Inorganic_2.pdf 554m5s

98

Prepared by V. Aditya vardhan

adichemadi(at)gmail.com

WARANGAL

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

http://www.adichemadi.com SOLVED CSIR UGC JRF NET CHEMICAL SCIENCES PAPER 1 (PART-B)

SERIES-2

NOTE: Related and additional questions appeared in previous GATE exams are also solved. 51 52 53 54 55 56 57 58 59 60

2. CO bonded to one metal 4. CO bridging three metals

V. W AD W IT .A Y DI A CH VA EM RD AD HA I.C N O M

51) CO bond order is lowest in 1. uncoordinated CO 3. CO bridging two metals

Explanation: * Carbon monoxide is a  -donor as well as  -acceptor. It not only forms  -bond and also forms 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 of carbon monoxide. Due to this  -back donation, the bond order of carbon monoxide decreases. * The bond order decreases with the number of metal centres with which carbon monoxide forms bonds. Hence the bond order in CO bridging three metals is lowest.

W

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

99



very much increase in energy

* * * 

2px 2py 2p z

WARANGAL

LUMO

HOMO

Energy

 *nb



2p z 2p y 2px



2s


very much decrease in energy



 nb

C

2s

CO

O

* 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 also increased 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 porbital of metal atom and thus acts as  -donor..

+

W

lobe of an empty p or d- orbital

C

O

M

..

M

..

σ - bond formation between Metal and CO

C

O

filled σ-orbital (HOMO)

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

100



WARANGAL

..

..

 - Back bond formation between Metal and CO

+

M

C

O

M

C

O

empty  * -orbital (LUMO)

filled d- orbital

W V. W AD W IT .A Y DI A CH VA EM RD AD HA I.C N O M

* Because of  -back bond formation, the C-O bond order decreases due to increase in the number of electrons in the antibonding orbitals of carbon monoxide. (Note: M-C bond order increases). 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:

Terminal carbonyl

1) Ni(CO)4

ν CO =2057 cm -1

µ 2 bridging carbonyl

Ir

2)

OC

Ir

CO

 CO = 2002 cm -1 (s), 1959 cm -1 (w), 1776 cm -1 (m)

C

O

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 for back-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 stretching frequency 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 for the 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)

101



WARANGAL

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-O bond and thus by decreasing the CO stretching frequency. Me Me

P

C

M

O

Me

and weaker the CO bond

greater the donation

greater is the pi back bonding


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

M

R

..

..

R

+

:

P

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 donating tendency of phosphorus. This strengthen the M-C bond and weakens the C-O bond. (as explained in the previous problem).

W

* 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 Metal to 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 of Ni(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 the ability of metal to back donate electrons to CO. This leads to more weaking of CO bond.


102


WARANGAL

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 at 1857 and 1886 cm-1 disappear. These changes in the infrared spectrum in hexane are due 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. (OC)3Co( 2 -CO) 2 Co(CO) 3   (OC) 4 C o-Co(CO) 4

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

Practice questions 1) The CO stretching frequency of which of the following is higher? 1) Mo(CO)3(PF3)3 2) Mo(CO)3(pyridine)3 3) Mo(CO)6 4) Mo(CO)3(PPh3)3 2) The HOMO in CO is 1)  -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]+ follow the trend: a) i > ii > iii b) ii > i > iii c) ii > iii > i d) iii > ii > i

W

4) Which of the following complexes will have the highest  CO stretching frequency in the IR? Why? A) [Mn(CO)3{P(OPh)3}3]+ B) W(CO)3(PEt3)3 C) [(PF3)Ag(CO)]+ 5) Which of the following complexes will have the lowest  CO stretching frequency in the IR? Why? A) Ni(CO)(PMe3)3 B) 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) H3B  CO and S) [V(CO)6]1) P > R > S > Q 3) Q > S > P > R

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

103



WARANGAL

Note: In borane carbonyl, the CO bond stretching frequency is 2165 cm-1. This is due to prominant  bonding from HOMO (which is non bonding but slightly antibonding) of carbon to LUMO on boron (the CO bond order increases). However, the  -bonding is also possible from B to C. But it does not affect the bond order of CO. The values of  CO Mn(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 CO group? a) 1700 cm-1 b) 1810cm-1 c) 1920cm-1 d) 2140 cm-1 52) The most unstable species among the following is 1. Ti(C2H5)4 2. Ti(CH2Ph)4 3. Pb(CH3)4 4. Pb(C2H5)4 Explanation: * Ti(C2H5)4 is kinetically most unstable due to possibility of  -elimination of hydride.





H2C

Ti

CH2

CH2

H

beta hydride elimination

H2C

H

Ti

The reaction is simply the transfer of a hydride (hydrogen atom) from the  -position on a

W

ligand to the metal center. Usually alkyl metal complexes containing a  -hydrogen decompose by this process. Hence metal complexes containing ethyl ligands are kinetically less stable. An important prerequisite for  -hydride elimination is the presence of an open coordination site on the metal complex. In above case the no. of VE on titanium is 8(electron deficient) 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 configuration in the valence shell of Pb and non availability of d-orbitals. Also Pb-C bond is relatively stronger. Additional information: Stability of alkyl metal complexes * Stability of alkyl metal complexes is influenced by many factors (both kinetic and thermodynamic), which are summarized below. (Most of them are also applicable to other type of complexes) 1) The M-C bond in case of transition metal complexes is less stable when compared to


104


WARANGAL

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 series 3d < 4d < 5d 4) In a given series, stability increases from left to right i.e., complexes of late transition metals(last five in a series) are more stable than those of early transtion metals(first five in a series). 5) Metals in high coordination number are more stable than those in low coordination numbers. 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 that


increases the energy gap (  ) will improve the stability. The following factors increase the o and 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.

W

d) Δ octahedral >Δ tetrahedral . Hence octahedral complexes tend to be more stable. 8) d0 configurations are more stable as there are no electrons to occupy antibonding orbitals. 9) In octahedral complexes, metals with substitution inert configurations (e.g. d3 and low spin 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 field ligands, 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) with d7 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 metals 11) 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 the following order. 1-norbornyl > benzyl > trimethylsilyl > neopentyl > Ph ~ Me >> Et (1oR) > 2o, 3o R In 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 violate Bredt's rule - which states that a double bond cannot be placed at the bridgehead of a bridged ring system, unless the rings are large enough).

M beta 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,

105



WARANGAL

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 the negative charge on carbon can improve the stability of M-C bond. Hence complexes with perfluoro 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 octahedral complexes 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 questions 52.1) 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 t2g3eg2 arrangement and has low CFSE value. Hence low stability and more reactivity (labile). Whereas Cr3+ with d3 configuration has t2g3 arrangement (exactly half filled) and has high CFSE 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.

Configuration t2g3eg2

CFSE -0.4 O

V2+

t2g3eg0

-1.2 O

Ni2+

t2g6eg2

-1.6 O

Ti2+

t2g2eg0

-0.8 O

Ion

2+

Mn

W

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 pressure of ethene, backward reaction (olefin insertion) is favored. (Le Chatelier’s principle)

106


adichemadi(at)gmail.com PPh3

Ph3P

H

+

Ru CH2

Cl

PPh3

Ph3P

Ru

WARANGAL H2C CH2

H

Cl

CH2

52.5) Statement: For the reaction, Ln MH   L n 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 forward reaction.


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)6 2. Mn2(CO)10 3. Fe2(CO)9 4. Re2Cl82Explanation:

Cl

Cl Cl

Cl

Re

Re

Cl

Cl

Cl Cl

W

* 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 a face-to-face manner. * Re2Cl82- (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 of bonding. Presence of  -bond favors eclipsed confirmation rather than expected staggered confirmation.

107


adichemadi(at)gmail.com Cl

Cl Re

Cl

Cl

Cl

Re

Cl

Cl

Cl

Cl

Cl

Cl

WARANGAL

Cl Re

Cl

Re Cl

Cl

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

Cl

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 orbitals corresponds to visible wavelengths of light and hence compounds with quadruple bond are brilliantly 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, d z 2 , d xz , d yz , and d xy orbitals are involved (assuming z-axis is along MM bond) d z 2 orbitals are involved in  bond formation.

W

d xz and d yz are involved in  bond formation. d xy orbitals are involved in  bond formation.

Note: d x 2  y 2 orbitals are involved in M-L bonding.


108


d z 2 -d z 2

WARANGAL

 bond

d yz - d yz

 bond


d xz - d xz

d xy - d xy

 bond

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

L

*

L L

M

L L

L

L

*

M

L

*

Energy d d d d yz xz xy z2

d xy d yz d xz d z 2





W



An M-M bond is formed, as each bonding orbital is filled with an electron pair. The number of bonds reaches 4 (maximum) with d4-d4 systems. But filling of antibonding orbitals cancel 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.

109



Electron Count

WARANGAL

Resulting M-M Bond

d1 - d 1 d2 - d 2

Single bond

d3 - d 3

Triple bond

d4 - d 4

Quadruple bond

d5 - d 5 d6 - d 6

Triple bond

d7 - d 7 d8 - d 8

Single bond

Double bond

optimum

Double bond (M-L bonding usually dominates)


No bond (symmetry interaction)

Additional questions 53.1) A) The number of M-M bonds in Ir4(CO)12 are: 1) four 2) six 3) eight 4) zero B) Explain the fluxionality of Ir4(CO)12 using any of its substituent. Ans:- A) Ir4(CO)12 - Tetrairidium dodecacarbonyl has following structure. OC

OC

Ir

CO

OC

CO

Ir

OC

Ir

OC

CO

CO

Ir

OC

CO

CO

Symmetry: Td

W

B) Fluxionality of Ir4(CO)12 can be explained by merry-go-round mechanism, just like in Co4(CO)12. Illustration: Ir4(CO)11PEt3 exists as two isomers in solution as shown below. This gives rise to fourteen different carbonyl signals. OC OC

Ir

OC CO

OC

OC OC

CO Ir

Ir

OC Ir OC

CO

CO CO

PEt 3

The all terminal structure

OC OC

Ir

CO

CO Ir

Ir

OC CO

Ir

CO

CO CO

PEt 3

The bridged structure

110



WARANGAL

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 [Re2Cl8]2[Re2Cl4(PMe2Ph)4]+ [Re2Cl4(PMe2Ph)4]

Oxidation state of Re +3 +1.5 +2

Configuration

Bond order

d4 d5.5 d5

4 2.5 3


53.3) Using the 18-electron rule, find out the Mo–Mo bond order in [( η 5–)Mo(CO)3]2. Draw the possible isomers for the molecule ( = 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) = 17e Hence there must be a single bond between Mo-Mo. The bond order is 1. OC

CO

CO

Mo

Mo

OC

CO

OC

CO

Mo

Mo

CO

OC

OC

CO

OC

CO

CO

CO

Mo

Mo

OC

CO

CO

W

Practice questions: 1) M-M bond is either not possible or only weak interactions are possible when configurations 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) σ 2 π 4 δ 2

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 σ 2 π 4 δ 2 ground state to σ 2 π 4 δ1 *1 excited state. Assertion: This transition is quantum mechanically allowed.


111


WARANGAL

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 is a) Mn2(CO)10 b) Cr2O72c) Re2Cl82-

d) Hg2(CH3COO)2


54) In which one of the following pairs the species have similar geometry? 1. CO2 and SO2 2. NH3 and BH3 3. CO32– and SO32– 4. SO42– and ClO4– Explanation: 1. CO2 is linear whereas SO2 is angular with a lone pair 2. NH3 is pyramidal with a lone pair whereas BH3 is trigonal planar 3. CO32– is trigonal planar whereas SO32– is pyramidal with a lone pair 4. Both SO42– and ClO4– are tetrahedral.

W

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 between the 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 pair 3) The electron pairs in the valence shell the repel each other and determines the shape of the molecule. 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 repulsion. 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 symmetrical structure, whereas the symmetry is distorted when there are lone pairs along with bond 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 more repulsion than single bonds. 8) The repulsion between electron pairs increases with increase electronegativity of central atom and hence the bond angle increases. 9) Shapes of molecules can be predicted from the number of electron pairs in the valence shell of central atom as follows:


112


WARANGAL

THE GEOMETRY OF MOLECULES CONTAINING ONLY BOND PAIRS IN THE CENTRAL ATOM Number of Formula Molecular geometry Examples bond pairs B A B 2 AB2 Linear BeCl2, BeF2 B

3

AB3

Trigonal planar

BF3, BCl3

A B

B

B

4

AB4

Tetrahedral

A B

CH4, CCl4

B


B

B

5

AB5

B

Trigonal bipyramidal

B

PCl5, PF5

A

B

B

B

B

6

AB6

Octahedral

B

B

B

B

W

SF6

A

113



WARANGAL


GEOMETRY OF MOLECULES CONTAINING CONTAINING ONE OR MORE LONE PAIRS IN CENTRAL ATOM Total number Number Number Shape of of of bond of lone Formula Bond angle Examples molecule electron pairs pairs pairs 120o 3 2 1 AB2E Angular SO2 Trigonal 107o48’ NH3 3 1 AB3E Pyramidal 102o30’ NF3 4 Angular 104o28’ H2O 2 2 AB2E2 (V- shaped) 103o F2O 4 1 AB4E See-Saw SCl4 , SF4 3 2 AB3E2 T-Shape 90o ClF3 5 o 2 3 AB2E3 Linear 180 XeF2, I3Square 5 1 AB5E 90o BrF5 pyramidal 6 Square 4 2 AB4E2 90o XeF4 planar 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 it is linear in shape with 180o bond angle.

180o

Cl

Be

Cl

Linear molecule

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

W

Cl

B Cl

120o

Cl

Trigonal planar shape

3) CH4: The valence shell of the central atom - carbon contains only four bond pairs. Hence it is tetrahadral in shape with 109o28' bond angle. The bond pairs are arranged tetrahedral symmetry so as to minimize repulsions. If the

114



WARANGAL

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

H 109o28' C H

H H

Tetrahedral structure of methane


4) NH3: There are three bond pairs and one lone pair in the central atom, nitrogen. The bond angle is decreased from 109o28' to 107o48' due to repulsion caused by lone pair.

N

H

H

H

107o48'

Trigonal pyramidal structure of ammonia molecule

5) H2O: There are two bond pairs and two lone pairs in the central atom, oxygen. The bond angle is decreased from 109o28' to 104o28' due to repulsion caused by two lone pairs.

O

H

H

104o28'

W

Angular shape of water molecule

Additional questions 53.1) The shape of BrF3 is: a) Trigonol pyramidal b) Trigonal plannar c) Trigonal bipyramidal d) T-shaped Ans:- BrF3 is a T-shaped molecule. The central ‘Br’ atom undergoes, sp3d hybridization in 1st excited state. It is an AX3 type of interhalogen compound. Other e.g. ClF3, BrCl3 etc., F

:

Br :

F

F

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

115



WARANGAL

Ans:- Z of Xe = 63. Ground state valence E.C of Xe = 5s2 5p6 5d0 Structure of XeF2 E.C in 1st ex.state: 5s2 5p5 5d1

5s

5p

5d

sp3d hybridization

Structure is linear (based on trigonal bipyramidal). :

F


:

Xe :

F

Logic: Xe is forming two  bonds which require two unpaired electrons. Hence sp3d hybridization in 1st excited state. Structure of XeF4 E.C in 2nd ex.state: 5s2 5p4 5d2

5s

5p

5d

sp3d2 hybridization

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

F

F

Xe

F

F

W

:

Logic: Xe is forming four  bonds which require four unpaired electrons. Hence sp3d2 hybridization in 2nd excited state. Structure of XeF6 E.C in 3rd ex.state: 5s2 5p3 5d3

5s

5p sp3d3 hybridization

5d

116



WARANGAL

F F F

Shape is distorted octahedral.

Xe

:

F

F F

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


Structure of XeO3 E.C in 3rd ex.state: 5s2 5p3 5d3

5s

5p

5d

sp3 hybridization

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. , the hybrid orbitals form  bonds and the remaining pure orbitals form  bonds. Structure of XeO4 E.C in 4th ex.state: 5s1 5p3 5d4

W

5s

5p

5d

sp3 hybridization

4 bonds

O Xe

Its shape is tetrahedral: 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.

117



WARANGAL

Structure of XeO2F2 E.C in 3rd ex.state: 5s2 5p3 5d3

5s

5p

5d

sp3d hybridization

2 bonds

4 bonds

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


Xe O

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 in the first excited state: ns2np3nd1. The hybridization is sp3d.

ns

np

nd

sp3d hybridization

Cl

Cl

:

Te

W

They have see-saw shapes: e.g.

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.

118



WARANGAL

:

Br O

O

-

O


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 are are arranged over eahc other. Each carbon undergoes sp2 hybridization. It is an electrical conductor due to presence of free 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 such that the B  and N   atoms lie on top of each other; this contrasts with the staggered arrangement found in graphite. Boron nitride is an insulator. It is thermally and chemically stable and find its use in high temperature equipment (ceramics) and nanotechnology.

N

B

B

B

N

N

N

B

B

N

N

B

N

N

N

B

W

N

B

B

B

N

N

B

B

N

B

B

N

N

B

B

N

N

N

N

N

B

B

B

B

B

N

N

B

N

B

B

B

N

B

N

N

N

B

N

B

B

N

N

N

B

B

N

B

53.6) The Lewis acid character of BF3, BCl3, BBr3 follows the order: a) BF3 < BBr3 < BCl3 b) BCl3 < BBr3 < BF3 c) BF3 < BCl3 < BBr3 d) BBr3 < BCl3 < BF3 Ans:- The effectiveness of formation of  -back bonding between halogen and boron atom decreases from B-F to B-Cl to B-Br. Smaller fluorine atom forms  -bond effectively with smaller boron atom. Hence the Lewis acidic nature increases from BF3 to BBr3. (This is contrary arguments using electronegativity.) Another reason is the energy required for the pyramidalization of geometry around boron during 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 pyramidalization energy.

119



WARANGAL

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 tetrahedral geometry requires more energy in case of BF3 (as it requires more energy to break pi-back bonding). Hence it will form complexes less readily. As a result, (CH3)3NBF3 is less stable. X

H3C

X

H3C

X

H3C

H3C N

:

+

X N

B H3C

H3C

B X

X

Note: (CH3)3NBBr3 is more stable.


53.8) The bond angle of Cl2O is 1) Smaller than that of F2O 3) Smaller than that of H2O

2) Greater than that of H2O 4) Same as that of F2O

Ans:Factors deciding the bond angle 1) Hybdrization of central atom. (e.g. sp3 - 109o28’) comment: The bond angle is equal to the angle between hybrid orbitals when there are no lone 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 angle decreases. 3) The electronegativity of central as well as ligand atoms (which are bonded to the central atom). comment: Greater the electronegativity of central atom, lower is the bond angle. When electronegativity of central atom is higher, then the bond pairs will be closer to it. It causes the 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.

W

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. , Cl2O is a paramagnetic molecule. But it does not dimerize as the unpaired electron (odd electron) is localized over the whole molecule (especially on O atoms). 53.8) Describe the structural features of oxyacids of phosphorus.

120



WARANGAL

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 H 3 PO 3

Name Phosphonic acid or Phosphorus acid

Structure O P

OH


H

OH

Preparation:

PCl 3 + 3H 2 O   H 3 PO 3 + 3HC1

Properties:

Dibasic; Oxidation state of ‘P’ = +3 One P-H bond & hence also a reducing agent; Can form salts containing H 2 PO 3 - or HPO 3 2- ; Salts are known as phosphonates or phosphites

H3PO2

O

Phosphinic acid or Hypophosphorus acid

P

OH

H

H

Preparation:

W

Properties:

Ba(H 2 PO2 ) 2 + H 2SO 4   2H 3PO 2 + BaSO4 Note: Hypophosphites salts are in turn are obtained when Phosphorus is treated with alkalis. e.g. P4 + 3NaOH + 3H 2 O   3NaH 2 PO 2 + PH3 Monobasic; Oxidation state of P = +1; Two P-H bonds & hence also a reducing agent; Forms salts containing H2PO2Salts are known as phosphinates or hypophosphites.

121


H3PO4

adichemadi(at)gmail.com O

Orthophosphoric acid IUPAC name is trihydroxidooxidophosphorus

Preparation: Properties:

P

OH

HO

OH

P4 O10 + 6H 2 O   4H 3 PO 4 Tribasic; Oxidation state of P = +5; Forms salts containing H2PO4-, HPO42- and PO43-

O

Pyrophosphoric acid or Diphosphoric acid

O

P

P


H4P2O7

WARANGAL

HO

OH

O

HO

OH



Preparation:

2H 3 PO 4   H 4 P2 O 7  H 2 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

O

O

P

HO

W

Properties:



P

P

O

O

HO

Preparation:

O

OH

OH

OH

H 3 PO 4  H 4 P2O 7   H 5P3O10  H 2O When phosphoric acid is heated, initially pyrophosphoric acid is formed, which upon further heating gives triphsophoric acid. Pentabasic; Oxidation state of P = +5; Na5P3O10 – sodiumtripolyphosphate is used in sulfonate detergents as sequestering agent for Ca2+ and Mg2+ ions. (softens water)

122


H4P2O6


Hypophosphoric acid

WARANGAL

O

O

P

P

HO

OH HO

OH

Preparation: Red P + NaOCl gives Na2H2P2O6 which upon acidification yields hypophosphoric acid. Tetrabasic; Properties: 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 polar molecule shown below. O

O

O

N

N

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 a bromine atom of another molecule and can accept an electron pair from it. As a result, it can dimerize. Br

Br

Al

Al

Br

Br

Br

Br

W

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

Cl B Cl

53.11) 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 central atom - no. of bonds (  &  ) formed by it. Hence the formula is:


No. of lone pairs =

123


WARANGAL

v-b 2

where v = 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 the central atom. For e.g. if this number is 5, then the hybridization is sp3d and the shape is based on trigonal bipyramidal structure. For PCl5 v-b 5-5 = 0 2 2


No. of lone pairs =

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 no distortion. For IF5

v-b 7-5 = 1 2 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 is one lone pair, the shape is square pyramidal. No. of lone pairs =

For SOF4

v-b 6-6 = 0 2 2 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 no distortion. No. of lone pairs =

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

W

For XeOF4

v-b 8-6 = 1 2 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 is one lone pair, the shape is square pyramidal. No. of lone pairs =

53.12) Write the strutures of the following compounds A ) 1,2-B 4C2H6 B) H3F4

C) SF4CH2

124



WARANGAL

H H

C

B

A)

B

B

H

H B

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

C H

H

F

H

F

H

P

B) F

tetrafluoro(methyl)-5-phosphane

H F

Note:The more electropositive (less electronegative) -CH3 group occupies the equitorial position. F

H


C)

F

S

F F

(tetrafluoro-6-sulfanylidene)methane

C

H

or methylene sulfurhexafluoride

Practice questions: 1) The hybridization in SF6 molecule is : a) sp3 d2 b) sp2d3

c) sp3d3

d) sp3

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

List – I XeF4 Ni(CN)53NO2+ Zn(CN)42-

List –II A. Angular B. Linear C. Square pyramid E. Tetrahedral F. Square planar

Note: NO2+ is angular; Ni(CN)53- is square pyramidal;

W

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 5) The square pyramidal molecular shape is adopted by 1) SOF4 2) XeOF4 3) ScOCl4 4) PF5 H.W: What are the structures of molecules given under other options?

4) IF7, XeF6


125


WARANGAL

6) The structure of XeO2F2 based on VSEPR theory is best described as 1) 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 H 3 N  BF3 is much _____ than in BF3. (ans: longer) 8) The molecule ClF3 has ____ non-bonded electron pairs. (ans: 2) (ans: pi-back bonding;


9) The planar structure of N(SiH3)3 is due to ______ 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, I3Note: All, except SCl2, are linear. 12) The minimum number of electrons needed to form a chemical bond between two atoms is: a) 1 b) 2 c) 3 d) 4 13) The pair of compounds having the same hybridization for the central atom is a) XeF4 and [SiF4]2b) [NiCl4]2- and [PtCl4]2c) 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

W

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. O

O

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

-

O Xe

O

-

O O

16) The series with the correct order of decreasing ionic size is a) K+ > Ca2+ > S2- > Clb) S2- > Cl- > K+ > Ca2+ + 2+ 2c) K > Cl > Ca > S d) Cl- > K+ > S2- > Ca2+ 17) The structure of SF4 is:

-

-

126



WARANGAL

a) Trigonal bipyramidal b) tetrahedral c) Octahedral d) square planar Note: According to some authors structure and shape are different . In above case, structure includes the lone pairs along with bond pairs. Whereas, shape includes only bond pairs 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 & pyramidal c) Both pyramidal d) Both trigonal planar Note: Trisilyl amine, N(SiH3)3 is trigonal planar due to sp2 hybridization of central nitrogen atom. There is a p  d bond formed due to overlapping of p-orbital with lone pair on nitrogen with empty d-orbital on silicon. Hence not only it is planar and also less basic. Its Lewi 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 XeO3 c) bent XeF2 and planar XeO3 d) linear XeF2 and tetrahedral XeO3 20) The chemical formula for hypophosphoric acid A) 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 C) 2 P-O and 4P = O bonds Structure:

B) 12 P-O and 4P = O bonds D) 6 P-P, 12 P-O and 4P = O bonds

O

P

O

O

O

O

O

P

O

W

O

P

P

O

O

23) Among the following isostructural compounds, identify the compound which has the highest lattice energy. A) LiF B) LiCl C) NaCl D) MgO Hint: Lattice energy is proportional to the product of charges on either ions and inversely related to the distance between them. Greater the charge and small the size, greater is the lattice 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

127



WARANGAL

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) SbF52Followup question: Theoretically how many isomeric forms are possible for PF3Cl2. Which among them is stable? Ans: Three isomers are possible. Among them, the isomer which has two Cl atoms present on equatorial position is more stable. Usually the electron pairs with larger volume tend to occupy equatorial positions. As Cl is bigger in size and less electronegative when compared to F, they tend to occupy two equatorial positions. F

F

Cl F

P

F

P

P


F

Cl

F

F

Cl

Cl

F

F

Cl

Cl

More stable

25) Which statement best describes the polarity of SF4Cl2? a. Always polar b. Always nonpolar c. Depending on the arrangement of outer atoms, this molecule could be polar or non polar. 26) The compound (SiH3)3 N is expected to be a) pyramidal and more basic than (CH3)3N b) planar and less basic than (CH3)3 N c) pyramidal and less basic than (CH3)3N d) planar and more basic than (CH3)3N

27) The nature of  -bond present in perchlorate ion is A) O  p   Cl  p 

B) O  d    Cl  p 

C) O  pπ  - Cl  dπ 

D) O  d    Cl  d  

W

O

-

Cl

O

O

O Perchlorate ion

28) The hybrid orbitals used by bromine atom in BrF3 are a) 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) ICl2d) PCl3


128


WARANGAL

30) The species which has a square planar structure is a) BF4– b) FeCl4– c) SF4 d) XeF4 31) In allene, hybridization of the central and terminal carbons, respectively, are a) 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 c) both bent

b) both linear d) bent and linear, respectively


33) Among the following molecules, the shortest bond length is to be found in a) C2 b) N2 c) O2 d) F2 o Note: C2 1.24 A 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 BF3 c) 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 pyramidal 2) [SiH3]–; trigonal planar 3) CS2; bent 4) [CMe3]+; trigonal planar 37) sp3d2 hybridization is observed in a) BrF b) ClF3

c) BrCl3

d) ICl5

W

38) The incorrect combination among following is a) ICl – sp3 – linear b) BrCl3 – sp3d – T-shape c) IF5 – sp3d2 – square pyramidal d) 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 planar B) trigonal planar, trigonal planar C) T-shaped, trigonal planar D) trigonal bipyramidal, T-shaped 41) Of the following, only ____ has sp2 hybridization of the central atom.

129



B) I3-

A) ICl3

WARANGAL

D) CO32-

C) PF5

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 the separation, r, in the absence of any other ions? a)  1/r b)  1/ r 2 c)  1/ r 3 d)  1/ r 6 Explanation: In the absence of any other ion, relation between the various types of interactions and the distance of separation are given below:  1/ r ---- for charge-charge interactions.  1/ r 2 ---- for ion dipole interactions.  1/ r 3 ---- for dipole dipole interactions.


 1/ r 6 ---- for dispersion interactions.

55) On oxidative addition of O2 to Ir(CO)Cl(PPh3)2, the oxidation state and coordination number of Ir changes, respectively, by 1. 1 and 3 2. 2 and 2 3. 3 and 1 4. 2 and 3 Explanation:

PPh3

O

C

+ O2 Oxidative Addition

Ir

Cl

179 pm

PPh3

C

190 pm

O

Ir

Cl

O

PPh3

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

W

Ir(CO)Cl(PPh3)2, trans-chloridocarbonylbis(triphenylphosphine) iridium(I)

PPh3

O

* Ir(CO)Cl(PPh3)2 is a square planar complex which is often referred to as Vaska’s compound. 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), and hence C.N is 6. (contains 18 VE) * The Dioxygen added is in peroxidic form (O22-) and not in superoxidic form (O2-). * Therefore both oxidation state and coordination number are changed by 2 (oxidation state from +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-

130



WARANGAL

trons in Iridium(I). Hence it can bind to ligands and expand its valence electrons to 18 by undergoing 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 during oxidative addition reactions. It is due to reduction of amount of  -back bonding from Ir to C 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 oxidative addition. Additional questions: 55.1) How do you prepare Co2(CO)8? Give equation only. Ans:- 2CoCO3  2 H 2  8CO   Co2 (CO )8  2 H 2O  2CO2

Ans:-


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

Ph3P

PPh3

Ir

Cl

PPh3

Cl2

Cl

PPh3

Ph3P

Ir

PPh3

Cl

Cl

+

PPh3

Ph3P

Ir

Cl

Cl

PPh3

meridional

facial

56) In linear metal nitrosyls NO acts as a/an: 1. One electron donor 2. Two electron donor 3. Three electron donor 4. Four electron donor

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

W

N

M

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

O

M

* 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, in both the cases, it is considered as two electron ligand.


131


WARANGAL

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


Neutral counting (covalent method): This approach assumes the molecule or fragment being studied consists of purely covalent bonds. It is usually considered easier especially for low-valent transition metals. * Locate the central atom and determine the number of its valence electrons by assuming it as a neutral atom. The counting of valence electrons for transition elements is different from main group elements. -In case of main group elements (s and p block) the electrons in the ‘s’ and ‘p’ orbitals of outer nth shell are considered for this purpose. -Whereas in case of transition metals, the electrons in ‘(n-1)d’ and ‘ns’ orbitals are taken into . * Add one for every halide or other anionic ligand which binds to the central through a sigma bond. * Add two for every lone pair bonding to the metal (e.g. each Lewis base binds with a lone pair). 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.

W

Ionic counting (Effective atomic number method): This approach assumes purely ionic bonds between atoms. * Calculate the number of valence electrons of the central atom by assuming an oxidation state 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 sigma bond. * 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 the metal. Each carbon atom contributes one electron.

132



WARANGAL

Number of electrons donated by some common ligands: Ligand Electrons contributed Electrons contributed (in neutral counting) (in ionic counting) X (Halide)

1

2

H

1

H

1

2 (as H-) 0 (as H+)

O

2

N NR3 (R = H, alkyl, aryl)

3

CR2

2

Ethylene

2

4 (as O2-) 6 (as N3-)

2

2 4 (as CR22-)


2 (as C2H4) 6 (as C5H5-) 6 (as C6H6)

cyclopentadienyl

5

benzene

6

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 following molecules / complexes. Also conclude whether the molecule is stable or not. a) Ferrocene

b)

c)

Ir

OC

d)

CO

O

Bu

Ta

Ir

C

t

Mn

OC

CO

OC

t

Bu

t

t

Bu

Bu

W

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 5 each: i.e., 8 + 2(5) = 18 valence electrons in iron Ionic counting: Fe2+ contributes 6 electrons, the two aromatic cyclopentadienyl rings contribute 6 each: i.e., 6 + 2(6) = 18 Conclusion: Ferrocene is a stable compound as the central metal atom contains 18 electrons (according to 18 electron rule)


133


WARANGAL

b) Ir

Ir CO

OC C O

c)

OC


* 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 assign only 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 CO contributes only one. i.e., 9 + 5 + 1+ 2 + 1 = 18 Ionic counting: Ir+ contributes 8 electrons, the cyclopentadienyl ring contributes 6 electrons, Ir-Ir bond contributes one electron, terminal CO contribute 2 electrons and bridged CO contributes only one. i.e., 8 + 6 + 1+ 2 + 1 = 18 Conclusion: It is also a stable compound. But this complex and its rhodium analog show fluxional behavior in solution that exchanges the bridging and terminal carbonyls.

Mn

CO

OC

* e.c. of Mn (Z = 25) is [Ar] 3d54s2 * oxidation state is +1.

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

t

Bu

t

W

Ta

t

Bu

t

Bu

Bu

* 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 counting formalism. In ionic counting formalism, it contributes 4 electrons. * The alkyl ligands are monoanionic. Each of alkyl contribute only 1 electron in neutral method, 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)

134



WARANGAL

Ionic counting: 0 + 1(4) + 3(2) = 10 Conclusion: It defies 18 electron rule and expected to be unstable. But it is kinetically stable due 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 it adopts a bent coordination geometry. Discuss (using Lewis dot-like figures) the distribution of electrons in both kinds of M-NO complexes and how these affect the structures (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 count for 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 positive charge work together to make NO+ the strongest  -backbonding ligand known. In fact, it can backbond enough to formally oxidize the metal center by two electrons to transform the NO+ ligand into a NO- ligand. This is evident in the transfer of a pair of electrons from the metal to the NO+ ligand to produce the bent NO- ligand in some of metal complexes. The lone pair that used to be on the metal center is now on the nitrogen of the NO- ligand. This is what is the meaning of hyper-  backbonding. When the NO+ ligand turns into a NO- ligand it has formally oxidized the metal center by 2 electrons. That represents the “ultimate” in  backbonding. 56.3) In the complex, [Ni2( η5 -)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-CO bond respectively are: Ans: CO stretching frequencies (two peaks around 1850 cm-1) indicate their bridging nature. VE count (ionic method): O

W

C

Ni

Ni

C O

- oxidation number of Ni = +1 - Each bridged CO contributes one electron to each metal. - contributes 6 electrons. - Ni-Ni bond contributes one electron. VE = 9 + 6 + (2 x 1) + 1 = 18 Practice questions:

135



WARANGAL

1) The N–O stretching frequency in linear transition metal nitrosyls is ______ that in nonlinear 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  C5 H 5 ) Fe(CO) 2

b) ( 5  C5 H 5 ) Mo(CO) 3

c) ( 5  C5 H 5 ) 2 Co

d) (η 5 - C 5 H 5 )Re(η 6 - C 6 H 6 )

4) Among the complexes (i) (C6H6)2Cr; (ii) [HMn(CO)5]; (iii) [(CH3CO)Rh(CO)I3]- and (iv) Fe(CO)2(CH3), the 18-electron rule is not followed by A) 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 is a) 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 for 1. NH3 2. trans-[PtCl2(NH3)2] 3. BF3 4. NF3 Explanation:

F

Cl

Pt

N

N

B

H

W

H

NH3

NH3

H

Cl

F

F

F

F

F

* Only NH3 and NF3 have dipole moments. NF3 has highest dipole moment, as the magnitude of N-F bond moment is larger. * trans-[PtCl2(NH3)2] and BF3have zero dipole moments due to symmetrical structure and as the 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 less symmetrical than trans isomer. * , BF3 is trigonal planar whereas, NH3 and NF3 are trigonal pyrimidal. Additional information:

136



WARANGAL

* Dipole moment (  ) is defined as the product of magnitude of charge(  ) on either of the poles (atoms) and the distance(d) between them.  =  .d * It is a vector quantity. It is represented by an arrow pointing towards negative pole (as shown in above diagrams). * The dipole moment of a bond is called bond moment. The final dipole moment of a molecule 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. % ionic character =

µ experimental µ theoretical

x 100

µ theoretical is calculated 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 is 1.275Å. Assuming 100% ionic character, the charge developed on H and Cl atoms would be 4.8 x 10-10e.s.u. Therefore, the theoretical dipole moment for 100% ionic character will be µ theoretical =  x d

% ionic character =

= 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.)

µ experimental µ theoretical

x 100 =

1.03 x 100 = 16.83% 6.12

* It is also used to determine the shape of a molecule. Usually, the symmetrical molecules have 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 resultant dipole moment are known. Following simplified equation can be used when the two bonds are same. net dipole moment = 2(bond moment) x cos

θ 2

where θ = bond angle

W

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 cos 0.95 = 2(0.72) x cos θ cos = 0.6597 2 bond angle = θ = 97 0

θ 2

 2

137



WARANGAL

O

O H

F

F


H

:

:

:

:

57.1) The dipole moment of the H2O molecule is 1.85 D; that of the F2O molecule is only 0.25 D. Explain. Ans:- Both molecules have angular structure, so differences in shape cannot for the difference in dipole moment (neglecting the slight bond angle difference here). Oxygen is more electronegative than hydrogen, so in H2O, the bond and non-bonding-pair dipoles reinforce each other (as they are acting in same direction). Hence more dipole moment. 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.

Practice questions: 1) The dipole moment of a hydrogen halide H-X is 1.8 D and the bond length is 1.5 Ao. The percent 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 is 1) NF3 2) BF3 3) CH2Cl2 4) NO2

W

5) Dipole moment of p-nitroaniline, when compared to nitrobenzene (X) and aniline (Y) will be 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 -NO2 group 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 on hydrogen and chlorine, respectively, are: A) +1.0 and -1.0 B) +0.85 and - 0.85 C) +0.356 and -0.356 D) +0.178 and -0.178 7) The molecule that does not possess a permanent dipole moment is a) NF3 b) BF3 c) CH2Cl2 d) NO2 58) An element 'X' emits successively two β particles, one α particle, one positron and one neutron. The mass and atomic numbers of the element are decreased by, respectively, 1. 4 and 1

138



WARANGAL

2. 5 and 1 3. 3 and 2 4. 3 and 1 Explanation: *  -particle is 2He4 nuclide with two positive charges. Hence loss of one  -particle results in decrease of atomic number(Z) by two and mass number(A) by four. (Hence the position of new element formed will be two groups left to the parent element) *  -particle is nothing but an electron. It can be represented by -1e0. It is emitted by the nuclide when a neutron is converted to proton. Hence the atomic number(Z) is increased by one. There is no change in mass number. (Therefore, the position of the new element formed will be one place right to the parent element) n1   1 p1  1e0  neutron proton electron ( particle)

0 0 e

 electron neutrino


0

* 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 again without any change in mass number. (Therefore, the the position of new element formed is shifted one group left to that of parent element) p1 proton

0   0 n1   00 e 1e neutron positron neutrino ( + )

1

*

Z

2    positron  neutron X A  Z  2 X A  Z X A4  Z 1 X A 4  Z 1 X A5

Therefore, the answer is 5 and 1

Additional information: TERMINOLOGY, CONCEPTS & FORMULAE OF NUCLEAR CHEMISTRY Nuclide: The atom with specific atomic and mass number. In nuclear reactions, different isotopes (same atomic number but with different mass numbers) of same element are considered as different nuclides with different nuclear properties. But for a chemical reaction, isotopes are considered to have same chemical properties and hence are not differentiated. Isotopes: The nuclides with same atomic number but different mass numbers. 12 e.g. 1H , 2 H , 3 H ; C , 13C ; 14 N , 15 N etc., 1

1

1

6

6

7

7

W

* 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 20

Ca,

40 19

K,

40 18

Ar ;

14 6

C,

14 7

N

Isotones: The nuclides with same number of neutrons. e.g.

40 20

Ca,

39 19

K,

38 18

Ar ;

32 16

S,

31 15

P,

30 14

Si

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


139


WARANGAL

Fermions: e.g. Neutrons, protons, electrons etc., * obey Pauli principle. * have anti symmetric wave functions. * have half integer spin values. * Fermions are again divided into 1) 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 in the range of 10-5 m. Radius of nucleus can be calculated empirically as follows: Radius of nucleus = R = R o A1/3 Where R o =1.2 x 10-13cm = 1.2 x 10-15 m = 1.2 fermi A = Mass number MeV: 1eV = 1.6 x 10-19 joules 1 MeV = 106 eV 1 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 due to spontaneous disintegration.

 particle = 24 He nuclide with 2+ charge  particle = -10 e (electron)  radiation = electromagnetic radiation

* 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 by Bq (or s-1)

W

Curie (Ci): 3.7 x 1010 disintigrations per second is called the Curie. It is an older and non SI unit. 1 Ci = 3.7 x 1010 Bq Kinetics: * Radioactivity is a first order reaction. The rate expression can be written as follows:  dN  rate = -     Nt  dt  Where:  = decay constant N t  No.of radioactive atoms at time 't'

140



WARANGAL

* The integrated form of above equation is: ln N t   t  ln N o or N  t   ln t or No or 1 No   ln t Nt or N 2.303  log o t Nt

Nt  e  t No

or

N t  N t e  t


Where : N o  initial no. of radioactive atoms

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

ln 2 2.303log 2 0.693     

Average life(  ):

Average life= =

1 t  1/ 2  1.44 x t1/ 2  0.693

Graphical method: The first order equation of radioactive decay can be written as follows: ln N t   t  ln N o

(it is in the form of y = mx + c)

Hence a graph of ln N t versus t will be a straight line, with a negative slope equals to  and an intercept, ln N o on y-axis.

ln N o

   slope

ln N t

W

t

t N  log o 2.303 Nt

OR (it is in the form of y = mx)

No Hence a graph of log N versus t is a straight line ing through the origin and has a t

slope of

 . 2.303

141



log

No N

slope 

WARANGAL

 2.303

t

STABILITY OF NUCLEUS Stability of nucleus is affected by the following factors:

number of neutrons


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 stability.

belt of stability

number of protons

* The nuclides present above and below this belt are unstable and tend to get stability by following 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. n1   1 p1  1e0  neutron proton electron ( particle) 0

0 0 e

 electron neutrino

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

W

p1 proton 1

0   0 n1   00 e 1e neutron positron neutrino + ( )

In case of electron capture (also called inverse beta decay), an electron within the atom is captured by the proton in the nucleus and is converted to neutron. 0 p1    1e proton electron (captured) 1

0 n1  + 0 e neutron neutrino 0

h

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

142



WARANGAL

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 they undergo 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 from individual nucleons. This energy liberated is called Binding energy.


Binding energy * Binding energy of nucleus is calculated by using Einstein’s equation. B.E = Δmc 2 * 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 with atomic number greater than 26 (atomic number of Fe) tend to undergo fission whereas those with 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 neutrons are exceptionally stable. e.g. 24 He,

16 8

O,

40 20

Ca,

208 82

Pb etc.,

* Nuclides with even number of protons or neutrons or total nucleons are also observed to be stable. RADIOACTIVE DISINTEGRATION SERIES Following four types of decay series are reported.

Starting nuclide 235 92 U

Type Natural 4n+3

Uranium series Neptunium series Thorium series

Natural

4n+2

Artificial

4n+1

237 93

natural

4n

232 92

W

Decay series Actinium series

238 92

Stable end product 207 82 Pb

U

206 82

Np

209 83

Th

208 82

Pb

No. of particles emitted Alpha Beta 7 4 8

6

Bi

7

4

Pb

6

2

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


143


WARANGAL

* In natural radioactive series, the final nuclide is lead. * The uranium series is also called as radium series. * The of Neptunium series are NOT found in nature because the half-life of the longest 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 and not 209 83

209 83

205 81

Tl

Bi .

Bi is discovered to be radioactive and further disingegrates to

205 81

Tl .


Nuclear Fission: * Splitting of heavy nucleus into two lighter nuclei of almost equal size. It usually accompanies the emission of neutrons and enormous amount of energy due to conversion of mass into energy. 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 chain reaction. The neutrons emitted further collide with the uranium nulei and help in continuation of the chain. Around 200 MeV of energy is liberated in each fission. 235 92

92 1 U  01n   141 56 Ba  36 Kr  3 0 n  200 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, density and purity. Note: A critical mass is the smallest amount of fissile material needed for a sustained nuclear chain reaction U235 is suitable for fission as it has high probability of undergoing fission, whereas U238 has less 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% of uranium is in the form of U235. Plutonium-239 has highest probability of fission.

W

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. 60 3) Co: 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, and kidney imaging 6) 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 the mechanisms 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

144



WARANGAL

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. A 4 Z 2

 

N daughter

A Z



 X parent  

A Z 1

M daughter

Note: 1) However this rule is not valid for emissions by f-block nuclides, as the parent and daughter 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)

23 11

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

β + emission. 23 23 24 (i) 10 Ne (ii) 12 Mg (iii) 22 11 Na (iv) 11 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 converted to a neutron and a positron(   ). The N/P ratios and the decay processes are given below. i)

N 13 23 ratio for 10 Ne =  1.3 ; Hence it undergoes   decay. P 10 23 0  23 10 Ne  11 Na + -1 e

ii)

iii)

N 11 23 ratio for 12 Mg =  0.92 ; Hence it undergoes   decay. P 12 23 23 0 Mg   12 11 Na + +1 e

N 11 ratio for 22  1 ; It should be stable. But it undergoes   decay ??? 11 Na = P 11 22 22 0 Na   11 10 Ne + +1 e Reason: Always the nuclides with N/P ratio =1 are not stable. The nuclides present in the

iv)

22 11

Na is present below the stability belt. Hence undergo positron

W

stability belt are only stable. decay.

N 13 ratio for 24  1.18 ; Hence it undergoes   decay. 11 Na = P 11 24 24  12 Mg + -10 e 11 Na 

58.3) Which isotope is produced by an (n,  ) reaction starting from 230Th? 1) 229Th 2) 229Ac 3) 231Th 4) 231Pa 231 230 Ans:- Th is formed along with the emission of  radiation, when Th is bombarded with neutron. In this reaction only mass number is increased. The short hand notation for this reaction is 230 Th (n,  ) 231Th It can also be represented as 230 Th + 0n1 -----------> 231Th + 

145


adichemadi(at)gmail.com -

58.4) How many α particles and β particles are emitted in the decay of

WARANGAL 238

U to

206

Pb?

Ans:- In the conversion of 238 U   206 92 82 Pb, Mass number is decreased by 238-206 = 32.

32  8. 4 Hence the decrement in atomic number should be 8 x 2 = 16. But the observed decrement is 92 - 82 = 10. Therefore, 6  particles are also released (which increase the Z value by 6). Hence the number of  -particles emitted will be

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


Ans:- rate of disintegrations = r =  N t

where  = decay constant;

N o x wt. of sample M Tc where N 0 =avogadro number; M Tc =molecular weightr of Technitium N t = number of 'Tc' nuclei =

6.02 x 1023 x 1.0 x 10-3 g = 6.08 x 1018 nuclei 99 g Hence r 6.3 x 105 nuclei . sec-1 =   1.036 x 10-13 sec-1 18 Nt 6.08 x 10 nuclei Nt =

58.6) The decay constant of radioactive disintegration of a nuclide is 3.010 x 10-3 s-1. What are the half life and average life of that nuclide? Ans:- Since radioactive disintegration is a first order process, ln 2 0.693 0.693    230.3 s   3.010 x 10-3 s 1 2 where λ=decay constant

Half life  t 1 

W

t1 1 Average life= =  2  1.44 x t 1  1.44 x 230.3 = 331.6 s  0.693 2

58.7) In how many days will a 12-gram sample of a radioactive isotope decay, leaving a total 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

146


Ans:-

2n =

N 0 W0 = N W


WARANGAL

where n = no. of half lives

12 =8=23 1.5 Hence the no. of half lives = 3

 2n =

to become Time taken for W0 (12g)   W(1.5g ) = n x t1/2  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  146 C  11H 7 N  0 n  14 7

N (n, p ) 146 C


Short hand notation

58.9) 10B, used in nuclear reactors, participates in two reactions. One reaction yields 7Li and another product X. The other reaction yields 3T and X. Identify X and write two balanced nuclear reactions. Ans:- In nuclear reactors, neutrons are produced due to nuclear fission. These neutrons collide with 10B. In both the reactions, shown below, the X is alpha particle(4He2).

1) 2) 58.10)

10 5

B + 01n   73 Li + 24 He

10 5

- one alpha particle is formed.

B + 01n   31T + 2 24 He - two alpha particles are formed.

201 81

Tl 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 (c) Given 10.0 grams of pure after 12.4 min? Ans:- (a)

201 81

Tl to be reduced to 0.20 gram by decay?

201 81

Tl initially, how many grams of pure

201 81

Tl will remain

201 81

0 Tl   201 82 Pb  1 e

ln 2 0.693 0.693    t1/ 2 3.1 min where  =decay constant

(b)

W



time taken = t =

W 2.303 2.303 x 3.1 2 log o  log  10.33min  W 0.693 0.2

(c) The number of half lives in 12.4 minutes = Hence the amount of

201 81

total time 12.4  =4 half life 3.1

Tl remained after 31 minutes = W 

Wo 10   0.625 g 2 4 16

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)


Ans:t

147


WARANGAL

N 2.303 x t1/ 2 2.303 Initial activity log o  log  Nt 0.693 Final activity 2.303 x 8.02 100 = log 0.693 1 2.303 x 8.02 100 = log 0.693 1 =53.4 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 carbon shows 15.0 disintegrations per minute, what will be its activity after 10,000 y?

N A t  log o  log o 2.303 Nt At 0.693t A  log o 2.303t1/ 2 At


Ans:-

or log

15 0.693 x 10000   0.56 At 2.303 x 5370

15  100.56  3.63 At 15 or At   4.13 dis/min 3.63 or

58.12) The mass of one atom of chromium-52 is 51.9405 amu. What is the binding energy per nucleon for the chromium-52 nucleus? Ans:- There are 24 protons and 52 - 24 = 28 neutrons in

56 24

Cr .

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

W

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 produces a three-dimensional image or picture of functional processes in the body. The system detects 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 by the nuclide encounters an electron and annihilated with it to produce a pair of gamma photons which move in opposite directions.

148



WARANGAL

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

d) 92U232


Practice questions: 1) The radioactive isotope used to locate brain tumors is a) 1D2 b) 16S31 c) 53I131

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. The elements after Uranium which are called trans-uranic elements are not only radioactive and also artificially 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.

W

5) Which of the following changes occurs in the nucleus when a positron particle is given off during 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 coronary artery disease is suspected. If after treatment the initial radioactivity of a person's blood is 20,000 counts per minute, how much time would have elapsed in order for the activity to decreast 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

149



ii. CAT (Computerized Axial Tomography) scan iii. MRI (Magnetic Resonance Imaging) scan iv. Iodine-131 treatment of hyperthyroidism Which of these procedures involves X-rays? 1) i and iii 2) ii and iv 3) ii only

WARANGAL

4) i only

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

14 7

N   168 O

b)

238 92

U   234 90Th

c)

27 13

Al   1531P

d)

28 14

31 Si   16 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 following naturally-occurring nuclides is radon-222 produced? a) carbon-12 b) potassium-40 c) uranium-238 d) thorium-232 Note: 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 be 87.5% 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 &

23

Na

4) 25Mg & 26Al

W

15) The chemical properties of atoms, molecules and ions is determined primarily by the number 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 a radioactive 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.

150



WARANGAL

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 8

O   157 N  __

A)

0 +1

e

B) -10 e

C) 42 He

D) 00 

21) What will be the resulting product, during the K-capture by a germanium-68 nucleus? 68 72 A) 68 B) 68 C) 32 D) 34 Ge Se 33 As 31 Ga Note: Electron capture, in which an inner-shell (K or L) electron is captured by a proton in the nucleus with the formation of a neutron. X-rays are emitted as the electrons cascade down to fill the vacancies in the lower energy levels.

22) If 87.5 percent of a sample of pure A) 6 days B) 8 days

99

Rh decays in 48 days, what is the half - life of C) 12 days D) 16 days

99

Rh ?

240 23) The first part of the decay series of 94 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)

228 90

Th

C)

228 89

Ac

232 D) 90 Th

24) What is the missing product in the following nuclear reaction? U   410 n + 136 53 I +__

a)

90 39

W

236 92

Y

b)

96 38

Sr

c)

96 39

Y

d)

98 40

Zr

25) The average number of neutrons produced during the nuclear fission of uranium-235 is a) 1 neutron

b) 2.5 neutrons

c) 3 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.

4) 4 neutrons


151


WARANGAL

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 are important 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 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 binding energy 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 on which one of the following processes? 1) neutron bombardment 2) fission 3) 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. The half-life of yttrium-90 is: a) 0.20 day b) 2.5 day d) 3.8 day e)14 day Hint: Do rough calculations.

d) Actinium series

W

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

34) The radioactive decay of 239 is an example of  239 92U  94 Pu a) zero order reaction b) photochemical reaction c) 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 the amount of 90Sr remaining after 5 years are: a) 1.44 years and 0.916 g b) 57.6 years and 0.75 g c) 28.8 years and 0.887 g d) 100 years and 0.982 g

152



WARANGAL

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 has 1. One singlet 2. One multiplet 3. Two singlets 4. Two multiplets


Explanation: * Ferrocene can exists in two confirmations: 1) more stable staggered and 2) less stable eclipsed. At high temperatures, the rotation is possible and the two forms will exist in equilibrium (mostly eclipsed). * Hence at high temperatures only one singlet is observed - as all the hydrogens are equivalent. Under NMR timescale all the protons on cyclopentadienyl ligand are equivalent at room temperature. * But at low temperatures, two signals are possible as the rotation is freezed and ferrocene prefers 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 rings readily.

W

60) In the estimation of Fe2+ by Cr2O72– using barium diphenylamine sulfonate as indicator, H3PO4 is added to 1. maintain the pH of the medium 2. decrease the Fe2+/3+ potential 3. increase the oxidizing power of Cr2O72– 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; and 3. the E 0 (elctrode potential difference and not internal energy change) is large enough to


153


WARANGAL

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+ + Cr2O72-+ 14H+-----------> 6Fe3+ + 2Cr3+ + 7H2O

E 0  0.56 V

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

E 0  0.77 V

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

E 0  1.33V

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


* As the dichromate ion is orange and chromic ion is green, the end point is masked. Hence this titration requires a redox indicator. ( A redox indicator is a compound that changes color over a potential range of interest due to redox reaction) * The common redox indicators used for this purpose are diphenylamine, diphenylbenzidine and Barium diphenylamine sulfonate. The color change for all these indicators is from green to 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 required standard potential (0.78 V) of redox indicators, phosphoric acid is added to lower the potential of Fe3+/Fe2+ couple. (As a result, the electrode potential of the overall reaction is increased 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 the concentration of Fe3+ ions is decreased. This lowers the potential of Fe3+/Fe2+ couple (, the electrode potential depends on the concentration). Note: Barium diphenylamine sulfonate is the best choice among the indicators as it is more soluble in water than others. Additional information: * Iron can also be determined by titrating Fe(II) with permanganate in acidic medium. The reduction half reaction of permaganate can be represented as

W

MnO4- + 8H+ + 5e- --------> Mn2+ + 4H2O E 0  1.51 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.HCl and Fe(III) is reduced to Fe(II). Hence there is likelihood of following oxidation of Cl- ions during the titration of Fe(II) ions against permanganate. This interferes the actual process. 2MnO4- + 10Cl- + 16H+ --------> 2Mn2+ + 5C12 + 8H2O E 0  1.36 V * To avoid above problem, Zimmermann and Reinhardt's solution (this is sometimes termed preventive solution) is added during the titration. This solution contains MnSO4 and phosphoric acid. * The manganese(II) sulphate lowers the reduction potential of the MnO4- - Mn(II) couple and thereby makes MnO4-, a weaker oxidising agent and the tendency of the permanganate ion 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.

154



WARANGAL

Note: Potassium permanganate is not a primary standard. It is difficult to obtain the substance perfectly pure and completely free from manganese dioxide. Moreover, ordinary distilled water is likely to contain reducing substances (traces of organic matter, etc.) which will react with the potassium permanganate to form manganese dioxide. The presence of the latter is very objectionable because it catalyses the autodecomposition of the permanganate solution on standing. GRAVIMETRIC DETERMINATION OF NICKEL * Hot and faintly acidic (dilute) solution of nickel salt is treated with 1% ethanolic solution of dimethylglyoxime (H2dmg) and then adding a slight excess of aqueous ammonia solution to get a red precipitate of bis(dimethylglyoximato)nickel(II), [Ni(Hdmg)2] quantitatively. Ni 2+ +

2H 2 dmg

ammonia   

Ni(Hdmg) 2 

+

2H +


Dimethylglyoxime H

H

O

N

+

Ni

2+

N

O

O H

N

Ni

N

H

O

N

+

N

H

O

O

N

N

O

H

O

hydrogen bonding

* The solution is buffered by adding ammonia. It prevents the pH of the solution to fall below 5. Otherwise the backward reaction is more favored. * DMG is soluble in alcohol and sparingly soluble in water. Hence avoid excess addition of DMG which may crystallizes out along with the chelate. * But if metals like cobalt are present, which may form soluble complexes with DMG, then greater amount of DMG must be added. * Tartarate or citrate ions are added before the precipitation of nickel solution to prevent the interference from other metals like Cr, Fe etc.,.

W

* If manganese, zinc, magnesium or other alkaline earth ions are present, ammonium chloride must be added to prevent the precipitation of their hydroxides along with ammonia. * To improve the compactness of the precipitate, homogeneous precipitation is often performed 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. of ammonia causes the pH to rise slowly. The result is the formation of a more dense, easily handled precipitate. Once the filtrate has been collected and dried, the nickel content of the solution is calculated stoichiometrically from the weight of the precipitate. * [Ni(Hdmg)2] can have hydrogen bonding in square planar geometry and hence prefers this structure 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 short Ni----Ni distances. Because of this, it has high lattice energy and insoluble in water.

Inorganic_2.pdf 554m5s

Overview 4q3b3c

More details 26j3b

More Documents from "Joel Joel" 711y49

Inorganic_2.pdf 554m5s

Map 244z5b

2c82t

Introduction To Brazing En4 735i4h

Proyect De Envestigacion Jota 43i5q

L'auditeur Logistique q5j1