0 + 5 = 5Hence the hybridization is sp3d and the shape is trigonal bipyramidal. No lone pair and nodistortion.

For IF5

v-b 7-5No. of lone pairs = = 12 2

No. of lone pairs + no. of ligand atoms = 1 + 5 = 6.Hence the hybridization is sp3d2 and the shape is based on octahedral structure. As there isone lone pair, the shape is square pyramidal.

For SOF4


No. of lone pairs + no. of ligand atoms = 0 + 5 = 5.Hence the hybridization is sp3d and the shape is trigonal bipyramidal. No lone pair and nodistortion.

Note: Less electronegative and bulky ‘O’ occupies equitorial position.

For XeOF4


No. of lone pairs + no. of ligand atoms = 1 + 5 = 6.Hence the hybridization is sp3d2 and the shape is based on octahedral structure. As there isone lone pair, the shape is square pyramidal.

53.12) Write the strutures of the following compoundsA) 1,2-B4C2H6 B) H3CPF4 C) SF4CH2


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A) B B

BB C

C H

HH

H

H

H

l,2-dicarba-closo-hexaborane(6).

B)P

F

FF

F H

H

H tetrafluoro(methyl)-5-phosphane

Note:The more electropositive (less electronegative) -CH3 group occupies the equitorialposition.

C) S C

H

HF

FFF

(tetrafluoro-6-sulfanylidene)methane

ormethylene sulfurhexafluoride

Practice questions:1) The hybridization in SF6 molecule is :

a) sp3d2 b) sp2d3 c) sp3d3 d) sp3

2) Match the compounds / species given in List - I with the structures in List - II

List – I List –II XeF4 A. Angular Ni(CN)5

3- B. Linear NO2

+ C. Square pyramid Zn(CN)4

2- E. Tetrahedral F. Square planar

Note: NO2+ is angular; Ni(CN)5

3- is square pyramidal;

3) Which one of the following statements is correct for XeO2F2?a) It has a square planar structure.b) It has a trigonal bipyramid based structure.c) It is isostructural with XeF4.d) It has a tetrahedral structure.

4) Which among the following pairs is square pyramidal in shape?a) BrF5, XeOF4 2) PCl5, XeO2F2 3) AsF5, BrF5 4) IF7, XeF6

5) The square pyramidal molecular shape is adopted by1) SOF4 2) XeOF4 3) ScOCl4 4) PF5

H.W: What are the structures of molecules given under other options?


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6) The structure of XeO2F2 based on VSEPR theory is best described as1) A square planar structure with the fluorines trans to each other.2) See-saw structure with F-Xe-F bond angle close to 120o.3) A perfect tetrahedral arrangement of substituents around Xe.4) See-saw structure with O-Xe-O bond angle close to 120o.

Note: F-Xe-F bond angle close to 90o.

7) The B-F bond distance in 3 3H N BF is much _____ than in BF3. (ans: longer)

8) The molecule ClF3 has ____ non-bonded electron pairs. (ans: 2)

9) The planar structure of N(SiH3)3 is due to ______ bonding. (ans: pi-back bonding;from nitrogen to silicon)

10) The hybridization of the central carbon atom in an alkene is ___ . (ans: sp2)

11) Which of the following pair is linear in shape and with three lone pairs?1) CO2, SO2 2) XeF2, SCl2 3) HgCl2, BeF2 4) XeF2, I3

-

Note: All, except SCl2, are linear.

12) The minimum number of electrons needed to form a chemical bond between two atomsis:

a) 1 b) 2 c) 3 d) 4

13) The pair of compounds having the same hybridization for the central atom isa) XeF4 and [SiF4]

2- b) [NiCl4]2- and [PtCl4]

2-

c) Ni(CO)4 and XeO2F2 d) [Co(NH3)6]3+ and [Co(H2O)6]

3+

14) The number of hydroxy (-OH) groups present in phosphorus acid is:a) one b) two c) three d) four

15) [XeO6]4- is octahedral whereas XeF6 is a disordered one, because:

a) Fluorine is more electronegative than oxygen.b) Xe has a lone pair in XeF6.c) XeF6 is neutral whereas [XeO6]

4- is anionic.d) Xe-F bond has more ionic character.

Note: Shape of perxenate ion ---- [XeO6]4- Xe

O

O

O-

O-

O-

O-

16) The series with the correct order of decreasing ionic size isa) K+ > Ca2+ > S2- > Cl- b) S2- > Cl- > K+ > Ca2+

c) K+ > Cl- > Ca2+> S2- d) Cl- > K+ > S2- > Ca2+

17) The structure of SF4 is:


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a) Trigonal bipyramidal b) tetrahedralc) Octahedral d) square planar

Note: According to some authors structure and shape are different terms. In above case,structure includes the lone pairs along with bond pairs. Whereas, shape includes only bondpairs and hence its shape is see-saw.

18) The structures of N(CH3)3 and N(SiH3)3 are respectively:a) Pyramidal & trigonal planar b) Trigonal planar & pyramidalc) Both pyramidal d) Both trigonal planar

Note: Trisilyl amine, N(SiH3)3 is trigonal planar due to sp2 hybridization of central nitrogenatom. There is a p d bond formed due to overlapping of p-orbital with lone pair onnitrogen with empty d-orbital on silicon. Hence not only it is planar and also less basic. ItsLewi basicity is decreased due to involvement of lone pair on nitrogen in -bonding.

This kind of -bonding is not possible with carbon as there are no available d-orbitals.

19) The xenon compounds that are isostructural with IBr2- and BrO3

- respectively are:a) linear XeF2 and pyramidal XeO3 b) bent XeF2 and pyramidal XeO3c) bent XeF2 and planar XeO3 d) linear XeF2 and tetrahedral XeO3

20) The chemical formula for hypophosphoric acidA) H3PO4 B) H3PO3 C) H4P2O5 D) H4P2O6

21) Which one of the following is a monobasic acid?a) H4P2O7 b) H3PO4 c) H3PO3 d) H3PO2

22) In a molecule of phosphorus (V) oxide there are:A) 4 P-P, 10 P-O and 4P = O bonds B) 12 P-O and 4P = O bondsC) 2 P-O and 4P = O bonds D) 6 P-P, 12 P-O and 4P = O bonds

Structure:

O

O

P

PO

P

P O

O

O

O

OOO

23) Among the following isostructural compounds, identify the compound which has thehighest lattice energy.

A) LiF B) LiCl C) NaCl D) MgOHint: Lattice energy is proportional to the product of charges on either ions and inverselyrelated to the distance between them. Greater the charge and small the size, greater is thelattice energy.Related question: Which among the above has least lattice energy? (ans: NaCl)

23) Which of the following species/molecules has a planar geometry?a) Ni(CO)4 b) SF4 c) CoCl2–

4 d) XeF4


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H.W: What are the structures of other species?

24) Which of the following will have the molecular shape of a trigonal bipyramid?a) PF3Cl2 b) IF5 c) BrF5 d) SbF5

2-

Followup question: Theoretically how many isomeric forms are possible for PF3Cl2. Whichamong them is stable?Ans: Three isomers are possible. Among them, the isomer which has two Cl atoms present onequatorial position is more stable.

Usually the electron pairs with larger volume tend to occupy equatorial positions. As Cl isbigger in size and less electronegative when compared to F, they tend to occupy two equato-rial positions.

PF

F

F

Cl

ClPF

Cl

Cl

F

F

PF

F

Cl

F

Cl

More stable

25) Which statement best describes the polarity of SF4Cl2?a. Always polarb. Always nonpolarc. Depending on the arrangement of outer atoms, this molecule could be polar or

non polar.

26) The compound (SiH3)3 N is expected to bea) pyramidal and more basic than (CH3)3Nb) planar and less basic than (CH3)3 Nc) pyramidal and less basic than (CH3)3Nd) planar and more basic than (CH3)3N

27) The nature of -bond present in perchlorate ion is

A) O Cp l p B) O Cd l p

C) O pπ -Cl dπ D) O Cd l d

Cl O

O-

OO

Perchlorate ion

28) The hybrid orbitals used by bromine atom in BrF3 area) sp2 b) sp3 c) sp3d d) sp3d3

29) Which of the following species has two nonbonded electron pairs on the central atom?a) TeCl4 b) CIF3 c) ICl2

- d) PCl3


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30) The species which has a square planar structure isa) BF4

– b) FeCl4– c) SF4 d) XeF4

31) In allene, hybridization of the central and terminal carbons, respectively, area) sp2 and sp2 b) sp2 and sp3 c) sp and sp2 d) sp and sp3

32) The structure of O3 and N3– are

a) linear and bent, respectively b) both linearc) both bent d) bent and linear, respectively

33) Among the following molecules, the shortest bond length is to be found ina) C2 b) N2 c) O2 d) F2

Note: C2 1.24 Ao

N2 1.10 Ao

O2 1.21 Ao

F2 1.42 Ao

34) Which ionic solid is expected to have the highest melting point?a) CaF2 b) NaCl c) CaO d) KBr

35) Which example below exhibits the largest bond angle?a) angle Cl–C–Cl in CHCl3 b) angle F–B–F in BF3c) angle H–O–H in H2O d) angle F–Be–F in BeF2

36) Match up the correct formula and shape of the central atom. Which pair is correct?1) [CO3]

2–; trigonal pyramidal2) [SiH3]

–; trigonal planar3) CS2; bent4) [CMe3]

+; trigonal planar

37) sp3d2 hybridization is observed ina) BrF b) ClF3 c) BrCl3 d) ICl5

38) The incorrect combination among following isa) ICl – sp3 – linearb) BrCl3 – sp3d – T-shapec) IF5 – sp3d2 – square pyramidald) None

39) The shape of ICl4- is:

1) octahedral 2) see saw 3) square planar 4) tetrahedral

40) The electron-domain geometry and molecular geometry of iodine trichloride are ____ and____, respectively.

A) octahedral, trigonal planarB) trigonal planar, trigonal planarC) T-shaped, trigonal planarD) trigonal bipyramidal, T-shaped

41) Of the following, only ____ has sp2 hybridization of the central atom.


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A) ICl3 B) I3- C) PF5 D) CO3

2-

42) Which of the following is non-linear according to VSEPR theory?a) CO2 b) [ClF2]

+ c) [N3]- d) [I3]

-

43) How does the strength of the charge-charge interaction between two ions depend on theseparation, r, in the absence of any other ions?

a) 1/r b) 21/ r c) 31/ r d) 61/ rExplanation: In the absence of any other ion, relation between the various types of interac-tions and the distance of separation are given below:

1/ r ---- for charge-charge interactions.21/ r ---- for ion dipole interactions.31/ r ---- for dipole dipole interactions.61/ r ---- for dispersion interactions.

55) On oxidative addition of O2 to Ir(CO)Cl(PPh3)2, the oxidation state and coordinationnumber of Ir changes, respectively, by1. 1 and 32. 2 and 23. 3 and 14. 2 and 3

Explanation:

Ir

PPh3

PPh3

C

Cl

O

Ir(CO)Cl(PPh3)2, Ir(CO)Cl(O2)(PPh3)2

Ir

PPh3

PPh3

C

Cl

O

O

O

trans-chloridocarbonylbis(triphenylphosphine)iridium(I)

+ O2

Oxidative Addition

179 pm190 pm

* Ir(CO)Cl(PPh3)2 is a square planar complex which is often referred to as Vaska’s com-pound. The oxidation state of Ir is +1. Coordination number is 4. (contains 16 VE)* During oxidative addition (not simple addition) of O2 the oxidation state is changed to +3.The compound formed is Ir(CO)Cl(O2)(PPh3)2. It is octahedral (a little bit distorted), andhence C.N is 6. (contains 18 VE)* The Dioxygen added is in peroxidic form (O2

2-) and not in superoxidic form (O2-).

* Therefore both oxidation state and coordination number are changed by 2 (oxidation statefrom +1 to +3 whereas C.N from 4 to 6).

Additional information:* Vaska’s complex is said to be “co-ordinatively unsaturated” as it contains 16 valence elec-


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trons in Iridium(I). Hence it can bind to ligands and expand its valence electrons to 18 byundergoing oxidative additions. During this process, the oxidation number of Ir is changed to+3.* The C-O stretching frequency is increased from 1967 cm-1 to greater than 2000 cm-1 duringoxidative addition reactions. It is due to reduction of amount of -back bonding from Ir toC and at the same time strengthening of C-O bond. (Refer to Qtn: 51)* The Ir-C bond length is decreased and C-O bond length is increased during oxidativeaddition.

Additional questions:55.1) How do you prepare Co2(CO)8? Give equation only.Ans:- 3 2 2 8 2 22 2 8 ( ) 2 2CoCO H CO Co CO H O CO

55.1) The square planar complex, [IrCl(PPh3)3] undergoes oxidative addition of Cl2 to givetwo products, which are:a) fac- and mer- isomers b) cis and trans isomersc) enantiomers d) linkage isomers

Ans:-

IrCl PPh3

Ph3P PPh3 Cl2 IrCl PPh3

Ph3P PPh3

Cl

Cl

IrCl Cl

Ph3P PPh3

Cl

PPh3

+

meridional facial

56) In linear metal nitrosyls NO acts as a/an:1. One electron donor2. Two electron donor3. Three electron donor4. Four electron donor

Explanation:* In linear metal nitrosyls, the M-N-O angle is 180o and NO is considered as a three electronligand within the neutral electron counting formalism.

M

N

O

* Whereas in "bent NO" complexes, NO is considered to be a one-electron, pseudohalide likeligand. The M-N-O angle is 120o

M

NO

* Whereas within ionic counting formalism, nitric oxide ligand is given one positive charge(NO+) in linear nitrosyls and is given one negative charge (NO-) in bent nitrosyls. However, inboth the cases, it is considered as two electron ligand.


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Additional information:Electron counting rules:* These rules are used to count the number of valence electrons in the central atom in amolecule. It is, like oxidation state assignments, a formalism and does not necessarily reflectthe distribution of electrons in the molecule. However, it serves a useful purpose in predictingstability and reactivity.* There are two methods of electron counting as described below and both give the sameresult.

Neutral counting (covalent method): This approach assumes the molecule or fragmentbeing studied consists of purely covalent bonds. It is usually considered easier especially forlow-valent transition metals.* Locate the central atom and determine the number of its valence electrons by assuming it asa neutral atom. The counting of valence electrons for transition elements is different frommain group elements.-In case of main group elements (s and p block) the electrons in the ‘s’ and ‘p’ orbitals ofouter nth shell are considered for this purpose.-Whereas in case of transition metals, the electrons in ‘(n-1)d’ and ‘ns’ orbitals are taken intoaccount.* Add one for every halide or other anionic ligand which binds to the central through a sigmabond.* Add two for every lone pair bonding to the metal (e.g. each Lewis base binds with a lonepair). Unsaturated hydrocarbons such as alkenes and alkynes are considered Lewis bases.Similarly Lewis and Bronsted acids (protons) contribute nothing.* Add one for each homoelement bond.* Add one for each negative charge, and subtract one for each positive charge.

Ionic counting (Effective atomic number method): This approach assumes purely ionicbonds between atoms.* Calculate the number of valence electrons of the central atom by assuming an oxidationstate on it.e.g. Fe2+ has 6 electrons

S2- has 8 electrons* Add two for every halide or other anionic ligand which binds to the metal through a sigmabond.* Add two for every lone pair bonding to the metal. Similarly Lewis and Bronsted acids(protons) contribute nothing.* For unsaturated ligands such as alkenes, count the number of carbon atoms binding to themetal. Each carbon atom contributes one electron.


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Number of electrons donated by some common ligands: Ligand

Electrons contributed (in neutral counting)

Electrons contributed (in ionic counting)

X (Halide) 1 2 H 1 2 (as H-) H 1 0 (as H+) O 2 4 (as O2-) N 3 6 (as N3-) NR3 (R = H, alkyl, aryl) 2 2 CR2 2 4 (as CR22-)

Ethylene 2 2 (as C2H4)

cyclopentadienyl 5 6 (as C5H5-)

benzene 6 6 (as C6H6)

linear NO 3 2 (as NO+) bent NO 1 2 (as NO-) Note: In case of ionic counting, the number of electrons contributed are made into even number.

Additional questions:56.1) Calculate the number of valence electrons in the central metal atom in the follow-ing molecules / complexes. Also conclude whether the molecule is stable or not.

a) Ferrocene b)

Ir Ir

C

O

OC CO

c)

COOC

MnOC

d)

Ta

tBu

tBu t

Bu

tBu

Ans: a) Ferrocene, (C5H5)2Fe;* e.c. of Fe (Z = 26) is [Ar] 3d64s2

* oxidation state is +2. (useful in ionic counting method)Neutral counting: Fe contributes 8 electrons, the two cyclopentadienyl-rings contribute 5each:

i.e., 8 + 2(5) = 18 valence electrons in ironIonic counting: Fe2+ contributes 6 electrons, the two aromatic cyclopentadienyl rings contrib-ute 6 each:

i.e., 6 + 2(6) = 18Conclusion: Ferrocene is a stable compound as the central metal atom contains 18 electrons(according to 18 electron rule)


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b)

Ir Ir

C

O

OC CO

* e.c of Ir (Z=77) is [Xe] 4f14 5d7 6s2 ; it belongs to Co group.* Metal metal bond does not affect the oxidation state.* The bridging carbonyl splits its electron donation between the two Ir centers. Hence assignonly one electron to one Iridium.Neutral counting: Ir contributes 9 electrons, the cyclopentadienyl ring contributes 5 electrons,Ir-Ir bond contributes one electron, terminal CO contribute 2 electrons and bridged COcontributes only one.

i.e., 9 + 5 + 1+ 2 + 1 = 18Ionic counting: Ir+ contributes 8 electrons, the cyclopentadienyl ring contributes 6 electrons,Ir-Ir bond contributes one electron, terminal CO contribute 2 electrons and bridged COcontributes only one.

i.e., 8 + 6 + 1+ 2 + 1 = 18Conclusion: It is also a stable compound. But this complex and its rhodium analog showfluxional behavior in solution that exchanges the bridging and terminal carbonyls.

c)

COOC

MnOC

* e.c. of Mn (Z = 25) is [Ar] 3d54s2

* oxidation state is +1.

Neutral counting: 7 + 5 + 3(2) = 18 (metal + Cp + three CO)Ionic counting: 6 + 6 + 3(2) = 18Conclusion: It is also a stable complex. This half-sandwitch is used as a gasoline(petrol) anti-knock agent.

d)

Ta

tBu

tBu t

Bu

tBu

* e.c of Ta (Z= 73) is [Xe] 4f14 5d3 6s2 ; it belongs to Vanadium group.* The alkylidene group is considered dianionic. It contributes 2 electrons in neutral countingformalism. In ionic counting formalism, it contributes 4 electrons.* The alkyl ligands are monoanionic. Each of alkyl contribute only 1 electron in neutralmethod, but 2 electrons in ionic method.* Hence the oxidation state is +5.

Neutral counting: 5 + 1(2) + 3(1) = 10 (metal + alkylidene ligand + three alkyl groups)


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Ionic counting: 0 + 1(4) + 3(2) = 10Conclusion: It defies 18 electron rule and expected to be unstable. But it is kinetically stabledue to bulky tert-Butyl groups.Note: This is the first example of Schrock alkylidene complex.

56.2) Calculate the number of valence electrons in the iron in sodium nitroprusside.Ans: Sodium nitroprusside: Na2[Fe(CN)5NO]

* e.c. of Fe (Z = 26) is [Ar] 3d64s2

56.3) The nitrosyl ligand usually coordinates as a cationic ligand, NO+. It can, however,occasionally act as an anionic NO- ligand. When it is behaving as an anionic ligand itadopts a bent coordination geometry. Discuss (using Lewis dot-like figures) the distri-bution of electrons in both kinds of M-NO complexes and how these affect the struc-tures (linear vs. bent). Assume in both cases that you are dealing with a [M-NO]+ unit(positive charge on the overall complex) where the metal has 2 or more d electrons.Clearly show the relative oxidation states of the metal and the relative d electron countfor each bonding case (linear vs. bent). In some ways NO- is the extreme case of NO+

acting as a hyper -backbonding ligand. What is the meaning of this statement?Ans:- NO+ is isoelectronic with CO, that is, it has the same bonding and electronic structure. The

difference is that the more electronegative nitrogen atom combined with the net positivecharge work together to make NO+ the strongest -backbonding ligand known. In fact, itcan backbond enough to formally oxidize the metal center by two electrons to transform theNO+ ligand into a NO- ligand. This is evident in the transfer of a pair of electrons from themetal to the NO+ ligand to produce the bent NO- ligand in some of metal complexes. The lonepair that used to be on the metal center is now on the nitrogen of the NO- ligand. This is whatis the meaning of hyper- backbonding. When the NO+ ligand turns into a NO- ligand it hasformally oxidized the metal center by 2 electrons. That represents the “ultimate” in -backbonding.

56.3) In the complex, [Ni2(5η -Cp)2(CO)2], the IR stretching frequency appears at 1857 cm-1

(strong) and 1897 cm-1 (weak). The valence electron count and the nature of M-CObond respectively are:

Ans: CO stretching frequencies (two peaks around 1850 cm-1) indicate their bridging nature.VE count (ionic method):

C

Ni

C

Ni

O

O

- oxidation number of Ni = +1- Each bridged CO contributes one electron to each metal.- Cp contributes 6 electrons.- Ni-Ni bond contributes one electron.

VE = 9 + 6 + (2 x 1) + 1 = 18

Practice questions:


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1) The N–O stretching frequency in linear transition metal nitrosyls is ______ that in non-linear nitrosyls. (ans: greater)

2) The number of electrons contributed by NO to metal in ionic counting method is _____ .a) 1 b) 2 c) 3 d) 4

3) The neutral complex that follows 18-electron rule is:a) 5

5 5 2( ) ( )C H Fe CO b) 55 5 3( ) ( )C H Mo CO

c) 55 5 2( )C H Co d) 5 6

5 5 6 6(η -C H )Re(η - C H )

4) Among the complexes (i) (C6H6)2Cr; (ii) [HMn(CO)5]; (iii) [(CH3CO)Rh(CO)I3]- and (iv)

CpFe(CO)2(CH3), the 18-electron rule is not followed byA) ii only B) ii & iii C) i & iv D) iii only

5) Which one of the following molecules does not obey the 18–electron rule?a) [Mn(CO)6]

+ b) Fe(CO)5 c) [Cr(CO)5]2– d) [Mn(CO)4Cl2]

2–

6) Which of the following does not obey 18 electron rule?a) Cr(CO)6 b) Fe(CO)5 c) V(CO)6 d) Mn2(CO)10

7) The complex which obeys the 18 electron rule isa) Fe(CO)4 b) Ni(CO)3(PPh3) c) Cr(CO)3 d) Cr(C5H5)2

57) Among the following molecules, the dipole moment is the highest for1. NH32. trans-[PtCl2(NH3)2]3. BF34. NF3

Explanation:

N

HH

H

Pt

NH3

NH3 Cl

Cl

B

F

FF

N

FF

F

* Only NH3 and NF3 have dipole moments. NF3 has highest dipole moment, as the magnitudeof N-F bond moment is larger.* trans-[PtCl2(NH3)2] and BF3have zero dipole moments due to symmetrical structure and asthe individual bond moments are cancelled out by each other.* cis-[PtCl2(NH3)2] has dipole moment. The bond moments are not cancelled out and it is lesssymmetrical than trans isomer.* Remember, BF3 is trigonal planar whereas, NH3 and NF3 are trigonal pyrimidal.

Additional information:


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* Dipole moment ( ) is defined as the product of magnitude of charge( ) on either of thepoles (atoms) and the distance(d) between them.

= .d* It is a vector quantity. It is represented by an arrow pointing towards negative pole (asshown in above diagrams).* The dipole moment of a bond is called bond moment. The final dipole moment of a mol-ecule is the vectorial addition of all the bond moments in it.* The unit of dipole moment is Debye (D) = 10-18 e.s.u cm* Dipole moment is used to determine the ionic nature of a polar covalent bond.

experimental

theoretical

theoretical

µ% ionic character = x 100

µµ iscalculated by assuming the bond is completely polar (ionic).

* For example, In HCl molecule, the observed dipole moment is 1.03 D and its bond length is1.275Å. Assuming 100% ionic character, the charge developed on H and Cl atoms would be4.8 x 10-10e.s.u.Therefore, the theoretical dipole moment for 100% ionic character will be

theoreticalµ = x d = 4.8 x 10-10e.s.u x 1.275 x 10-8cm= 6.12 x 10-18e.s.u.cm= 6.12 D (1D = 10-18 e.s.u. cm.)

experimental

theoretical

µ 1.03% ionic character = x 100 = x 100 = 16.83%µ 6.12

* It is also used to determine the shape of a molecule. Usually, the symmetrical moleculeshave zero dipole moment. For example CO2 has zero dipole moment as it has linear shape,whereas SO2 has net dipole moment as it has angular structure.

* Bond angle between two bonds can be calculated if their bond moments and resultantdipole moment are known. Following simplified equation can be used when the two bondsare same.

θnet dipole moment = 2(bond moment) x cos2

where θ = bond angle

Additional questions:57.1) The dipole moment of H2S is 0.95 D. If the S-H bond moment is 0.72 D, find the bond

angle.

Ans:- net dipole moment = 2(bond moment) x cos2

0

θ0.95 = 2(0.72) x cos2

θcos = 0.65972

bond angle = θ = 97


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57.1) The dipole moment of the H2O molecule is 1.85 D; that of the F2O molecule is only0.25 D. Explain.

Ans:- Both molecules have angular structure, so differences in shape cannot account for thedifference in dipole moment (neglecting the slight bond angle difference here).

Oxygen is more electronegative than hydrogen, so in H2O, the bond and non-bonding-pairdipoles reinforce each other (as they are acting in same direction). Hence more dipole mo-ment.

Fluorine is more electronegative than oxygen, and hence in F2O, the two kinds of dipoles(bond moments and dipole due to lone pair) are opposed. Therefore less dipole moment.

O

H H::

O

F F

::

Practice questions:1) The dipole moment of a hydrogen halide H-X is 1.8 D and the bond length is 1.5 Ao. Thepercent ionic character of H-X bond will be:

a) 60% b) 25% c) 40% d) 80%

2) In a non polar AX2, the bond moment of AX is 0.56 D. The bond angle is:a) 900 b) 450 c) 1800 d) 300

Hint: No need of mathematics.

3) Which of the following molecules will have a permanent dipole moment?.A) SiF4 B) XeF4 C) SF4 4) BF3

4) The molecule that does not possess a permanent dipole moment is1) NF3 2) BF3 3) CH2Cl2 4) NO2

5) Dipole moment of p-nitroaniline, when compared to nitrobenzene (X) and aniline (Y) willbe

A) Greater than (X) and (Y) B) Smaller than (X) and (Y)C) Greater than (X) but smaller than (Y) D) Equal to zero

Hint: In p-nitroaniline, electron flow due to +M effect of -NH2 group and -I effect of -NO2group are in same direction i.e. towards the nitro group.

6) The dipole moment of HCl is 1.08 D and the bond length is 1.27 Ao. The partial charge onhydrogen and chlorine, respectively, are:

A) +1.0 and -1.0 B) +0.85 and - 0.85C) +0.356 and -0.356 D) +0.178 and -0.178

7) The molecule that does not possess a permanent dipole moment isa) NF3 b) BF3 c) CH2Cl2 d) NO2

58) An element 'X' emits successively two β particles, one α particle, one positron and oneneutron. The mass and atomic numbers of the element are decreased by, respectively,1. 4 and 1


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2. 5 and 13. 3 and 24. 3 and 1

Explanation:* -particle is 2He4 nuclide with two positive charges. Hence loss of one -particle resultsin decrease of atomic number(Z) by two and mass number(A) by four. (Hence the position ofnew element formed will be two groups left to the parent element)* -particle is nothing but an electron. It can be represented by -1e

0. It is emitted by thenuclide when a neutron is converted to proton. Hence the atomic number(Z) is increased byone. There is no change in mass number. (Therefore, the position of the new element formedwill be one place right to the parent element)

1 1 0 00 1 1 0 neutron proton electron electron ( particle) neutrino

en p e

* A positron(+1e0) is just like an electron but with a positive charge. It is formed when a

proton is converted to neutron. Hence the atomic number (Z) is decreased by one and againwithout any change in mass number. (Therefore, the the position of new element formed isshifted one group left to that of parent element)

1 1 0 01 0 1 0

+

proton neutron positron neutrino ( )

ep n e

* 2 4 4 52 1 1

positron neutronA A A A AZ Z Z Z ZX X X X X

Therefore, the answer is 5 and 1

Additional information:TERMINOLOGY, CONCEPTS & FORMULAE OF NUCLEAR CHEMISTRYNuclide: The atom with specific atomic and mass number.

In nuclear reactions, different isotopes (same atomic number but with different massnumbers) of same element are considered as different nuclides with different nuclear proper-ties. But for a chemical reaction, isotopes are considered to have same chemical propertiesand hence are not differentiated.

Isotopes: The nuclides with same atomic number but different mass numbers.e.g. 1 2 3 12 13 14 15

1 1 1 6 6 7 7, , ; , ; , etc.,H H H C C N N* They belong to same element.* They possess same chemical properties but different physical and nuclear properties.* They possess same number of protons but different number of neutrons.

Isobars: The nuclides with different atomic numbers but same mass numbers,e.g. 40 40 40 14 14

20 19 18 6 7, , ; , Ca K Ar C N

Isotones: The nuclides with same number of neutrons.e.g. 40 39 38 32 31 30

20 19 18 16 15 14, , ; , , Ca K Ar S P Si

Types of particles: Elementary particles can be divided into fermions and bosons.


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Fermions: e.g. Neutrons, protons, electrons etc.,* obey Pauli principle.* have anti symmetric wave functions.* have half integer spin values.

* Fermions are again divided into1) Leptons: e.g. electron, which cannot interact by strong nuclear interactions.2) Hadrons: e.g. neutron and proton, which can have strong nuclear interactions.

Bosons: e.g. Photons.* do not obey Pauli principle.* have symmetric wave functions.* have integer spin values.

Radius of nucleus: Radius of nucleus is in the order of 10-15 m, whereas that of atom is inthe range of 10-5 m. Radius of nucleus can be calculated empirically as follows:

o -1

/

3 -15

o 1 3RadWhereR =1.2 x 10 cm =

ius of n

1.2 x 10 m = 1.2 fermi

ucleus = R

A = Mass nu

= R A

mberMeV:

1eV = 1.6 x 10-19 joules1 MeV = 106 eV1 amu = 931.46 MeV (you can use approximate value of 930 MeV for quick calculations)

RADIOACTIVITY* Unstable heavy nuclides undergo disintegration and emit , particles and radiation dueto spontaneous disintegration.

420

-1

particle = nuclide with 2+ charge particle = (electron) radiation = electromagnetic radiation

Hee

* Artificially prepared lighter and unstable nuclides show or positron emissions.

* Radioactivity is measured as follows:Becquerel (Bq): One nuclear disintegration per second. It is an SI unit. It is represented byBq (or s-1)

Curie (Ci): 3.7 x 1010 disintigrations per second is called the Curie. It is an older and non SIunit.

1 Ci = 3.7 x 1010 Bq

Kinetics:* Radioactivity is a first order reaction. The rate expression can be written as follows:

t

Where: = decay constant

N No.of radioa

dNrate = -

ctive atoms at time 't'

dt tN


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* The integrated form of above equation is:

:initial no. o

ln ln

ln

1 ln

2.303 log

f radioactive atoms

t o

t tt tt t

o o

o

t

o

t

o

N t N

N Nt e N N eN N

Nt N

Nt

oror or

or

or

Whe

N

reN

Half life(t1/2): The time taken for initial number of radioactive atoms to become half.

1/ 2ln 2 2.303log 2 0.693t

Average life( ):

1/ 21/ 2

1Average life= = 1.44 x 0.693

t t

Graphical method: The first order equation of radioactive decay can be written as follows:ln lnt oN t N (it is in the form of y = mx + c)

Hence a graph of ln tN versus t will be a straight line, with a negative slope equals to

and an intercept, ln oN on y-axis.

t

slope ln tN

ln oN

OR

log2.303

o

t

NtN

(it is in the form of y = mx)

Hence a graph of log o

t

NN versus t is a straight line passing through the origin and has a

slope of 2.303

.


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t

No

Nlog

2.303slope

STABILITY OF NUCLEUSStability of nucleus is affected by the following factors:

N/P Ratio:* Nuclides with N/P = 1 are stable upto atomic number(Z) =20.* Above atomic number 20, the nuclides with N/P ratio slightly greater than 1 are stable.* In the following graph, the zone, in which stable nuclides are found, is called belt of stabil-ity.

num

ber o

f neu

trons

number of protons

belt of stability

* The nuclides present above and below this belt are unstable and tend to get stability byfollowing types of radioactive decay:

1) The nuclides above the belt (with high N/P ratio) get stability by undergoing -decay..In this process, a neutron is converted to a proton.

1 1 0 00 1 1 0 neutron proton electron electron ( particle) neutrino

en p e

2) The nuclides below the belt (with low N/P ratio) get stability by undergoing positrondecay or by electron capture. In positron decay, a proton is converted to neutron.

1 1 0 01 0 1 0

+

proton neutron positron neutrino ( )

ep n e

In case of electron capture (also called inverse beta decay), an electron within the atom iscaptured by the proton in the nucleus and is converted to neutron.

1 0 1 01 1 0 0 +

proton electron neutron neutrino (captured)

ep e n h

Electron capture will occur when the energy difference between the parent atom and thedaughter atom is less than 1.022 MeV. The electrons may be captured from either K or Lshells (K-capture or L-capture). During this process, electrons are reorganized from outer to


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inner shells and the excess energy is emitted as photon.However both the processes lead to increase in the N/P ratio by decreasing the proton

number.

* The nuclides with atomic numbers greater than 83 are too large to be stable and theyundergo alpha emission to decrease the size.

Mass defect( m )* Δm=theoretical mass of nucleus - actual mass of nucleus* Greater the mass defect greater is the stability.* It is because some of the mass is converted to energy during the formation of nucleus fromindividual nucleons. This energy liberated is called Binding energy.

Binding energy* Binding energy of nucleus is calculated by using Einstein’s equation.

2B.E = Δmc* Average binding energy (B.E/no. of nucleons) is measure of nuclear stability.* The top three nuclides with highest average binding energies are Ni-62, Fe-58 and Fe-56(almost 8.8 MeV).* Since fusion of iron nuclides is energy consuming process (destabilizing), the nuclides withatomic number greater than 26 (atomic number of Fe) tend to undergo fission whereas thosewith atomic number less than 26 tend to undergo fusion.* 1 amu = 931.46 MeV (you can use approximate value of 930 MeV for quick calculations)

Magic numbers* The nuclides with 2, 8, 20, 28, 50 and 82 protons or 2, 8, 20, 28, 50, 82 and 126 neutronsare exceptionally stable.e.g. 4 16 40 208

2 8 20 82, , , etc., He O Ca Pb

* Nuclides with even number of protons or neutrons or total nucleons are also observed to bestable.

RADIOACTIVE DISINTEGRATION SERIESFollowing four types of decay series are reported.

No. of particles emitted

Decay series Type Starting nuclide

Stable end

product Alpha Beta Actinium series Natural 4n+3 235

92U 20782 Pb 7 4

Uranium series Natural 4n+2 23892U 206

82 Pb 8 6 Neptunium

series Artificial 4n+1 23793 Np 209

83 Bi 7 4

Thorium series natural 4n 23292Th 208

82 Pb 6 2

* The type of series is based on the remainder obtained when the masses of nuclides in thegiven series are divided by 4.* Remember these series from 4n+3 to 4n in the order of letters in ‘AUNT’.


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* In natural radioactive series, the final nuclide is lead.* The uranium series is also called as radium series.* The members of Neptunium series are NOT found in nature because the half-life of thelongest lived isotope in the series is short compared to the age of the earth.* It is recently discovered that, in neptunium series, the actual stable and final product is 205

81Tl

and not 20983 Bi .

20983 Bi is discovered to be radioactive and further disingegrates to 205

81Tl .

Nuclear Fission:* Splitting of heavy nucleus into two lighter nuclei of almost equal size. It usually accompa-nies the emission of neutrons and enormous amount of energy due to conversion of mass intoenergy.

It may occur spontaeously or when the heavy nucleus is bombarded with neutrons.

E.g. U235 undergo nuclear fission when bombarded with slow moving neutrons. It is a chainreaction. The neutrons emitted further collide with the uranium nulei and help in continuationof the chain. Around 200 MeV of energy is liberated in each fission.

235 1 141 92 192 0 56 36 03 200U n Ba Kr n MeV

Other reactions are also possible and the number of neutrons emitted may be 2 or 3 or 4.On average 2.6 neutrons are emitted.

Probability of fission depends on the critical mass, shape of the fission material, densityand purity.Note: A critical mass is the smallest amount of fissile material needed for a sustained nuclearchain reaction

U235 is suitable for fission as it has high probability of undergoing fission, whereas U238 hasless probability of fission as it tends to capture the neutron by forming U239.Note: Most of the naturally occuring Uranium is in the form of U238, whereas only 0.7% ofuranium is in the form of U235.

Plutonium-239 has highest probability of fission.

APPLICATIONS OF RADIO ISOTOPES:1) 131I: is used in the diagnosis and treatment of thyroid disorders; in brain scanning.2) 32P: is used in i) bone metastasis to control the pain,

ii) in agriculture and synthesis of nucleotides.iii) in studying blood circulation.

3) 60Co: is used in treatment of brain tumours.4) 24Na: is used in detecting the location of pipeline leaks.5) 99Tcm: Brain, heart, lung, thyroid, gall bladder, skin, lymph node, bone, liver, spleen, andkidney imaging6) 11C, 13N, 15O, 18F are used in PET (Positron Emission Tomography)7) 2H, 13C, 15N, 18O are used as tracers in chemical and biological reactions to find out themechanisms and pathways.8) 35S: in heart diagnosis.

Additional questions:58.1) State Soddy-Fajan's Group Displacement Law.Ans:- i) When an -particle is emmited by the parent nuclide, the position of daughter nuclide is

shifted to the left by two groups from the position of parent nuclide. i.e. Z is decreased by


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two units.ii) When a -particle is emitted, the position is shifted to the right by one group. i.e. Z is

increased by one unit.4

2A

daughterZ N

AparentZ X 1

AdaughterZ M

Note: 1) However this rule is not valid for emissions by f-block nuclides, as the parent anddaughter nuclides belong to the same group (IIIB) in the periodic table.

2) The daughter nuclide formed by -decay is the isobar to the parent nuclide.

58.2) 2311Na is a stable nucleus. Predict which of the following unstable nuclei decay by β or

+β emission.23 23 22 2410 12 11 11(i) Ne (ii) Mg (iii) Na (iv) Na

Ans:- - decay occurs when the N/P ratio is greater than 1. During this decay, a neutron is

converted to a proton and an electron( ). - decay occurs when the N/P ratio is less than 1. During this decay, a proton is con-

verted to a neutron and a positron( ).The N/P ratios and the decay processes are given below.

2310

23 23 010 11 -1

N 13i) ratio for Ne = 1.3 ; Hence it undergoes decay.P 10

Ne Na + e

2312

23 23 012 11 +1

N 11ii) ratio for Mg = 0.92 ; Hence it undergoes decay.P 12

Mg Na + e

2211

22 22 011 10 +1

N 11iii) ratio for Na = 1 ; It should be stable. But it undergoes decay ???P 11

Na Ne + e

Reason: Always the nuclides with N/P ratio =1 are not stable. The nuclides present in thestability belt are only stable. 22

11Na is present below the stability belt. Hence undergo positrondecay.

2411

24 24 011 12 -1

N 13iv) ratio for Na = 1.18 ; Hence it undergoes decay.P 11

Na Mg + e

58.3) Which isotope is produced by an (n, ) reaction starting from 230Th?1) 229Th 2) 229Ac 3) 231Th 4) 231Pa

Ans:- 231Th is formed along with the emission of radiation, when 230Th is bombarded withneutron. In this reaction only mass number is increased. The short hand notation for thisreaction is

230Th (n, ) 231ThIt can also be represented as

230Th + 0n1 -----------> 231Th +


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58.4) How many α particles and -β particles are emitted in the decay of 238U to 206Pb?

Ans:- 238 20692 82In the conversion of U Pb,

Mass number is decreased by 238-206 = 32. 32Hence the number of -particles emitted will be 8.4

Hence the decrement in atomic number should be 8 x 2 = 16.But th

e observed decrement is 92 - 82 = 10.Therefore, 6 particles are also released (which increase the Z value by 6).

58.5) 1.0 x 10-3 g of a sample of Tc-99 is showing 6.3 x 105 disintegrations per second. Cal-culate its decay constant.

Ans:-

Tc

ot

T

t

c

0

23

t

where = decay constant;

N x wt. of sampleN = number of 'Tc' nuclei = M

where N =avogadro number; M =molecular weightr of Tech

rate of disintegrations =

nitium

6.02 x 1

r =

0 x 1.0 x N =

N

-318

5 -1-13 -1

18t

10 = 6.08 x 10 nuclei99 g

Hencer 6.3 x 10 nuclei . sec = 1.036 x 10 sec

N 6.08 x 10 nuclei

g

58.6) The decay constant of radioactive disintegration of a nuclide is 3.010 x 10-3 s-1. Whatare the half life and average life of that nuclide?

Ans:- Since radioactive disintegration is a first order process,

1 -3 12

ln 2 0.693 0.693Half life 230.3 s3.010 x 10

where λ=decayconstant

ts

12

12

1Average life= = 1.44 x 1.44 x 230.3 = 331.6 s0.693

tt

58.7) In how many days will a 12-gram sample of a radioactive isotope decay, leaving atotal of 1.5 grams of the original isotope? (The half-life of isotope is 8.07 days)

a) 8.0 b) 16 c) 20 d) 24


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Ans:- n 0 0

n 3

to become0

1/2

N W2 = = where n = no. of half livesN W

122 = =8=21.5

Hence the no. of half lives = 3

Time taken for W (12g) W(1.5g ) = n x t 3 x 8.07 days = 24 days

58.8) C14 is formed when neutrons, produced by cosmic rays from space, collide with N14.Write the balance equation for this process. Also represent in short hand notation?

Ans:- 14 1 14 17 0 6 1

14 147 6

Short hand notation ( , )

N n C H

N n p C

58.9) 10B, used in nuclear reactors, participates in two reactions. One reaction yields 7Liand another product X. The other reaction yields 3T and X. Identify X and write twobalanced nuclear reactions.

Ans:- In nuclear reactors, neutrons are produced due to nuclear fission. These neutrons collidewith 10B. In both the reactions, shown below, the X is alpha particle(4He2).

10 1 7 45 0 3 2

10 1 3 45 0 1 2

1) B + Li + He - one alpha particle is formed.

2) B + T + 2 He - two alpha particles are formed.

n

n

58.10) 20181Tl undergoes -decay. The half-life is 3.1 min.

(a) Write the nuclear equation.(b) How long will it take for 2.00 gram of 201

81Tl to be reduced to 0.20 gram by decay?

(c) Given 10.0 grams of pure 20181Tl initially, how many grams of pure 201

81Tl will remainafter 12.4 min?

Ans:- (a) 201 201 081 82 1Tl Pb e

(b)

1/ 2

ln 2 0.693 0.6933.1 min

where =decay constant

2.303 2.303 x 3.1 2time taken = t = log log 10.33min0.693 0.2

o

t

WW

(c) The number of half lives in 12.4 minutes = total time 12.4half life 3.1

= 4

Hence the amount of 20181Tl remained after 31 minutes = 4

10 0.6252 16

oWW g

58.11) Calculate the time necessary to reduce the activity of 131I to 1% of its initial value.(The t1/2 of 131I = 8.02 days)


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Ans:- 1/ 22.303 x 2.303 Initial activitylog log

0.693 Final activity2.303 x 8.02 100 = log

0.693 12.303 x 8.02 100 = log

0.693 1 =53

o

t

N ttN

.4 days

Note: Activity is proportional to the number of atoms at that instance of time.

58.12) 14C decays to 14N by -β decay with a half-life of 5730 y. If a 1.0 g sample of carbonshows 15.0 disintegrations per minute, what will be its activity after 10,000 y?

Ans:-

1/ 2

0.56

log log2.3030.693 log

2.303

15 0.693 x 10000 log 0.562.303 x 5370

15 10 3.63

15 4.13 dis/min3.63

o o

t t

o

t

t

t

t

N AtN A

Att A

orA

orA

or A

58.12) The mass of one atom of chromium-52 is 51.9405 amu. What is the binding energyper nucleon for the chromium-52 nucleus?

Ans:- There are 24 protons and 52 - 24 = 28 neutrons in 5624Cr .

Hence the theoretical mass would be = 24(1.00728) + 28(1.00866) = 52.4172 amuObserved mass = 51.9405 amu

Mass defect ( m ) = 52.4172 - 51.9405 = 0.4767 amu

Binding energy = 0.4767 x 930 = 443.3 MeV

Binding energy per nucleon = 443.3/52 = 8.525 MeV (not exact figure but an approx.)

58.12) Write the principle involved in Positron Emission Tomography.Ans:- Positron emission tomography (PET) is a nuclear medicine imaging technique which pro-

duces a three-dimensional image or picture of functional processes in the body. The systemdetects pairs of gamma rays emitted indirectly by a positron-emitting radionuclide (tracer),which is introduced into the body on a biologically active molecule. The positron emitted bythe nuclide encounters an electron and annihilated with it to produce a pair of gamma photonswhich move in opposite directions.


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Radionuclides used in PET scanning are typically isotopes with short half lives such ascarbon-11 (~20 min), nitrogen-13 (~10 min), oxygen-15 (~2 min), and fluorine-18 (~110min).These radionuclides are incorporated either into compounds normally used by the bodysuch as glucose (or glucose analogues), water or ammonia.

58.13) Write a short note on activation analysis.Ans:- It is used to determine the amount of an element ‘X’ in a compound. Following steps are

employed in activation analysis.- Irradiate the compound so that ‘X’ is converted to radioactive isotope X*. Usually it is

done by bombarding the sample with neutrons.- Separate X* from the sample and determine the activity which gives idea about the

amount of element X in the compound.

Practice questions:1) The radioactive isotope used to locate brain tumors is

a) 1D2 b) 16S

31 c) 53I131 d) 92U

232

2) The relative penetrating powers of and -particles; -radiation and neutrons (n) are:1) n 2) n γ > β > α 3) n 4) n

3) Which element in Group 18 is naturally radioactive and has no stable isotopes?a) Ar b) Kr c) Xe d) Rn

Note: The elements with atomic number, Z>83 have no stable isotopes. All are radioactive. Theelements after Uranium which are called trans-uranic elements are not only radioactive and alsoartificially prepared.

4) Which of the following best describes the operation of a cyclotron?a) an uncharged particle is accelerated to great speeds by a fast moving current of air

which is generated by a voltage of not less than 500 Bev.b) a charged particle is accelerated by alternating the charge on adjacent Dees,

while being subjected to a magnetic force which causes the particle to move in a spiral.c) a charged particle is accelerated by changing charges in a series of pipes or tubes.d) an uncharged particle is accelerated by alternating charges on adjacent Dees.

5) Which of the following changes occurs in the nucleus when a positron particle is given offduring a nuclear decay process?

(1) the production of a neutron;(2) the loss of neutron;(3) the production of a proton;(4) the splitting of a neutron into a proton and an electron;

6) 31Ga67 (half life = 67 hrs) is used in nuclear medicine to determine blood flow when coro-nary artery disease is suspected. If after treatment the initial radioactivity of a person's bloodis 20,000 counts per minute, how much time would have elapsed in order for the activity todecreast to 4066 counts per minute?

a) 67 hours b) 60 hours c) 154 hours d) 28 days

7) Consider the following types of procedures used in medicine:i. PET (Positron Emission Tomography) scan


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ii. CAT (Computerized Axial Tomography) scaniii. MRI (Magnetic Resonance Imaging) scaniv. Iodine-131 treatment of hyperthyroidism

Which of these procedures involves X-rays?1) i and iii 2) ii and iv 3) ii only 4) i only

8) Which of the following transmutations entails an absorption of an alpha particle and releaseof a neutron?

a) 14 167 8N O b) 238 234

92 90U Th c) 27 3113 15Al P d) 28 31

14 16Si S

9) Nuclides that have too HIGH N/P (compared with stable nuclides) undergo:a) alpha decay b) beta decay c) positron decay d) electron capture.

10) Radon-222 is a potentially serious problem in many homes. From which of the followingnaturally-occurring nuclides is radon-222 produced?

a) carbon-12 b) potassium-40 c) uranium-238 d) thorium-232Note: Radon-222 is produced from Radium-226 which inturn is produced from uranium-238.

11) The half life of Sr-90 is 28 years. How long will it take for a given sample of Sr-90 to be87.5% decomposed.

a) 2 x 28 years b) 4 x 28 years c) 3 x 28 years d) 5 x 28 years

12) What is the radius of nucleus of 27Al?a) 1.2 fermi b) 3.6 fermi c) 2.4 fermi d) 2.7 fermi

13) Which of the following statements is INCORRECT?A) Fermions obey Pauli principle and have symmetric wave functions.B) Bosons have only integer spin values.C) Leptons have antisymmetric wave functions and cannot interact directly by strong

nuclear interactions.D) Bosons have antisymmetric wave functions and do not obey Pauli principle.

14) Which of the following pairs of nuclides are isotones?1) 15O & 15N 2) 18O & 17O 3) 21Ne & 23Na 4) 25Mg & 26Al

15) The chemical properties of atoms, molecules and ions is determined primarily by thenumber and arrangement of which of the following?

a) nucleons b) electrons c) neutrons d)quarks

16) Which of the following can be detected by a Geiger counter?1) alpha particles 2) beta particles 3) gamma rays 4) All of these.

17) Which of the following statement is correct about the behavior of radiations, emitted by aradioactive nuclide, in electric field?

a) -particles are deflected towards the negatively charged plate.b) -particles are deflected towards the positively charged plate.c) -radiations are undeflected in the electric field.d) All are correct.


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18) Which of the following statement is NOT correct?a) The ratio of penetration powers of , & radiations is 1:100:10000.b) The ratio of ionization power of , & radiations is 10000:100:1.c) The , & radiations affect the photographic plate.d) The atomic number and mass number are changed during the emission of -rays.

19) Choose the FALSE statement.A) Nuclei with an even number of protons and an even number of neutrons tend to be

stable.B) Light nuclides are stable when the atomic number (Z) equals the mass number minus

the atomic number (A–Z).C) Nuclei with too many neutrons per proton tend to undergo positron emission.D) Nuclei with too few neutrons per proton tend to undergo positron emission.

20) What is the missing product in this reaction?15 158 7 __O N

A) 0+1e B) 0

-1 e C) 42 He D) 0

0

21) What will be the resulting product, during the K-capture by a germanium-68 nucleus?A) 68

33 As B) 6831Ga C) 68

32 Ge D) 7234 Se

Note: Electron capture, in which an inner-shell (K or L) electron is captured by a proton in thenucleus with the formation of a neutron. X-rays are emitted as the electrons cascade down to fillthe vacancies in the lower energy levels.

22) If 87.5 percent of a sample of pure 99 Rh decays in 48 days, what is the half - life of 99 Rh ?A) 6 days B) 8 days C) 12 days D) 16 days

23) The first part of the decay series of 24094 Pu involves three alpha emissions followed by two

beta emissions. What nuclide has been formed at this intermediate stage of the decay series?A) 228

88 Ra B) 22890Th C) 228

89 Ac D) 23290 Th

24) What is the missing product in the following nuclear reaction?236 1 13692 0 53U 4 n + I +__

a) 9039 Y b) 96

38Sr c) 9639 Y d) 98

40 Zr

25) The average number of neutrons produced during the nuclear fission of uranium-235 isa) 1 neutron b) 2.5 neutrons c) 3 neutrons 4) 4 neutrons

26) The function of the CONTROL RODS in a nuclear power plant is to:1) slow the neutrons down so that they can cause fission.2) absorb the heat produced so that it can be carried to an external turbine.3) absorb neutrons to control the amount of fission that is occuring.4) provide the fuel needed for fission to occur.


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27) Which of the following statements about food irradiation is TRUE?a) The irradiated food is radioactive.b) Irradiation of food produces chemical degradation products similar to those

produced from cooking food using heat.c) Food irradiation employs positrons.d) The radioactive nuclide that is used in food irradiation is thorium-230.

28) Which of the following characteristics of particles produced by radioactive decay areimportant for assessing the potential for biological damage to living systems?1) mass 2) charge 3) penetrating ability 4)kinetic energy 5) All of these.

29) The function of the MODERATOR in a nuclear power plant is to:1) carry the heat produced from the fission reactions to an external turbine.2) absorb neutrons and thereby control the fission reactions.3) increase the speed of neutrons produced during fission reactions.4) slow down the neutrons produced from the fission reactions.

30) In a graph of binding energy per nucleon vs. atomic mass, Fe, has the highest bindingenergy per nucleon of all nuclei. This means that:

1) Fe is more stable than any other nuclide.2) Nuclei lighter than Fe become more stable by fission processes.3) Fe decays by positron emission.4) Nuclei heavier than Fe become more stable by fusion processes.

31) Positron Emission Tomography (PET) is a medical diagnostic technique that is based onwhich one of the following processes?

1) neutron bombardment 2) fission3) annihilation 4) electron scattering

32) A freshly prepared sample of yttrium-90 undergoes 7.6 x 105 disintegrations per minute(dpm) at a certain time. Exactly 14 days later, the same sample undergoes 1.6 x 104 dpm. Thehalf-life of yttrium-90 is:

a) 0.20 day b) 2.5 day d) 3.8 day e)14 dayHint: Do rough calculations.

33) Which of the following is ‘4n’ radioactive decay series?a) Thorium series b) Uranium series c) Neptunium series d) Actinium series

34) The radioactive decay of 239 23992 94U Pu is an example of

a) zero order reaction b) photochemical reactionc) chain reaction d) consecutive reaction

Note: The daughter nuclide is also radioactive and hence consecutive.

35) One gram of 90Sr gets converted to 0.953 g after 2 years. The half life of 90Sr, and theamount of 90Sr remaining after 5 years are:

a) 1.44 years and 0.916 g b) 57.6 years and 0.75 gc) 28.8 years and 0.887 g d) 100 years and 0.982 g


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36) Which one of the following statements concerning radioactive decay is true?A) The half-life of a radioactive isotope depends on the amount of radioactive material

present.B) The activity of a radioactive isotope is inversely proportional to its decay constant.C) The half-life increases as more of the isotope decays.D) The activity of a radioactive isotope decays hyperbolically with time.E) The half-life is inversely proportional to the decay constant.

59) The 1H NMR spectrum of ( 5-C5H5)2Fe recorded at room temperature has1. One singlet2. One multiplet3. Two singlets4. Two multiplets

Explanation:* Ferrocene can exists in two confirmations: 1) more stable staggered and 2) less stableeclipsed. At high temperatures, the rotation is possible and the two forms will exist in equilib-rium (mostly eclipsed).* Hence at high temperatures only one singlet is observed - as all the hydrogens are equiva-lent. Under NMR timescale all the protons on cyclopentadienyl ligand are equivalent at roomtemperature.* But at low temperatures, two signals are possible as the rotation is freezed and ferroceneprefers staggered confirmation.

Fe Fe

Practice questions:1) The false statement about ferrocene is:

1) It obeys the 18-electron rule. 2) It is diamagnetic.3) It is an orange solid. 4) It resists electrophilic substitution.

Note: It undergoes electrophilic substitution on Cp rings readily.

60) In the estimation of Fe2+ by Cr2O72– using barium diphenylamine sulfonate as indicator,

H3PO4 is added to1. maintain the pH of the medium2. decrease the Fe2+/3+ potential3. increase the oxidizing power of Cr2O7

2–

4. stabilize the indicator

Explanation:* Potassium dichromate is an excellent oxidizing agent for iron(II) since:1. dichromate and iron(II) react quantitatively and with a known stoichiometry;2. the reaction is sufficiently fast to be practical for a titration; and3. the 0E (elctrode potential difference and not internal energy change) is large enough to


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produce a well-defined endpoint.4. dichromate is obtainable in a state of high purity and can be used as primary standard.5. solutions of dichromate in water are stable indefinitely.

In an acid medium, the reaction between iron(II) and dichromate ion (the oxidizing agent) is:6Fe2+ + Cr2O7

2-+ 14H+-----------> 6Fe3+ + 2Cr3+ + 7H2O0 0.56E V

Standard potentials for the half-reactions are (both are written as reduction reactions)Fe3+ + e- --------------> Fe2+ 0 0.77E V

Cr2O72 + 14H + + 6e---------->2Cr3+ + 7H2O

0 1.33E V

(Can you show how 0E for the complete reaction becomes 0.56 V, from the given 0Evalues of above half reactions?)

* As the dichromate ion is orange and chromic ion is green, the end point is masked. Hencethis titration requires a redox indicator. ( A redox indicator is a compound that changes colorover a potential range of interest due to redox reaction)* The common redox indicators used for this purpose are diphenylamine, diphenylbenzidineand Barium diphenylamine sulfonate. The color change for all these indicators is from greento violet and the required standard potential is around 0.78 V.* As the electrode potential (0.56 V) of the overall reaction is not coinciding with the re-quired standard potential (0.78 V) of redox indicators, phosphoric acid is added to lower thepotential of Fe3+/Fe2+ couple. (As a result, the electrode potential of the overall reaction isincreased from 0.56 V to 0.78 V and the end point will be clear.)* Phosphoric acid combines and stabilises the Fe3+ ion by forming [Fe(HPO4)]

+. Thus theconcentration of Fe3+ ions is decreased. This lowers the potential of Fe3+/Fe2+ couple (Remem-ber, the electrode potential depends on the concentration).

Note: Barium diphenylamine sulfonate is the best choice among the indicators as it is moresoluble in water than others.

Additional information:* Iron can also be determined by titrating Fe(II) with permanganate in acidic medium. Thereduction half reaction of permaganate can be represented as

MnO4- + 8H+ + 5e- --------> Mn2+ + 4H2O

0 1.51E V * Sulfuric acid is the most suitable acid to carry out oxidation permanganate ions.* But during the estimation of iron in ores, the ore sample is often digested with conc.HCland Fe(III) is reduced to Fe(II). Hence there is likelihood of following oxidation of Cl- ionsduring the titration of Fe(II) ions against permanganate. This interferes the actual process.

2MnO4- + 10Cl- + 16H+ --------> 2Mn2+ + 5C12 + 8H2O

0 1.36E V * To avoid above problem, Zimmermann and Reinhardt's solution (this is sometimestermed preventive solution) is added during the titration. This solution contains MnSO4 andphosphoric acid.* The manganese(II) sulphate lowers the reduction potential of the MnO4

- - Mn(II) coupleand thereby makes MnO4

-, a weaker oxidising agent and the tendency of the permanganateion to oxidise chloride ion is thus reduced.* Phosphoric acid, as usual, complexes with Fe(III) ions and reduces the potential of Fe(III)-Fe(II) couple.


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Note: Potassium permanganate is not a primary standard. It is difficult to obtain the substanceperfectly pure and completely free from manganese dioxide. Moreover, ordinary distilledwater is likely to contain reducing substances (traces of organic matter, etc.) which will reactwith the potassium permanganate to form manganese dioxide. The presence of the latter isvery objectionable because it catalyses the autodecomposition of the permanganate solutionon standing.

GRAVIMETRIC DETERMINATION OF NICKEL* Hot and faintly acidic (dilute) solution of nickel salt is treated with 1% ethanolic solution ofdimethylglyoxime (H2dmg) and then adding a slight excess of aqueous ammonia solution toget a red precipitate of bis(dimethylglyoximato)nickel(II), [Ni(Hdmg)2] quantitatively.

ammonia2+ +2 2Ni(HNi + 2H dmg + 2H

Dimethyd

lglyox m)

img

e

N

N

O

OH

N

N

O

O

Ni

H

N

N

O

OH

H

N

N

O

O H

H

Ni2++ +

hydrogen bonding

* The solution is buffered by adding ammonia. It prevents the pH of the solution to fall below5. Otherwise the backward reaction is more favored.

* DMG is soluble in alcohol and sparingly soluble in water. Hence avoid excess addition ofDMG which may crystallizes out along with the chelate.

* But if metals like cobalt are present, which may form soluble complexes with DMG, thengreater amount of DMG must be added.

* Tartarate or citrate ions are added before the precipitation of nickel solution to prevent theinterference from other metals like Cr, Fe etc.,.

* If manganese, zinc, magnesium or other alkaline earth ions are present, ammonium chloridemust be added to prevent the precipitation of their hydroxides along with ammonia.

* To improve the compactness of the precipitate, homogeneous precipitation is often per-formed by adjusting the pH to 3 or 4. This is done by adding urea and heating the solution.Urea undergoes hydrolysis with water by liberating ammonia. A slow increase in the conc. ofammonia causes the pH to rise slowly.

The result is the formation of a more dense, easily handled precipitate. Once the filtratehas been collected and dried, the nickel content of the solution is calculated stoichiometricallyfrom the weight of the precipitate.

* [Ni(Hdmg)2] can have hydrogen bonding in square planar geometry and hence prefers thisstructure over tetrahedral. In tetrahedral geometry, H-bonding is not possible.

* In the solid state, molecules of [Ni(Hdmg)2] pack in vertical columns with relatively shortNi----Ni distances. Because of this, it has high lattice energy and insoluble in water.

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