INORGANIC CHEMISTRY REASONING QUESTIONS

1. There is a considerable increase in covalent radius from N to P. However, from As to Bi only small increase in covalent radius is observed.

Ans. This is due to the presence of completely filled d and/or f orbital in heavier members.

2. The ionization enthalpy of the group 15 elements is much greater than that ofgroup 14 elements in the corresponding periods.

Ans. Because of the extra stable half-filled p orbital electronic configuration and smaller size.

3. R3P=O exist but R3N=O does not.

Ans. Due to the absence of d orbitals in  valence shell of nitrogen and because of inability of Nitrogen to expand its covalency beyond  four, nitrogen cannot form d π–pπ bond.

4. In solid state PCl5 exists as Ionic compound.

Ans. Since solid phosphorous pentachloride exists as [PCl4]+ [PCl6]-and  hence exhibit some ionic character. [PCl4]+ is tetrahedral and the anion [PCl6]– is octahedral.

5. Tendency to show –2 oxidation state diminishes from Sulphur to polonium in group 16.

Ans. The outer electronic configuration of group 16 elements is ns2 np4. These elements therefore have the tendency to gain two electrons to complete the octet. Since elctronegativity and I.E. decrease on going down the group, tendency to show –2 oxidation state diminishes.

6. Ozone (O3) act as a powerful oxidising agent.

Ans. Due to the ease with which it liberates atoms of nascent oxygen (O3 → O2 + O), it acts as a powerful oxidizing agent.

7. Halogens are coloured.

Ans. Halogens are coloured. This is due toabsorption of radiations in visible region which results in the excitation of outer electrons to higher energy level. By absorbing different quanta of radiation, they display different colours. For example, F2 has yellow, Cl2 greenish yellow, Br2 red andI2 violet colour.

8. Cl2 bleaches a substance permanently but SO2 does it temporarily.

Ans. Cl2 bleaches a substance permanently because it is due to oxidation but SO2 does it temporarily because it is due to reduction.

9. Xe does not forms compounds such as XeF3 and XeF5.

Ans. By the promotion of one, two or three electrons from filled p-orbital to the vacant d-orbital in the valence shell, 2,4 or 6 half filled orbitals are formed. Thus Xe can combine only with even number of fluorine atoms and not odd. 

10. Helium is used for inflating aeroplane tyres & filling balloons for meteorogical observations.

Ans. Helium is a non-inflammable and lightgas.

Why does nitrogen does not form pentahalides?

Nitrogen with n=2 has s and p orbitals only.It does not have d orbitals to expand its coelent beyond four.Thats why it does not form pentahalides.

Why does PH3 has lowrer boiling point than NH3?

Unlike NH3 PH3 molecules are not associated through hydrogen bonding in liquid state.That’s why the Boling point of PH3 is lower than NH3.

Why are Pentahalides more covalent than Trihalides ?

Higher the positive oxidation state of central atom ,more will be it’s polarising power which in turn increases the covalent character of bond formed between the central atom and the other atom.

Why is BIH3 the strongest reducing agent amongst the hydrides of group 15 elements?

BIH3 is the strongest reducing agent amongst all the hydrides of group 15 element.Because BIH3 is the least stable amongst the hydrides of group 15.

Why does NH3 acts as a Lewis Base?

Nitrogen atom in NH3 has one lone pair of electron which is available for donation.Therefore it acts as a Lewis base.

Why is the highest oxidation state of metal exhibited in its oxides or fluorides only?

The highest oxidation state of a metal exhibited in its oxides or fluorides because of small size and higher electronegatively oxygen or fluorine can oxidise the metal to its higher oxidation state.

Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?

Mn2+ compounds are more stable than Fe2+ towards oxidation to their +3 state because Mn2+ has 3d configuration which has extra stabilty.

Why is Actinoids contraction is greater from elements to elements than lanthanoids contraction?

The 5F electrons are more effectively shielded from nuclear charge.In other words the 5F electrons themseles provide poor shielding from element to element in the series.

Why Cu+ ions is not stable in aqueous solution?

Cu+ ions in aqueous solution undergoes disproportionation ie

2Cu+ -> Cu2+ (aq) + Cu(s)

Why is Cr2+ reducing and Mn3+ oxidising when both have d4 configuration?

Cr2+ is reducing as its configuration changes from d4 to d3 later having half filled t2g level.On the other hand ,the change from Mn2+ to Mn3+ results in the half filled d5 configuration which has extra stability.

Q.Why is first ionisation energies of boron and aluminium are lower than that of beryllium

Ans. In B and Al,the first electron to be removed lies in p-orbitals while in beryllium and magnesium the valency electrons of an orbital are strongly attracted by nucleus and hence, difficult to remove tan the p-orbitals.

Q. Boric acid can be titrated against sodium hydroxide using phenolphthalein indicator only in presence of polyhydroxy compounds like catechal. Explain .

Ans. Boric acid,B(OH)3 is a very weak lewis acid. It forms a stable complex with polyhydroxy compounds. Due to formation of this stable complex, boric acid acts as a strong acid and hence can be titrated against NaOH

Q. Why (SiH3)3N is a weaker base than (CH3)3N.

Ans. In (SiH3)3N , lone pair of electrons on nitrogen is involved in pπ-dπ back bonding, while in (CH3)3N pπ-dπ back bonding is not possible because of absence of d-orbitals. Hence, (CH3)3N is more basic than (SiH3)3N

Q. The experiment determined N-F bond length in NF3 is greater than the sum of the single bond covalent radii of N ans F. why

Ans. Nitrogen and flourine both are small and have high electron density and hence they repel the bonded pair of electrons leading to larger bond length than expected.

Q. Why Nitrogen is a gas while other members of this group are solids.

Ans. Due to smaller size of nitrogen atom,it can undergo sideways overlap with the result two π bonds are formed between two nitrogen atoms (N≡N). The discrete nitrogen molecules are held together by weak Van Der Waal’s forces. On the other hand,other atoms are bigger, so sideways overlap is not strong and hence multiple bonds are not formed . Thus more number of atoms are linked together by single covalent bonds with the result their molecular weight become high ans hence these are solids.

Q. Nitrous oxide supports combustion more vigorously than air.why

Ans. Nitrous oxide decomposes by heat of burning substances to produce gases having one third oxygen by volume

2N2O2N2 + O2

while air has one fifth volume of oxygen. Oxygen being a supporter of combustion, keeps combustion on.

Q. Nitric acid acts only as an oxidising agent while nitroous acid can act both as an oxidising and reducing agent.

Ans. In HNO3, N is in its highest oxidation state ,hence, it can only be converted to lower oxidation state,i.e., it can act as an oxidising agent. In HNO2, N is in +3 oxidation state hence, it can be raised to higher oxidation state or lowered to lower oxidation state .thus HNO2 can act as an oxidising as well as reducing agent.

Q. The wooden shelf under reagent bottle containing conc.H2SO4 blackens after sometime.

Ans. Conc. H2SO4, being a strong dehydrating agent ,when trickles down on the bench removes water from wood (cellulosic material ) leaving behind black carbon.

Q. Chlorine displaces iodine from potassium salts.

Ans. Due to more electronegativity,chlorine takes up an electron from the I- ion forming Cl- ion and iodine .

Q. Bleaching of flowers by Cl2 is permanent while by SO2 it is temporary.

Ans. Cl2 bleaches by oxidation ,while SO2 bleaches by reduction. Hence product bleached by SO2 is reoxidised by air to its original form.

Q. Out of cobalt and zinc salts, which is attracted in a magnetic field . Explain with reason.

Ans. Cobalt has d7 electrons in the outer orbit I.e., 3 unpaired electrons,hence it will be attracted in a magnetid field ,while zinc having d10 electrons will not be attracted.

Q.Why is first ionisation energies of boron and aluminium are lower than that of beryllium

Ans. In B and Al,the first electron to be removed lies in p-orbitals while in beryllium and magnesium the valency electrons of an orbital are strongly attracted by nucleus and hence, difficult to remove tan the p-orbitals.

Q. Boric acid can be titrated against sodium hydroxide using phenolphthalein indicator only in presence of polyhydroxy compounds like catechal. Explain .

Ans. Boric acid,B(OH)3 is a very weak lewis acid. It forms a stable complex with polyhydroxy compounds. Due to formation of this stable complex, boric acid acts as a strong acid and hence can be titrated against NaOH

Q. Why (SiH3)3N is a weaker base than (CH3)3N.

Ans. In (SiH3)3N , lone pair of electrons on nitrogen is involved in pπ-dπ back bonding, while in (CH3)3N pπ-dπ back bonding is not possible because of absence of d-orbitals. Hence, (CH3)3N is more basic than (SiH3)3N

Q. The experiment determined N-F bond length in NF3 is greater than the sum of the single bond covalent radii of N ans F. why

Ans. Nitrogen and flourine both are small and have high electron density and hence they repel the bonded pair of electrons leading to larger bond length than expected.

Q. Why Nitrogen is a gas while other members of this group are solids.

Ans. Due to smaller size of nitrogen atom,it can undergo sideways overlap with the result two π bonds are formed between two nitrogen atoms (N≡N). The discrete nitrogen molecules are held together by weak Van Der Waal’s forces. On the other hand,other atoms are bigger, so sideways overlap is not strong and hence multiple bonds are not formed . Thus more number of atoms are linked together by single covalent bonds with the result their molecular weight become high ans hence these are solids.

Q. Nitrous oxide supports combustion more vigorously than air.why

Ans. Nitrous oxide decomposes by heat of burning substances to produce gases having one third oxygen by volume

2N2O2N2 + O2

while air has one fifth volume of oxygen. Oxygen being a supporter of combustion, keeps combustion on.

Q. Nitric acid acts only as an oxidising agent while nitroous acid can act both as an oxidising and reducing agent.

Ans. In HNO3, N is in its highest oxidation state ,hence, it can only be converted to lower oxidation state,i.e., it can act as an oxidising agent. In HNO2, N is in +3 oxidation state hence, it can be raised to higher oxidation state or lowered to lower oxidation state .thus HNO2 can act as an oxidising as well as reducing agent.

Q. The wooden shelf under reagent bottle containing conc.H2SO4 blackens after sometime.

Ans. Conc. H2SO4, being a strong dehydrating agent ,when trickles down on the bench removes water from wood (cellulosic material ) leaving behind black carbon.

Q. Chlorine displaces iodine from potassium salts.

Ans. Due to more electronegativity,chlorine takes up an electron from the I- ion forming Cl- ion and iodine .

Q. Bleaching of flowers by Cl2 is permanent while by SO2 it is temporary.

Ans. Cl2 bleaches by oxidation ,while SO2 bleaches by reduction. Hence product bleached by SO2 is reoxidised by air to its original form.

Q. Out of cobalt and zinc salts, which is attracted in a magnetic field . Explain with reason.

Ans. Cobalt has d7 electrons in the outer orbit I.e., 3 unpaired electrons,hence it will be attracted in a magnetid field ,while zinc having d10 electrons will not be attracted.

Question:

Out of noble gases, only xenon is known to form chemical compounds.

Answer:

Except radon, which is radioactive, xenon has lowest ionisation enthalpy among noble gases and hence it readily forms chemical compounds particularly with O2  and F2

Question:

Fluorine is anomalous in many properties.

Answer:

The anomalous behaviour of fluorine is due to its small size, highest electronegativity, low F-F bond dissociation enthalpy, and non availability of d orbitals in valence shell.

Question:

SCl6 is not known but SF6 is known.

Answer:

Because of high electronegativity of fluorine, sulphur exhibits its maximum oxidation state (+6) in SF6

Question:

N2 is less reactive at room temperature.

Answer:

 Because of strong pπ–pπ overlap Nitrogen has triple bond between two nitrogen atoms N≡N which has high bond dissociation energy. So it is less reactive

Question:

NO2 is coloured but N2O4 is colourless.

Answer:

NO2 has unpaired electrons therefore it absorbs light from visible and

radiate brown colour whereas N2O4 does not have unpaired electrons so

it does not absorb light from visible region

Question:

Tendency to show –2 oxidation state diminishes from Sulphur to polonium in group 16.

Answer:

The outer electronic configuration of group 16 elements is ns2 np4. These elements therefore have the tendency to gain two electrons to complete octet. Since electronegativity and I.E. decrease on going down the group, tendency to show –2 oxidation state diminishes.

Question:

There is large difference between the melting point of Oxygen & Sulphur.

Answer:

The large difference between the melting and boiling points of oxygen and sulphur may be explained on the basis of their atomicity; oxygen exists as diatomic molecule (O2) whereas

sulphur exists as polyatomic molecule (S8)

Question:

All the bonds in the molecules of PCl5 are not equal

Answer:

PCl5 has a trigonal bipyramidal shape in the gas space. A trigonal

bipyramidal is an irregular structure in which some bond angles are 90 degree and others of 120 degree resulting in unequal P-Cl bond lengths 

Question:

In solid state PCl5 exists as Ionic compound.

Answer:

 Since solid phosphorous pentachloride exists as [PCl4]+ [PCl6]

- and  hence

exhibit some ionic character. [PCl4]+ is tetrahedral and the anion, [PCl6]

–

octahedral.

Question:

There is a considerable increase in covalent radius from N to P. However, from As to Bi only small increase in covalent radius is observed.

Answer:

This is due to the presence of completely filled d and/or f orbital in heavier members

Question:

Nitrogen does not form pentahalides.

Answer:

Nitrogen with n = 2, has s and p orbitals only. It does not have d orbitals to expand its covalence beyond four. That is why it does not form pentahalides.

Question:

 H3PO2 and H3PO3 act as as good reducing agents while H3PO4 does not.

Answer:

In H3PO2, two H atoms are bonded directly to P atom & in H3PO3 one H atom is

bonded directly to P atom which imparts reducing character to the acid, whereas in H3PO4 there is no H atom bonded directly to P atom

Question:

Dioxygen is a gas but Sulphur is a solid.

Answer:

Because O2 is Diatomic molecules, hence weak Vander Waal’s force of

attraction thus is a gas whereas S8 is octaatomic hence Stronger Vander

Waal’s force of attraction thus it is solid.

Question:

OF2 should be called oxygen fluoride and not fluorine oxide.

Answer:

OF2 should be called oxygen fluoride and not fluorine oxide. Since oxygen

is less electronegative than fluorine, OF2 should be called oxygen

diflouride.

Question:

Xe does not forms compounds such as Xe F3 and XeF5.

Answer:

By the promotion of one, two or three electrons from filled p-orbital to the vacant d-orbital in the valence shell, 2,4 or 6 half filled orbitals are formed. Thus Xe can combine only with even number of fluorine and not odd. 

• Q:Boron forms no compound in unipositive state while thallium is quite stable in unipositive state. Explain.

• A:m+ ionic state of grp 13 elements exists due to inert pair effect when ns2 electrons penetrate in (n-1)d subshell to become inert . the inert pair effect begins from n>4 and increases with increase in value of n. The n n i.E., Outermost shell for boron is 2 and thus it does not form B+ ion.

• Q:Although the ionization energy of boron (8.30ev) is less than gold(9.22ev), yet former is semi-metal while later is metal.

• A: Here metallic and non-metallic characters are explained in terms of the lattice structure of the solid. Boron has 6 or less atoms as nearest neighbors in solid state while gold has 12 atoms. In general, good metals have a large number of neighbors atoms, while non-metals have relatively less atoms.

• Q:Pure BBr3 and BI3 are colorless, but they become colored on exposure to light, why?

• A: Due to liberation of free halogen by photolysis

• Q: Why is aluminium used in in making electrical cables despite its low conductivity as compared to copper.

• A: it is cheaply available. However, Since it has lesser conducting nature than copper, thicker aluminium wires are used.

• Q: Why are cryolites used in the electrolytic manufacturers of Al from alumina?

• A: Alumina has very high m.pt. and fuses at about 2000 degree Celsius. Addition of cryolite lowers the fusion temperature of mixture as well as increases the conductance of fusion mixture.

• Q: Alum is often used to stop bleeding from cuts. why?

• A: Al3+ ions liberate from alum are highly effective in coagulating negatively charged colloids like blood.

• Q: Aluminium is a very reactive metal, but is still used for making pans. Explain.

• A: Due to formation of a protective oxide film on the surface. It does not corrode easily.

• Q: Aluminium containers can be used to store conc. HNO3. Explain.

• A: Al on contact with conc. HNO3 becomes passive due to the coating of aluminium oxide on its surface and thus it can store the acid.

• Q: Molten Albr3 is a poor conductor of current. Explain.

• A: AlBr3 is covalent in nature and thus during fusion does not produce Al3+ and Br- ions. Hence it is poor conductor of current.

• Q: It is difficult to obtain pure crystalline boron, why?

• A: Its melting point is very high and liquid is corrosive

WHY IS N22 LESS REACTIVE AT ROOM

TEMPERATURE ?

IT IS BECAUSE OF THE STRONG pл-pл OVERLAP RESULTING

INTO THE TRIPLE BOND, (N≡N), CONSEQUENTLY HIGH BOND

ENTHALPY.

NOBLE GASES HAVE VERY LOW BOILING POINT. GIVE REASON ?

NOBLE GASES ARE MONOATOMIC. THEIR ATOMS ARE HELD TOGETHER BY WEAK DISPERSION FORCES AND HENCE CAN BE LIQUIFIED AT VERY LOW TEMPERATURES. SO, THEY HAVE LOW BOILING POINTS.

WHY DO THE TRANSITION ELEMENTS EXHIBIT HIGHER ENTHALPIES OF ATOMISATION ?

BECAUSE OF LARGE NUMBER OF UNPAIRED ELECTRONS IN THEIR ATOMS THEY HAVE STRONGER INTERATOMIC INTERACTION AND HENCE STRONGER BONDING BETWEEN ATOMS RESULTING IN HIGHER ENTHALPIES OF ATOMISATION.

PH33 HAS LOWER BOILING POINT THAN NH3. WHY ?

UNLIKE NH33, PH, PH33 MOLECULES ARE MOLECULES ARE NOT ASSOCIATED THROUGH NOT ASSOCIATED THROUGH HYDROGEN BONDING IN LIQUID HYDROGEN BONDING IN LIQUID STATE. THAT IS WHY BOILING STATE. THAT IS WHY BOILING POINT OF PHPOINT OF PH33 IS LOWER THAN IS LOWER THAN NHNH33..

WHY DOES V2O5 ACT AS CATALYST ?

V2O5 ACT AS CATALIST BECAUSE IT HAS A LARGE SURFACE AREA. IT CAN FORM UNSTABLE INTERMEDIATES WHICH READILY CHANGE INTO PRODUCTS.

TETRAHEDRAL COMPLEXES DO NOT SHOW GEOMETRICAL ISOMERISM, WHY ?

IT IS BECAUSE THE RELATIVE POSITION OF THE LIGANDS ATTACHED TO THE CENTRAL METAL ATOM ARE THE SAME WITH RESPECT TO EACH OTHER.

THE TRANS ISOMERS OF COMPLEX CoCl2 (en)2 IS OPTICALLY INACTIVE , WHY ?

IT IS BECAUSE THE TRANS ISOMER HAS A PLANE OF SYMMETRY AND CAN BE DIVIDED IN TWO EQUAL HALVES.

WHY CHELATE COMPLEXES ARE MORE STABLE THAN UNCHELATE COMPLEXES?

WHEN A CHELATING LIGAND ATTACHES TO THE CENTRAL METAL ATOM THE PROCESS IS ACCOMPANIED BY THE INCREASE IN ENTROPY RESULTING IN THE FORMATION OF THE STABLE COMPLEX.

EXPLAIN WHY K33[Fe(CN)66] IS MORE STABLE THAN K44[Fe(CN)66]?

IT IS BECAUSE THE STABILITY OF COMPLEX DEPENDS UPON THE CHARGE DENSITY ON CENTRAL ION. MORE IS THE CHARGE DENSITY GREATER IS THE STABILITY.

THE SPECIES [CuCl44]³ ˉEXISTS WHILE [CuI4]³ ˉDOES NOT WHY?

THIS IS BECAUSE IODINE ATOM IS MUCH LARGER IN SIZE AND COPPER ATOM IS NOT ABLE TO ACCOMMODATE FOUR SUCH BIG ATOMS AROUND IT.

WHY DOES NH3 FORM HYDROGEN BOND BUT

PH3 DOES NOT?

• Because of high electro-negativity and small size of nitrogen, ammonia forms hydrogen bonds. On the other hand phosphorous has low electro-negativity and large size and hence cannot form hydrogen bonds.

WHY IS NH3 BASIC WHILE BiH3 FEEBLY BASIC?

• Both N and Bi have a lone pair of electron in NH3 and BiH3 respectively. They can donate the electron pair and therefore behave as lewis base.

• In NH3, nitrogen has small size and the lone pair is concentrated on a small region and the electron density on it is maximum.

• Consequently it has greater electron releasing tendency. But the size of Bi is large and the electron density of the lone pair is less. As a result it has lesser tendency to donate electron pair. Hence, NH3 is basic while BiH3 is feebly basic.

WHY DOES NITROGEN SHOW CATENATON PROPERTIES LESS

THAN PHOSPHOROUS?

• The single N-N bond is weaker than the single P-P bond because of high inter-electronic repulsion of the non-bonding electrons, owing to the small bond length. As a result the catenation tendency is weaker in nitrogen.

WHY DOES R3P=O EXIST BUT R3N=O DOES NOT?

• R3N=O does not exist because N cannot have covalency more than 4. Moreover, R3P=O exists because P can extend its covalency more than 4 as well as can form dπ-pπ bond whereas N cannot.

WHY ARE HALOGENS STRONG OXIDISING AGENTS?

• Halogens have a strong tendency to accept electrons and therefore act as strong oxidizing agents.

WHY DOES FLUORINE FORM ONLY 1 OXOACID, HOF?

• Due to small size and high electro-negativity fluorine cannot act as central atom, it cannot form a higher oxo-acid.

WHY IS H2O A LIQUID WHILE H2S A GAS?

• Because of small size and electro-negativity of oxygen, molecules of water are highly associated through hydrogen bonding resulting in its liquid state.

WHY DOES O3 ACT AS A POWEFUL OXIDIZING AGENT?

• Ozone is thermodynamically unstable with respect to oxygen since its decomposition into oxygen results in the liberation of heat and an increase in large negative Gibbs energy change for its conversion into oxygen.

• Therefore it easily liberates atoms of nascent oxygen and acts as a powerful oxidizing agent.

H2S IS LESS ACIDIC THAN H2Te. WHY?

• Due to the decrease in bond dissociation enthalpy down the group, acidic character increases down the group.

• Hence, H2Te is more acidic than H2S.

FLUORINE EXHIBITS ONLY -1 OXIDATION STATE WHEREAS OTHER

HALOGENS EXHIBIT +1, +3, +5 AND +7 OXIDATION STATES ALSO. EXPLAIN.

• Fluorine is the most electro-negative element and cannot exhibit any positive oxidation state. Other halogens have d orbitals and therefore, can expand their octets and show +1 , +3, +5 and +7 oxidation states also.

Q .INCREASING ORDER OF ELECTRONEGATIVITY 0F

O- < O < O+

THE POSITIVE CHARGE ON ATOM INCREASES ITS ELECTONEGATIVITY

WHILE NEGATIVE CHARGE DECREASES ITS

ELECTRONEGATIVITY

Q. INCREASING BOND LENGTH OF N2 < O2 < F2 < Cl2

NITROGEN CONTAINS TRIPLE BOND , OXYGEN CONTAINS DOUBLE BOND

AND FLUORINE AND CHLORINE CONTAIN A SINGLE BOND EACH .

CHLORINE INVOLVES BONDING OF 3p ORBITALS WHILE FLUORINE INVOLVES

2p ORBITALS.

INCREASING BASICITY OF I-< Br- < Cl- < F-

STRONGER THE ACID , WEAKER ITS CONJUGATE BASE

INCREASING ORDER OF BOILING POINT OF PH3 < AsH3 < NH3 < SbH3

BOILING POINT INCREASES WITH INCREAES IN MOLECULAR MASS

WITH EXCEPTION IN NH3 BECAUSE OF HRDROGEN BONDING

WHY IN MOIST AIR COPPER CORRODES TO PRODUCE A GREEN

LAYER ON THE SURFACE

IN THE PRESENCE OF MOIST AIR A THIN FILM OF GREEN BASIC COPPER

CARBONATE IS FORMED ON ITS SURFACE AND HENCE COPPER

CORRODES AND FORM CuCO3.Cu(OH)2 WHICH IS

GREEN

WHY DOES AgNO3 PRODUCE A BLACK STAIN ON THE SKIN

IN THE PRENSENCE OF ORGANIC MATTER (SKIN) AND LIGHT , AgNO3

DECOMPOSES TO PRODUCES A BLACK STAIN OF METALLIC SILVER

INCREASING IONIC CHARACTER OF LiBr < NaBr < KBr < RbBr < CsBr

THE LARGER THE DIFFERENCE BETWEEN THE

ELECTRONEGATIVITY ,GREATER THE IONIC CHARACTER

INCREASING ORDER OF HYDRATION ENERGY

Ba+2 < Sr+2 < Ca+2 < Mg+2 < Be+2

THE SMALLER THE SIZE , MORE THE HYDRATION ENERGY

INCREASING BOND ANGLE NF3 < NH3

THERE IS LESSER REPULSION IN BONDING PAIRS IN NH3

INCREASING BOILING POINT IN HCl < HBr < HI < HF

THE LARGER THE DIFFERENCE BETWEEN THE

ELECTRONEGATIVITY ,GREATER THE IONIC CHARACTER

AND HENCE GREATER THE BOILING POINT. ANOMALOUS BEHAVIOUR

OF HF IS DUE TO HYDROGEN BONDING

1. WHAT IS THE COVALENCE OF NITROGEN IN

N2O5 ?

Ans- COVALENCY DEPENDS UPON THE NUMBER OF

SHARED PAIR OF ELECTRONS. NOW IN N2O5, NITROGEN ATOM HAS FOUR SHARED PAIR OF ELECTRONS.

THERFORE THE VALENCY OF N IN N2O5 IS 4.

2. WHY ARE PENTAHALIDES MORE COVALENT

THAN TRIHALIDES?

Ans- AS WE KNOW, HIGHER THE POSITIVE

OXIDATION STATE OF CENTRAL ATOM, MORE WILL BE ITS POLARISING POWER WHICH, IN TURN, INCREASES THE COVALENT CHARATER OF BOND FORMED BETWEEN THE CENTRAL ATOM AND THE OTHER ATOM.

3.BOND ANGLE IN PH4+ IS HIGHER IN PH3.

WHY?

Ans-

BOTH ARE sp3 HYBRIDISED. IN PH4+ ALL THE FOUR

ORBITALS ARE BONDED WHEREAS IN PH3 THERE IS A LONE PAIR OF ELECTRONS ON P, WHICH IS RESPONSIBLE FOR LONE PAIR-BOND PAIR REPULSION IN PH3 REDUCING THE BOND ANGLE TO LESS THAN 109°28’.

4.WHAT HAPPENS WHEN PCl5 IS HEATED?

Ans-

PCl5 HAS THREE EQUATORIAL (202 pm) AND TWO AXIAL (240 pm) BONDS. SINCE AXIAL BONDS ARE WEAKER THAN EQUATORIAL BONDS, THEREFORE WHEN PCl5 IS HEATED, THE LESS STABLE AXIAL BONDS BREAK TO FORM PCl3.

5.WHY DOES OZONE ACT AS A POWERFUL OXIDISING AGENT?

Ans-

DUE TO THE EASE WITH WHICH IT LIBERATES ATOMS OF NASCENT OXYGEN, IT ACTS AS A POWERFUL OXIDISING AGENT.

O3 → O2 + O (NASCENT OXYGEN)

6.WHY DOES THE REACTIVITY OF NITROGEN DIFFER FROM PHOSPHORUS?

Ans- NITROGEN EXISTS AS DIATOMIC MOLECULE. DUE

TO THE PRESENCE OF A TRIPLE BOND BETWEEN THE TWO N ATOMS, THE BOND DISSOCIATION ENERGY IS LARGE (941.4 KJ/ mol). THUS NITROGEN IS INERT AND UNREACTIVE IN ITS ELEMENTAL STATE. PHOSPHORUS EXISTS AS A TETRATOMIC MOLECULE(P4). SINCE THE P—P SINGLE BOND IS MUCH WEAKER (213 KJ/mol),THERFORE, PHOSPHORUS IS MUCH MORE REACTIVE THAN NITROGEN.

7.WHY DOES NH3 FORM HYDROGEN BONDS BUT PH3 DOES NOT?

Ans- THE ELECTRONEGATIVITY OF N(3.0) IS MUCH

HIGHER THAN THAT OF H(2.1). HENCE,N—H BOND IS QUITE POLAR AND HENCE NH3 UNDERGOES INTERMOLECULAR H—BONDING. ON THE OTHER HAND, BOTH P AND H HAVE AN ELECTRONEGATIVITY OF 2.1 THEREFORE, P—H BOND IS NOT POLAR AND HENCE PH3 DOES NOT UNDERGO H—BONDING.

8.WHY DOES R3P=0 EXIST BUT R3N=O DOES NOT (R IS ALKYL GROUP)?

Ans- N DUE TO THE ABSENCE OF d-orbitals, CANNOT

FORM p-d MULTIPLE BONDS. THUS, N CAN’T EXPAND ITS COVALENCY BEYOND FOUR BUT IN R3N=O, N HAS A COVALENCY OF 5. SO THE COMPOUND R3N=O DOES NOT EXIST. P DUE TO THE PRESENCE OF d-orbitals FORMS p-d MULTIPLE BONDS AND HENCE CAN EXPAND ITS COVALENCY BEYOND 4. THEREFORE, P FORMS R3P=O IN WHICH THE COVALENCY OF P IS 5.

9.WHY DOES NITROGEN SHOW CATENATION

PROPERTIES LESS THAN PHOSPHORUS?

Ans- THE PROPERTY OF CATENATION DEPENDS UPON

THE BOND STRENGTH OF THE ELEMENT. SINCE THE N—N (159 KJ/mol) BOND STRENGTH IS MUCH WEAKER THAN P—P (213 KJ/mol) BOND STRENGTH, HENCE, NITROGEN SHOWS LESS CATENATION PROPERTIES THAN PHOSPHORUS.

10.EXPLAIN WHY INSPITE OF NEARLY THE SAME ELECTRONEGATIVITY, OXYGEN FORMS HYDROGEN BONDING WHILE CHLORINE DOES NOT?

Ans- ALTHOUGH O AND Cl HAVE THE SAME

ELECTRONEGATIVITY, YET THEIR ATOMIC SIZE ARE MUCH DIFFERENT i.e. O=66pm AND Cl=99pm. THUS, ELECTRON DENSITY PER UNIT VOLUME ON OXYGEN ATOM IS MUCH HIGHER THAN THAT OF ON CHLORINE ATOM. HENCE, OXYGEN FORMS HYDROGEN BONDS WHILE CHLORINE DOES NOT, EVEN THOUGH BOTH HAVE APPROX THE SAME ELECTRONEGATIVITY.

Why is nitrous acid oxidant as well as reductant?

• The oxidation state of N in nitrous acid is +3 which lies in between its lowest oxidation state of -3 and highest oxidation state of +5. Since its oxidation state +3 to any lower value to -3,therefore it acts as oxidizing agent .

• Further since the oxidation state of N in HNO2 can be increased from +3 to +5 , therefore it acts as a reducing agent.

All bonds in PCl5 are equivalent. Justify.

• PCl5 has trigonal bipyrimidal structure. It has 3 equivalent equatorial and two equivalent axial P-Cl bonds. However due to greater bond pair – bond pair repulsions , the two axial P-Cl bonds are longer and hence different from the three equatorial bonds

• Why is OF6 not known?

• Due to absense of d-orbitals in its valence shell, oxygen shows a maximum oxidation state of +2

• Why does Mn(||) shows maximum paramagnetic character amongst the bivalent ions of the first transition state?

• Mn2+ has maximum no. of unpaired electrons .i.e. 3d5

• Give reasons indicate which of the following would be colored?

• Cu+, VO2+, Sc3+, Ni2+• (at. No. Cu=29 V=23 Sc=21 Ni =28 )

• Ni2+ due to incompletely filled d-orbitals.

• Zinc, Cadmium and Mercury are generally not considered as transition metals. Why?

• These metals in their most common oxidation state of +2 have completely filled d orbitals.

• Why KMnO4 is used in cleaning surgical instruments in hospitals?

• This is because KMnO4 has a germicidal effect.

• How would you account for the increasing oxidising power in the series

VO2+ < Cr2O7 2- <MnO4- ?

• This is due to the increasing stability of the lower species to which they are reduced.

• Why Cr, Mn and Fe has nearly same atomic radii?

• As effective nuclear charge is nearly same in all 3 cases.

• Why in permanganate ion, there is a covalency between Mn and Oxygen?

• In MnO4- ion Mn is in highest oxidation state. (+7). In higher oxidation state transition metals forms covalent bonds.

Q1 On what ground you can say that scandium (z=21) is a transition element but zinc (z=30) is not?

• Ans : On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state (3d) , it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d ) in its ground state as well as in its oxidised sate , hence it is not regarded as a transition element.

1

10

Q2 Silver atom has completely filled d orbitals (4d ) in its ground state . How can you say that it is a transition element ?

10

• Ans: Silver (z=47) can exhibit +2 oxidation state wherein will have incompletely filled d-orbitals (4d) , hence a transition element.

Q3 In the series Sc (z=21) to Zn (z=30), the enthalpy of atomisation of zinc is the lowest , i.e., 126 kJ mol . Why?-1

• Ans : In the formation of metallic bonds , no electrons from 3d-orbitals are involved in case of zinc, while in all other metals of 3d series , electrons from the d-orbitals are always involved in the formation of metallic bonds.

Q4 Which of the 3d series elements of the transition metals exhibits the largest number of oxidation states and why?

• Manganese (z=25) , as its atom has the maximum number of unpaired electrons.

Q5 Why is Cr reducing and Mn oxidising when both have d configuration ?

2+ 3+ 4

• Cr is reducing as its configuration changes from d to d , the latter having a half-filled t level . On the other hand , the change from Mn to Mn results in the half-filled (d ) configuration which has extra stability.

2+ 4 3

2g3+

2+

5

Q6 How would you for the irregular variation of ionisastion enthalpies (first and second) in the first series of the of the transition elements ?

• Ans : Irregular variation of ionisation enthalpies is mainly attributed to varying degree of stability of different 3d-configurations ( e.g. d , d , d ,are exceptionally stable)

0 510

Q7 Why is the oxidation state of a metal exhibited in its oxide or fluoride ?

• Ans: Because of small size and high electronegativity oxygen or fluorine can oxisidise the metal to its highest oxidation state.

Q8 Which is a strong oxidising agent Cr or Fe and why ? 2+2+

• Ans : Cr is stronger reducing than Fe .• Reason : d d occurs in case of Cr to Cr • But d d occurs in case of Fe to Fe • In a medium (like water) d is more stable as compared to d

4 3

6 5 2+ 3+

2+ 3+

Q9 Explain why Cu ion is not stable in aqueous solutions ? +

• Ans : Cu in aqueous solution undergoes disproportionation, i.e. , • 2Cu (aq) Cu (aq) + Cu(s)• The E value for this is favourable.

+

+ 2+

Q10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

• The 5f electrons are effectively shielded from nuclear charge . In other words the 5f electrons themselves provide poor shielding from element to element to element in the series.

Why are pentahalides more covalent than trihalides?

Higher the positive oxidation state of the central atom, more will be its polarisingpower which, in turn , increases the covalent character of bond formed between the central atom and the other atom. Thusthe pentahalides, which have a more positive central atom than trihalides are more covalent than trihalides.

Though nitrogen exhibits +5 oxidation state, it does notform a pentahalide. Give reason.

Nitrogen with principle quantum no. n=2, has s and porbitals only. It does not have d orbital to expand itscovalence beyond four. That is why it dos not formpentahalides.

PH3 has lower boiling point than NH3. Why?

Unlike NH3, PH3 molecules are not associated throughhydrogen bonding in liquid state. That is why the boilingpoint of PH3 is lower than that of NH3.

Why is BiH3 the strongest reducing agent amongst all thehydrides of Group 15 elements?

The stability of group 15 hydrides decreases from NH3 to BiH3. Consequently the reducing character increases fromNH3 to BiH3.These makes BiH3 the strongest reducing agent of all the hydrides of group 15 elements.

Why is N2 less reactive at room temperature?

N2 is inert at room temperature because of the high bondenthalpy N N triple bond but its reactivity increases rapidlywith temperature.

Why does NH3 act as a Lewis base?

Nitrogen atom in NH3 has one lone pair of electrons whichis available for donation. Therefore, it acts as a Lewis base.

N

H H H

..Lone pair

PH3 is basic in nature. Give reason.

The lone pair on phosphorus of PH3 makes it a Lewis base.Further the following reaction proves that PH3 is a basebecause it reacts with an acid. PH3 + HI PH4I

Bond angle in PH4 is higher than that in PH3. Why?+

Both are sp hybridised. In PH4 all the four orbitals arebonded but in PH3 only three orbitals are bonded and the fourth one carries a lone pair which brings the bondedorbitals closer due to bond pair-lone pair repulsion, thisresults in a smaller angle than expected for sphybridisation.

3 +

3

Why does PCl3 fume in moisture?

PCl3 reacts readily with moisture to form orthophosphorousacid and gives out HCL fumes.

PCl3 + 3H2O H3PO3 + 3HCl

Why is H2O a liquid and H2S a gas?

Due to small size and high electronegativity of oxygen H2Omolecules are highly associated with inter-molecularhydrogen-bonding which results in it’s liquid state. There is no such bonding in H2S and so it’s a gas.

• Q. Nitrogen shows anomalous behaviour ?

• Ans : Nitrogen differs from the rest of the members of this group due to its smaller size, high electro negativity, high ionization enthalpy and non-availability of d orbitals.

• Q. Pentahalides of group 15 are more covalent than trihalides ?

• Ans: Higher the positive oxidation state of central atom, more will be its polarizing power which, in turn, increases the covalent character of bond formed between the central atom and the other atom.

• Q. Bond angle in PH4+ is higher than that in PH3 ?

• Ans: Both are SP3 hybridised. In PH4+ all the four orbitals are bonded whereas in PH3 there is a lone pair of electrons on P, which is responsible for lone pair-bond pair repulsion in PH3 reducing the bond angle to less than 109° 28′.

• Q.NO2 dimerises to form N2O4 ?

• Ans:NO2 contains odd number of valence electrons. It behaves as a typical odd molecule. On dimerisation, it is converted to stable N2O4 molecule with even number of electrons.

• Q. Nitrogen is fairly inert gas ?

• Ans: Nitrogen exists as triply bonded diatomic non polar molecule. Due to short internuclear distance between two nitrogen atoms the bond strength is very high. It is, therefore, very difficult to break the bond.

• Q. Nitrogen exists as diatomic molecule & phosphorus as P4 ?

• Ans: Because N2 is Diatomic molecules, hence weak Vander Waal’s force of attraction thus is a gas whereas P4 is  tetraatomic hence Stronger Vander Waal’s force of attraction thus it is solid.

• Q. NO2 is coloured but N2O4 is colourless ?

• Ans. NO2 has unpaired electrons therefore it absorbs light from visible and radiate brown colour whereas N2O4 doesnot have unpaired electrons so it does notabsorb light from visible region.

• Q. Tendency to show –2 oxidation state diminishes from sulphur to polonium in group 16 ?

• Ans: The outer electronic configuration of group 16 elements is ns2 np4. These elements therefore have the tendency to gain two electrons to complete octet. Since elctronegativity and I.E. decrease on going down the group, tendency to show –2 oxidation state diminishes.

• Q. Oxygen generally exhibit oxidation state of –2 only whereas other members of the family exhibit +2, +4, +6 oxidation states also ?

• Ans: Oxygen is a electronegative element thus exhibit oxidation state of –2. .Other members of the family have d orbitals and therefore, can expand their octets and show + 2, + 4, + 6 oxidation states also.

• Q. Sulphur vapours exhibits paramagnetism ?

• Ans :In vapour state sulphur partly exists as S2 molecule which has two unpaired electrons in the antibonding π* orbitals like O2 and, hence, exhibits paramagnetism.

Ques1.

The solubility of calcium acetate decreases while that of the lead nitrate

increases with increase in temperature?

Ans1.

Because dissolution of calcium acetate is a exothermic process while that of lead nitrate is a endothermic process

Ques2.

H3PO2 and H3PO3 act as as good reducing agents while

H3PO4 does not?

Ans2:

In H3PO2, two H atoms are bonded directly to P atom & in H3PO3 one H atom is bonded directly to P atom

which imparts reducing character to the acid, whereas in H3PO4 there is no

H atom bonded directly to P atom

Ques3.

NO2 is coloured but N2O4 is colourless?

Ans3.

NO2 has unpaired electrons therefore it absorbs light from visible and radiate brown colour whereas N2O4 doesnot

have unpaired electrons so it does notabsorb light from visible

region.

Ques4.

Sulphur vapours exhibits paramagnetism

Ans4.

In vapour state sulphur partly exists as S2 molecule which has two unpaired

electrons in the antibonding π* orbitals like O2 and, hence, exhibits

paramagnetism.

Ques5.

NOis paramagnetic in the gaseous state but dimagnetic in liquid and

solid state ?

Ans5.

NO = 5 + 6 = 11 e–, it has odd pair of e– and hence paramagnetic in gaseous state, but in liquid and solid state, it

exists as dimer

Ques6.

Why is Cr2+ reducing and Mn3+ oxidising when both have d4

configuration ?

Ans6.Cr2+ is reducing as its configuration changes from d4

to d3, the d3 has half-filled t2g level.n the other hand, the change from Mn2+ to Mn3+

results in the half filled (dS)configuration which has extra stability.

Cr2+ = 3 d4 4 s0 Mn3+ = 3 d4 4 s0

Cr3+ = 3 d3 4 s0Cr3+ = 3 d3 4 s0have half-filled t2g level.

Mn2+ = 3 d5 4 s0

half-filled extra stable.

Ques7.

Although Cu+ has configuration 3 d10 4 s0 (stable) and Cu2+ has

configuration 3 d9 (unstable configuration) still Cu2+ compounds are more

stable than Cu+?

Ans7.

It is due to much more (–) D Hydration H– of Cu2+ (aq) than Cu+, which ismore than compensates for the II

ionization enthalpy of Cu.

Ques8.

When mercuric iodide is added to an aqueous solution of KI, the freezing

point is raised?

Ans8.

Due to the fomation of complex K2 (Hg I4), number of particles in the solution

decreases and hence thefreezing point is raised

Ques9.

redox couple has less positive electrode potential than

couple?

Ans9.

In Mn2+ d5 configuration leads to extrastability of half filled configuration, so

Mn3+ / Mn2+ (d4) tends to getconverted to stable d5, configuration of

Mn2+, by accepting an electron so Mn3+/Mn2+ redox couple has more

positive potential than couple.

Ques10.

ClF3 exists but FCl3 does not. Why?

Ans10.

Due to unavailability of d-orbitals in fluorine atom it cannot expand its valence

shell. Therefore it is unable to form FCl3 whereas Chlorine has vacant d-

orbitals. Hence it can promote one of the 3 p electrons to the 3-d sub shell and shows

+3 oxidation state and forms ClF3

• Give reason: The maximum no. of covalent bonds formed by nitrogen is 4.

• Ans. Nitrogen has 3 unpaired electrons & 1 lone pair of electrons , therefore it can form 3 covalent bonds & 1 coordinate bond therefore its covalency is 4 e.g. in NH4 .

• Explain why oxygen is a gas while sulphur is a solid.

• Oxygen is diatomic molecule & has double bond between 2 oxygen atoms, therefore there is weak van der Waals force of attraction between molecules ,therefore it is a gas. Sulphur is octaatomic molecule (S8 ) & has more intermolecular force of attraction , that is why it exists as solid.

• Explain why fluorine always exhibit an oxidation state of -1 only.

• Ans. Fluorine shows only -1 oxidation state because it is most electronegative element 7 does not have d- orbitals.

• Assign reasons for the following observations.• (1) Hydrogen iodide is a stronger acid than

hydrogen fluoride in aq. solution.• (2) The basic character among the hydrides of

group 15 elements decreases with increasing atomic no.’s.

• Ans. (1) HI has lower bond dissociation energy than HF.

• (2) It is due to increase in atomic size of group 15 elements , lone pair of electrons is less available for bond formation.

The d and f – Block Elements

• Why do transition metals have high enthalpy of hydration?

• Ans. Transition metal ions are smaller in size & have higher charge, therefore , have higher enthalpy of hydration.

• Explain why do transition elements show variable oxidation states. Write all the possible oxidation states of an element having atomic no. 25.

• Ans. They show variable oxidation state because electrons from ‘s’ as well as d – orbitals take part in bond formation. Mn (25) can show +2, +3, +4, +6, & +7 oxidation states.

• Give reasons for each of the following :• (1)Size of trivalent lanthanoid cations

decreases with increase in the atomic no.• (2) chemistry of all the lanthanoids is quite

similar.

• Ans. (1) It is due to poor shielding effect of f – electrons , effective nuclear charge increases , ionic size decreases.

• (2) It is due to similar ionic size which is due to lanthanoid contraction , they resemble in their properties.

• K2PtCl6 is well known compound whereas corresponding compound of Ni is not known, Why?

• Ans. K2PtCl6 is known because energy required to remove 4 electrons in Pt is less as compared to Ni, therefore , corresponding compound of Ni is not known.

• Explain the following :• (1) transition elements tend to be unreactive

with increasing atomic no. in the series.• (2) d – Block elements exhibit more oxidation

states than f – block elements.

• Ans. (1) Transition metals form layer of oxides on their surface due to which they become unreactive.Secondly, reactivity decreases with increase in atomic no. due to decrease in size & increase in ionisation energy.

• (2)In d- block elements, e- of s- orbital & d-orbital both take part in bond formation. In f-block elements due to poor shielding effect of f-electrons effective nuclear charge increases, therefore lesser no. of oxidation states are shown.

• Give reasons :• (1) Cr2+ is a strong reducing agent whereas

Mn2+ is not.• (2) The transition metal ions such as Cu+, Ag+,

& Sc3+ are colourless.

• Ans. (1) Cr2+ is less stable than r3+ , therefore it is good reducing agent whereas Mn2+ is stable due to half filled d –orbitals therefore it is not reducing agent.

• (2) Cu+, Ag+, & Sc3+ are colourless because they do not have unpaired electrons.

• 1) Bond angle PH4 + is higher than in PH3. Why?

• Ans: Both PH4 + and PH3 involve sp3 hybridisation of P atom. In PH4 + all the four orbitals are bonded, whereas in PH3 there is a lone pair of electron on P. In PH4 +, the HPH bond angle is tetrahedral angle of 109.5°. But in PH3, lone pair-bond pair repulsion is more than bond pair-bond pair repulsion so that bond angles become less than normal tetrahedral angle of 109.5°. The bond angle in PH3 has been found to be about 93.6°.

• 2) Out of noble gases, only Xenon is known to form chemical compounds. Explain.

• Ans: Except Radon, which is radioactive, Xenon has lowest ionisation enthalpy among noble gases and hence it readily forms chemical compounds particularly, with O2 and F2.

• 3) HCl when reacts with finely divided iron forms ferrous chloride and not ferric chloride. Why?

• Ans: HCl reacts with finely divided iron and produces H2 gas.

• 2Fe + 6HCl → 2FeCl3 + 3H2• Liberation of hydrogen prevents the formation

of ferric chloride.

• 4) Halogen have maximum negative gain

enthalpy in the respective periods of the periodic table. Why?

• Ans: . The halogens have the smallest size in their respective periods and therefore, high effective nuclear charge. Moreover, they have only one electron less than the stable noble gas configuration (ns²np6).

• Therefore, they have strong tendency to accept one electron to acquire noble gas electronic configurations and hence have maximum negative electron gain enthalpy in their respective periods.

• 5) Explain why inspite of nearly same electronegativity, Oxygen forms hydrogen bonding while chlorin does not.

• Ans: Oxygen has smaller size than chlorine. The smaller size of oxygen favours hydrogen bonding. In other words, though electronegativity of Cl is same as that of O, it does not form hydrogen bonding because of its larger size.

• 6) Why is europium (II) more stable than cerium (II)?

• Ans: Europium (II) has electronic configuration [Xe] 4f7 5d0 while cerium (II) has electronic configuration [Xe] 4f1 5d1. In Eu2+, 4f subshell is half filled and 5d-subshell is empty. Since half filled and completely filled electronic configurations are more stable, Eu2+ ion is more stable than Ce2+ in which neither 4f subshell nor 5d subshell is half filled or completely filled.

• 7) Why are Mn2+ compounds more stable than Fe2+ towards oxidation to their +3 state?

• Ans: Mn2+ compounds are more stable than Mn3+ due to stable 3d5 (half filled) configuration. On the other hand, Fa3+ is more stable than Fe2+ because of half filled configuration of Fe3+. Therefore, Mn2+ compounds are more stable than Fe2+ towards oxidation to their +3 state.

• 8) K2PtCl6 is known but Ni compound is not known. State reason for it?

• Ans: This is because Pt4+ is more stable than Ni4+ as a sum of four ionisation energies of Pt is less than that of Ni.

• 9) Why do transition elements exhibit higher enthalpies of atomization?

• Ans: The high enthalpies of atomization are due to large number of unpaired electrons in their atoms. Therefore, they have stronger interatomic interactions and hence, stronger bonding between atoms. Thus, they have high enthalpies of atomization.

• 10) Why do Zr and Hf exhibit similar properties?

• Ans: Due to lanthanoid contraction, Hf and Zr have almost similar size and therefore, their properties are similar.