P- Block elements (BOARD Q/A)
Q. 1.
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Q.2.
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Q.3.
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Q.4.
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Q.5.
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Q.6.
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Q.7.
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Q.8.
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Q.9.
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Q.10.
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Q.11.
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Q.12.
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Q.13.
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Q.14.
Why is Bi (V) a stronger oxidant than Sb (V)?
Because Bi (III) is more stable than Sb (III) due to inert pair effect.
Why is red phosphorus less reactive than white phosphorus?
This is due to polymeric structure of red phosphorus or angular strain in P4 molecule of white
phosphorus where the angle is only 600.
The maximum number of covalent bonds formed by nitrogen is 4. Why?
Nitrogen has three unpaired electrons and one lone pair of electrons; therefore, it can form three
covalent bonds and one co-ordinate bond.
Give reasons for the least reactivity of nitrogen molecule?
Due to presence of a triple bond between the two N-atoms, the bond dissociation enthalpy (941.4
KJ mol-1) is very high. Hence N2 is least reactive.
Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Give reason.
Nitrogen does not haved orbitals to expand its covalence beyond four. That is why it does not
form pentahalide.
In trimethyl amine, the nitrogen has a pyramidal geometry whereas in trisilylamine N
(SiH3)3 ,it has a planar geometry.
(CH3)3 N is pyramidal due to sp3 hybridization and has a lone pair of electrons. (SiH3)3 N has sp2
hybridization because lone pair of nitrogen is donated to vacant d-orbitalof Si.
BH4 and NH4 are isolobal. Explain.
Both BH4 and NH4 have tetrahedral shapes, i.e., four lobes of sp3 –hybrized orbitals, Hence they
are isolobal
Why is NH3 a good complexing agent?
NH3 is a good complexing agent because nitrogen has a lone pair of electrons which it can donate
to form co-ordinate bond.
On being slowly passed through water, PH3 forms bubbles but NH3 dissolves. Why is it so?
N-H bond is more polar than P-H bond. Hence, NH3 forms hydrogen bonds with H2 Omolecules
and hence dissolves in it whereas PH3 does not dissolve and forms bubbles.
Why does NO2 dimerise?
NO2 contains odd number of valence electrons. On dimerisation, it is converted to a stable N2O4
molecule with even number of electrons.
Why does PCI3 fume in moisture?
PCI3 hydrolysis in the presene of moisture giving fumes of HCL.
Why is nitrous acid oxidant as well as reductant?
The oxidation state of N in nitrous acid (H-O-N) is +3 whick lies in between its lowest oxidation
state of -3 and highest oxidation state of +5. Since the oxidation state of N in (HNO2) can be
decreased from +3 to any lower value, therefore, it acts as an oxidizing agent.
Further since the oxidation state of N in HNO2 can be increased from +3 to +4 or +5, therefore, it
acts as a reducing agent. Thus, nitrous acid acts both as an oxidant as well as a reductant.
Why are the Group 16 elements called chalcogens?
Chalcogens means ore forming. The elements of Group 16 are called chalcogens because many
metals are found as oxides and sulphides and a few as selenides and tellurides.
Oxygen molecule has formula O2 while sulphur S8- Why?
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Q.15.
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Q.16.
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Q.17.
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Q.18.
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Q.19.
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Q.20.
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Q.21.
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Q.22.
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Q.23.
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Oxygen atom being small in size form multiple bond while S atom being large in size form single
bond with other S atom, the puckered ring structure S8 is most stable.
Sulphur disappears when boiled with sodium sulphite. Why?
When sodium sulphite is heated with sulphur, we get sodium thiosulphate which is soluble in
water that is why sulphur disappears.
Na2SO3 +S heat Na2S2O3.
Why is I2 more soluble in KI than in water?
It is due to formation of soluble complex KI3.
I2+KI
KI3.
Write an example of a neutral molecule which is iso-electronic with CIO-.
OF2 and CIF are iso-electronic with CIO-.
Why is F2O referred to as a fluoride but CI2O is an oxide?
F2 Ois called oxygen fluoride because fluorine is more electronegative than oxygen whereas CI2O
is called chlorine oxide because oxygen is more electronegative than chlorine.
Noble gases have largest radii. Explain.
In noble gases, we can measure only van der Waals’ radii which are larger than covalent radii.
How does xenon atom form compounds with fluorine even though the xenon atom has a
closed shell configuration?
This is because 1, 2 or 3 electrons from the 5p-orbitals can be excited to empty 5d-orbitals thus
making 2, 4 or 6 half-filled orbitals available for bond formation.
Give the reason which prompted Bartlett to prepare first noble gas compounds.
Barlett observed ionization of energy of O2 is 1180 kJ mol-1 whereas that of Xe is 1170 kJ mol-1.
He could prepare O2 pt F6 which prompted him to prepare first noble gas compound Xe+ [Pt F6].
Noble gases have very low boiling points. Why?
Noble gases being monoatomic have no interatomic forces except weak dispersion forces and
therefore, they are liquefied at very low temperatures. Hence, they have low boiling points.
Account for the following observations
(i)
Amoung the halogens F2 is the strongest oxidizing agent?
(ii)
fluorine exhibits only – 1 oxidation state whereas other halogens exhibit higher
positive oxidation states also.
(iii)
acidity of oxo acid of chlorine is HOCI <HOCIO<HOCIO2<HOCIO3
(i)
This is due to the
(a) Low enthalpy of dissociation of F-F bond.
(b) High hydration enthalpy of F.
(ii)
Due to its high electronegativity.
(iii)
Higher the oxidation state of chlorine in oxo acid, stronger the acid.
Q.24. Account for the following:
(a)
Chlorine water has both oxidizing and bleaching properties.
(b)
H3PO2 and H3PO3 act as as good reducing agents while H3PO4 does not.
(c)
On addition of ozone gas to KI solution, violet vapours are obtained.
Ans. (a)
Chlorine water produces nascent oxygen which is responsible for bleaching action and
oxidation.
C12 + H2O
2HC1 [O]
(b)
no P
(c)
Both H3 PO2 and H3PO3 have P H bonds, so they act as reducing agents, but H3PO4, has
H bonds, so it cannot act as a reducing agent.
Ozone gas acts as a strong oxidizing agent, so it oxidizes iodide ions to Iodine.
2I (aq)+H2O(J)+O3 (g)
2 OH(aq) +I2 (g)+O2 (g)
I2 vapours evolved have violet colour.
Q.25. Give reasons for the following:
(a)
CN- ion is known but CP- ion is not known.
(b)
NO2 demerises to form N2O4
(c)
ICI is more reactive than I2.
Ans. (a)
Nitrogen being smaller is size forms pπ- pπ multiple bonding with carbon, so CN ion is
known, but phosphorus does not form pπ- pπ bond as it is larger in size.
(b)
NO2 is an odd electron molecule and therefore gets dimerised to stable N2O4.
(c)
Because ICI has less bond dissociation enthalpy than I2.
Q.26. Account for the following:
(i)
NH3 is a stronger base than PH3.
(ii)
Sulphur has a greater tendency for catenation than oxygen.
(iii)
Bond dissociation energy of F2 is less than that of CI2.
Ans. (i)
NH3 is stronger base than PH3. This is because the lone pair of electrons on N atom in
NH3 is directed and not diffused as it is in PH3 due to larger size of phosphorus and hence more
available for donation.
(ii)
Sulphur has a greater tendency for catenation than oxygen because W-S bond is stronger
than O-O bond due to less interelectronic repulsions.
(iii)
Bond dissociation energy of F2 is less than CI2 this is due to relatively large electronelectron repulsion among the lone pairs in F2 molecule where they are much closer to each other
than in case of CI2.
Q.27. Explain the following situations:
(i)
In the structure of HNO3 molecule, the N-O bond (121pm) is shorter than N-OH
bond (140 pm).
(ii)
SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed.
(iii)
XeF2 has a straight linear structure and not a bent angular structure.
Ans. (i)
ö:
ö:
+
H O N
HO N
ö:
ö:
As a result of resonance, N-O bond length is average of single bond and double bond whereas NOH bond has purely single bond character. Therefore, N-O bond is shorter than N-OH bond in
HNO3.
(ii)
S atom in SF4 is not sterically protected as it is surrounded by only four F atoms, so
attack of H2O molecules can take place easily and hence hydrolysis takes place easily. In
contract, in SF6, S is sterically protected by six F atoms. Therefore does not allow H2O molecules
to attack S atoms. As a result of this, SF6 does not undergo hydrolysis.
(iii)
In Xe F2, Xe is sp3d hybridized having 2 bond pair and 3 lone pair of electrons. The
presence of 3 lone of electrons in Xe F2 at equidistance to have minimum repulsion is responsible
for its linear structure.
Q.28. Give reasons for the following:
(i)
NCl3 gets readily hydrolysed while NF3 does not.
(ii)
Elemental nitrogen exists as a diatomic molecule whereas elemental phosphorus is a
tetraatomic molecule.
Ans. (i)
In NCI3, CI has vacant d-orbitals to accept the lone pair of electrons donated by O-atom
of H2O molecule but in NF3, F does not have d-orbitals.
NCI3+3H2O
NH3 +3HOCI ; NF3 +H2O
No reaction
(ii)
Nitrogen because of its small size forms pπ- pπ multiple bonds with other nitrogen atoms
and thus it exists as a diatomic molecule (N=N or N2 ). Phosphorus, on the other hand, because of
its large size usually does not form pπ- pπ multiple bonds with other phosphorus atoms but
instead forms single bonds. Consequently, it exists as a stable tetraatomic (P4)
Molecule.
Q.29. Account for the following:
(i)
PH3 has lower boiling point than NH3.
(ii)
PF5 is known but NF5 is not known.
Ans. (i)
The electronegativity of N (3.0) is much higher than that of P(2.1). So, NH3 undergoes
extensive intermolecular H-bonding and hence it exists as an associated molecule. To break these
K-bonds, a large amount of energy is needed. On the other hand, PH3 does not undergo Hbonding and thus exists as discrete molecules. Therefore, the boiling point of PH3 is much
lower than that of NH3.
(ii)
P has vacant 3d-orbitals in its valence shell while N does not have. As a result, P can
form additional bonds to give PF5 while N cannot extend its covalency beyond three and hence it
forms only NF3 but not NF5.
Q.30. Give reasons for the following:
(i)
Unlike phosphorus, nitrogen shows little tendency for catenation.
(ii)
Bismuth is a strong oxidising agent in the pentavalent state.
Ans. (i)
Catenation (i.e., linking of atoms .of the same kind with one another) is related to the
atom-atom bond energy. Greater the atom-atom bond energy, greater is the catenation. Because of
low N—N bond energy (1638 kJ mol-1) nitrogen shows little tendency for catenation; P—P bond
energy (2O1.6 kJ mol-1 ) is quite high, hence, it shows more tendency for catenation than
nitrogen.
(ii)
As the inert pair effect is very prominent in Bi, its + 5 oxidation state is less stable than
its + 3 oxidation state. In other words, bismuth in the pentavalent state can easily accept two
electrons and thus gets reduced to trivalent bismuth.
Bi5+2eBi3+
Thus, it acts as a strong oxidising agent.
Q.31. Give reasons for the following:
(i)
When NaBr is heated with conc. H2S04, Br2 is produced but when NaC1 is heated
with conc. H2S04 is, produced.
(ii)
Oxygen generally exhibits an oxidation state of — 2 only whereas other members
of its family show oxidation states of +2, +4 and +6 as well.
(iii)
Among the hydrides of Group 16, water shows unusual physical properties.
Ans. (1)
When NaBr is heated with conc. H2SO4, HBr is first produced which being a reducing
agent reduces H2SO4 to SO2 while HBr itself gets oxidised to Br2.
NaBr + H2 SO4
NaHSO4—+ HBr
2 HBr+ H2SO4
2H2O+ SO2 +Br2,
As a result, only Br2 is produced.
Similarly, NaCl reacts with conc. H2SO4 to form HCI but since HC1 does not act as a reducing
agent, it does’ not get oxidised to Cl2.
NaCl + H2SO4
NaHSO4 + HCI
HC1 + H2SO4
No action
As a result, only HCI is evolved.
(ii)
The electronic configuration of oxygen is ls2 2s2 2px2 2py1, 2pz1, i.e., it has two half-filled
orbitals and there is no d-orbital available for excitation of electrons. Further, it is the most
electronegative element of its family. Hence, it shows oxidation state of -2 only. Other elements
like sulphur have d-orbitals available for excitation, thereby giving four and six half-filled
orbitals. Moreover, they can combine with more electronegative elements. Hence, they show
oxidation states of +2, +4 and +6 also.
(iii)
Because of high electronegativity of O, the O—H in H2O forms strong intermolecular Hbonds. Thus, water exists as an associated molecule while other hydrides of Group 16 do not form
H-bonds and hence exist as discrete molecules. Hence, water shows unusual physical properties,
i.e., high boiling point, high thermal stability and weaker acidic character as compared to other
hydrides of Group 16.
Q.32. Account for the following:
(i)
H2S acts only as a reducing agent but SO2 acts both as a reducing agent as well as
an oxidising agent.
(ii)
SF6 is known but SH6 is not known.
(iii)
Compounds of fluorine with oxygen are called fluorides of oxygen and not the
oxides of fluorine.
Ans. (i)
The minimum oxidation number (O.N.) of S is — 2 while its maximum O.N. is + 6. In
SO2, the O.N. is +4, hence, it can not only increase its O.N. by losing electrons but also reduce its
O.N. by gaining electrons. Thus, it acts both as a reducing agent as well as an oxidising agent. In
contrast, in. H2S, S has an O.N. of — 2. Thus, it can only increase its U.N by losing electrons and
hence acts only as a reducing agent.
(ii)
Fluorine being the strongest oxidising agent oxidises sulphur to its maximum oxidation
state of + 6 and thus forms SF6. In contrast hydrogen being a very weak oxidising agent cannot
oxidise S to its maximum oxidation state of + 6 and hence does not form SH6.
(iii)
This is because fluorine is more electronegative than oxygen.
Q.33. Assign a reason for each of the following:
(i)
SCI6 is not known but SF6 is known.
(ii)
Sulphur hexafluoride is used as a gaseous electrical insulator.
Ans. (i)
Fluorine is a much stronger oxidising agent than chlorine, therefore, it can easily oxidize
sulphur to its maximum oxidation state of + 6 and hence forms SF6. Chlorine being a weaker
oxidising agent can oxidise sulphur at the maximum to its + 4 oxidation state and hence can form
SCI4 but not SCI6.
(ii)
SF6, is a colourless, odourless and non-toxic gas at room temperature. It is thermally
stable and chemically inert. Because of its inertness and high tendency to suppress internal
discharges, it is used as a gaseous electrical insulator in high voltage generators and switch gears.
Q.34. Give reasons for the following observations:
(i)
SF6 is inert towards hydrolysis.
(ii)
Sulphur exhibits greater tendency for catenation than selenium.
Ans. (i)
In SF6, S is sterically proctected by six F atoms and hence does not allow H2O molecules
to attack the S atom. Further, F does not have d-orbitals to accept the electrons donated by H2O
molecules. As a result of these two reasons, SF6 does not undergo hydrolysis. On the other hand,
in SF4, S is not sterically protected since it is surrounded by only four F atoms. Thus, attack of 11,0 molecules can take place easily and hence hydrolysis occurs.
(ii)
As we move from S to Se, the atomic size increases and hence the strength of E—E bond
decreases. Thus, S—S bond is much stronger than Se—Se bond consequently, S shows greater
tendency for catenation than selenium.
Q.35. Account for the following:
(i)
Bond dissociation energy of F2 is less than that of CI2.
(ii)
Both NO and CIO, are odd electron species but NO dimerises while CIO2 does not.
(iii)
Bleaching of flowers by chlorine is permanent while that by sulphur dioxide is
temporary.
Ans. (i)
Fluorine atom being smaller in size, the interelectronic repulsions between the nonbonding electrons present in the 2p-orhitals of fluorine atoms are much larger than similar
interelectronic repulsion in the 3p-orhitals of chlorine atoms.
(ii)
In NO, the odd electron on N is attracted b only one O-atom but in ClO2, the odd electron
on CI is attracted by two O-atoms. Thus, the odd electron on N in NC) is localised while the odd
electron on CI in CIO2 is delocalised. Consequently, NO has a tendency to dimerise but CIO2
does not.
(iii)
Cl2 bleaches coloured material by oxidation:
Cl2 +H2O
2HC1+0
Coloured material + [O]
Colourless
and hence bleaching is permanent. In contrast, SO2 bleaches coloured material by reduction and
hence bleaching is temporary since when the bleached colourless material is exposed to air, it gets
oxidised and the colour is restored,
SO2+2H2O
H2S04+2H
aerial
Coloured material + H
Colourless material
Coloured material.
oxidation
Q.36. What are interstitial compounds? Why are such compounds well known for transition
metals?
Ans. Interstitial compounds are those in which small atoms occupy the interstitial sites in the crystal
lattice. Interstitial compounds are well known for transition metals because small-sized atoms of
H, B, C, N, etc., can easily occupy positions in the voids present in the crystal lattices of
transition
metals.
Q.37. How is the variability in oxidation states of transition metals different from that of the nontransition metals?
Ans. The oxidation states of transition elements differ from each other by unity (due to incomplete
filling of d-orbitals) whereas oxidation states of non-transition elements normally differ by two
units.
Q.38. Account for the following:
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Q.39.
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Q.40.
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Q.41.
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Q.42.
(i)
Of the d4 species, Cr 2+ is strongly reducing while manganese (III) is strongly
oxidising.
(ii)
Cobalt (III) is stable in aqueous solution but in the presence of complexing reagents,
it is easily oxidised.
(iii)
The d1 configuration is very unstable in ions.
(i)
E° value for Cr 3+ /Cr2+ is negative (— 0.41 V) while as E° value for Mn3+ /Mn 2+ is
positive (+ 1.57 V). Thus, Cr2+ ions can easily undergo oxidation to give Cr3+ ions and, therefore,
act as strong reducing agent. On the other hand, Mn 2+ can easily undergo reduction to give Mn2+
and hence act as oxidising agent.
(ii)
Co (III) has greater tendency to form coordination complexes than Co (II). Thus, in the
presence of ligands, Co (II) changes to Co (III), i.e., is easily oxidised.
(iii)
The ions with d1 configuration have the tendency to lose the only electron present in dsubshell to acquire stable d° configuration. Therefore, they are unstable and undergo oxidation or
disproportionation.
What are disproportion reactions? Give two examples.
Disproportion reactions are those reactions in which the same substance undergoes oxidation as
well as reduction. In disproportionation reaction, oxidation number of an element increases as
well decreases to form two different products. For example,
VI
VII
IV
2+
3MnO4 +4H
2MnO4 +MnO2 +2H2O
Give example and suggest reasons for the following features of the transition metal
chemistry.
(i)
The lowest oxide of transition metal is basic, the highest is acidic.
(ii)
A transition metal exhibits higher oxidation states in oxides and fluorides.
(iii)
The highest oxidation state is exhibited in oxo-anions of a metal.
(i)
The lower oxide of transition metal is basic because the metal atom has low oxidation
state whereas highest is acidic due to highest oxidation state. For example, MnO is basic whereas
Mn2O7 is acidic.
(ii)
A transition metal exhibits higher oxidation states in oxides and fluorides because
oxygen and fluorine are highly electronegative elements, small in size (and strongest oxidising
agents). For
example, osmium shows an oxidation states of +6 in OsF6 and vanadium shows an oxidation
states of +5 in VO5
(iii)
Oxometal anions have highest oxidation state, e.g., Cr in Cr2O72- has an oxidation state
of +6 whereas Mo in MnO4- has an oxidation state of +7 This is again due to the combination of
the metal with oxygen, which’ is highly electronegative and oxidising element.
What are inner- transition elements? Decide which of the following atomic numbers are the
numbers of the inner-transition elements 29, 59, 74, 95, 102, 104.
The f-block elements, i.e., in which the last electron enters into f-subshell are called innertransition elements. These include lanthanoids (58-71) and actinoids (90—103). Thus, elements
with atomic numbers 59, 95 and 102 are inner-transition elements.
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justfy
this statement by giving some examples from the oxidation state of these elements.
Ans.
Q.43.
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Q.44.
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Q.45.
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Q.46.
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Q.47.
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Q.48.
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Q.49.
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Q.50.
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Q.51.
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Q.52.
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Q.53.
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Q.54.
Lanthanoids show a limited number of oxidation state viz., +2, +3 and +4 (out of which +3 is
most common). This is because of a large energy gap between 4f, 5d and 6s subshells. The
dominant oxidation state of actionoids is also +3 but they show a number of other oxidation states
also, e.g., uranium (Z=92) and plutonium (Z=94), show +3, +4 and +6, neptunium (Z=94) shpws
+3, +4, +5 and +7, etc. This is due to small energy difference between 5f, 6d and 7s sibshells of
the actinoids.
Why do the transition elements show variable oxidation states?
Transition elements show variable oxidation states because electrons in ns and (n —1) d-orbitals
are available for bond formation.
First ionization energies of 5d elements are higher than those of 3d and 4d elements. Give
reason.
Because of weak shielding (or screening) effect of 4f electrons, the effective nuclear charge
acting on the valence electrons in 5d elements is quite high. Hence, the first ionization energies of
5d elements are higher than those of 3d and 4d elements.
Transition elements show paramagnetic behaviour. Give reason.
Substances containing unpaired electrons are said to be paramagnetic. Transition elements
contain unpaired electrons in their (n —1) d-orbitals. Hence, they are paramagnetic.
Name a transition element which does not exhibit variable oxidation state.
Scandium (Z = 21) does not exhibit variable oxidation states.
Why are transition elements known as d-block elements?
The last electron enters (n —1) d-orbital, i.e , d-orbital of the penultimate shell Hence these are
known as d-block elements.
Why is copper (At no. 29) considered a transition metal?
Copper is considered a transition metal because copper in oxidation state +2, i.e., Cu2+ has
incompletely filled d-subshell (3d9).
write the formula of a compound where transition metal is in +7 oxidation state.
KMnO4.
write any one use of pyrophoric alloys.
pyrophoric alloys emit sparks when struck. Hence, they are used in making flints for lighters.
What type of magnetic behaviour is shown by the molecules containing unpaired electrons?
Paramagnetism.
In the transition series with an increase in atomic number, the atomic radius does not
change very much. Why?
As we move from left to right along a transition series, the nuclear charge increases which tends
to decrease the size of the atom but the addition of electrons in the d-subshell increases the
screening effect which counter balances the effect of increased nuclear charge.
What is the basic difference between the electronic configurations of transition and inner
transition elements?
General electronic configuration of transition elements = [Noble gas] (n—1) d1-10 ns1-2 and for
inner transition elements = (n-2)f1-14 (n-1)d0-1 ns0-2. Thus, in transition elements, last electron
enters d-orbital of the penultimate shell while in inner transition elements, it enters f-orbital of the
penultimate shell.
Why do transition elements show similarities along the horizontal period?
Ans.
Q.55.
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Q.56.
Ans.
Q.57.
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Q.58.
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Q.59.
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Q.60.
Ans.
Q.61.
Ans.
Q.62.
Ans.
Q.63.
Ans.
Q.64.
all transition elements contain incompletely filled d-subshell whereas the outer shell electronic
configuration remains the same.
What is the most characteristic oxidation state of lanthanoides?
+3 oxidation state.
Why is k2Cr2O7 generally preferred over Na2 Cr2 O7 in volumetric analysis although both
are oxidising agents?
Na2 Cr2 O7 is hygroscopic, i.e., absorbs moisture from atmosphere, whereas K2Cr2 O7 does not.
Hence K2Cr2O7 is preferred over Na2Cr2O7.
The size of the trivalent cations in the lanthanoid series decreases steadily as the atomic
number increases. What is this known as?
This is known as Lanthanoid contraction.
Why do transition metals have high enthalpy of hydration?
Transition metal ions are smaller in size and have higher charge, therefore, have higher enthalpy
of hydration.
Why do Zr and Hf exhibit almost similar properties?
Zr and Hf have similar ionic size.
What are the two important oxidation states of group 6 elements of the periodic table?
+3 and +6.
What is the general formula by which the electronic configuration of the transition elements
is represented?
(n-1)d1-10ns1-2
Explain the following facts
(a)
transition metals act as catalysts.
(b)
chromium group elements have the highest melting points in their respective series.
(c)
transition metals form coloured complexes.
(a) The catalytic activity of transition metals is attributed to the following reasons:
(i)
Because of their variable oxidation states transition metals form unstable intermediate
compounds and provide a new path with lower activation energy for the reaction.
(ii)
In some cases, the transition metal provides a suitable large surface area with free
valencies on which reactants are adsorbed.
(b)
Because they have strong metallic bonds due to greater number of unpaired d electrons.
(c)
This is due to d-d transition.
Discuss the relative stability in aqueous solutions of +2 oxidation state among the elements:
Cr, Mn, Fe and Co. How would you justify this situation?
(At. Nos. Cr = 24, Mn = 25, Fe = 26, Co = 27)
On the basis of electrochemical series the standard electrode potential shows the following order
EåMn2+/Mn < EoCx2+/Cr < EoFe2+/Fe < EoCo2+/Co
Therefore Co 2+ gets easily reduced to metallic cobalt while it is difficult to reduce Mn 2+ Hence
Mn 2+ will be most stable and the increasing stability order will be
Co2+ < Fe2+ < Cr2+ < Mn 2+
Assign a reason for each of the following:
(i) The third ionization energy of Mn (Z = 25) is higher than that of either
Cr (Z = 24) or Fe (Z = 26).
(ii) Simple copper (I) salts are not stable in aqueous solutions.
Ans.
Q.65.
Ans.
Q.66.
Ans.
Q.67.
Ans.
Q.68.
Ans.
(i)
This is because Mn2+ is more stable as it has exactly half filled configuration 3d5 4s0.
(ii)
Cu2+ (aq) is much more stable than Cu+(aq). This is because, although second ionization
enthalpy of copper is large but for Cu2+(aq) is much more negative than that of Cu+ (aq) and
therefore, it more compenstate for the second ionization enthalpy of copper. Therefore, Cu+ion
aqueous solution undergoes disproportionation.
2Cu2+ (aq)
Cu2+ (aq) + Cu(s)
Write IUPAC name of Fe4 [Fe(CN)6]3.
Iron (III) hexacyanidoferrate (II).
Write IUPAC name of COCI2(en)2]C1.
Dichloribis (ethane-1, 2-diamine) cobalt (III) Chloride.
Write the formula of Pentammine Chloridoplatinum (IV).
[Pt(NH3)5C1]C13.
Which compound is used to estimate the hardness of water volumetrically?
EDTA.
Q.69. Write all the possible isomers of [Co (NH3)5 (SCN)] Cl.
Ans. [Co(NH3)5(SCN)]CI,[Co(NH3)5 (NCS)]CI, [Co(NH3)5CI] SCN,[Co(NH3)CI]NCS
MA5B type complexes do not show geometrical or optical isomerism.
Q.70. Square planar complexes with coordination number of four exhibit geometrical isomerism
whereas tetrahedral complexes do not. Why ?
Ans. Tetrahedral complexes do not show geometrical isomerism because the relative positions of the
ligands attached to the central metal atom are same with respect to each other.
Q.71. Write the chemical formula of potassium hexacyanoferrate (III).
Ans. K3 [Fe (CN)6].
Q.72. Giving a suitable example for each, explain the following:
(i) Crystal field splitting, (ii) Linkage isomerism, (iii) Ambidentate ligand
Ans. (i) Crystal field spliting: When the ligands approach the central metal ion, the electrons in the dorbitals of central metal ion will be repelled by the lone pairs of the ligands. Because of these
interactions the degeneracy of d orbitals of the metal ion is lost and these split into two sets of
orbitals having different energies. This is known as crystal field splitting, e.g., for d4,
configuration is t32g e1g in the presence of weak field ligand.
(ii) Linkage isomerism: The isomers which have same molecular formula but differ in the
linkage of ligand atom to the central metal atom are called linkage isomers, e.g.,
[CO(NH3)5 NO2] CI2 and [Co(NH3)5ONO]Cl2
(iii) Ambidentate ligand: A unidentate ligand which can bind to the central metal atom through
any of the two donor atoms present in it is called ambidentate ligand, e.g., NO2 can bind to
O
metal either through nitrogen, i.e., as nitrito-N (
N ) or through oxygen atom, i.e., as
O
nitrito —O (
O—N=O)
Q.73. Explain the following:
(i) Low spin octahedral complexes of nickel are not known.
(ii) The π—complexes are known for transition elements only.
(iii) CO is a stronger ligand than NH3 for many metals.
Ans.
(i) Ni in its atomic or ionic state can not afford two vacant 3d orbitals hence d2SP3 hybridisation
is not possible.
(ii) Transition metals have vacant d orbitals in their atoms or ions into which the electron pairs
can be donated by ligands containing m electrons, e.g., C6H6, CH2 = CH2, etc. Thus, dπ—Pπ
bonding is possible.
(iii) Because in case of CO back bonding takes place in which the central metal uses its filled d
orbital with empty anti bonding m molecular orbital of CO.
Q.74. The spin only magnetic moment of [MnBr4]2- is 5.9 BM. Predict the geometry of the
complex ion?
Ans. Since the coordination number of Mn2+ ion in the complex ion is 4, it will be either tetrahedral
(sp3 hybridisation) or square planar (dsp2 hybridisation). But the fact that the magnetic moment of
the complex ion is 5.9 BM, it should be tetrahedral in shape rather than square planar because of
the presence of five unpaired electrons in the d orbitals.
Q.75. Differentiate between weak field and strong field coordination entity.
Ans.
1.
2.
3.
4.
Weak field
coordination entity
They are formed when the crystal 1e1d
stahilisation energy ( 0 ) in octahedral
complexes is less than the energy required
for an electron pairing in a single orbital
(P).
They are also called high spin complexes.
They are mostly paramagnetic in nature.
Never formed by CN ligands:
1.
2.
3.
4.
Strong field
coordination entity
The’ are formed when the crystal
field stabilisation energy ( 0 ) is
greater than the P.
They are called low spin complexes.
They are mostly diamagnetic or less
paramagnetic than weak field.
Formed by CN like ligands.