d and f

Question 1

What are d and f block elements called?

Answer 1

They are called transition and inner transition elements.

They are placed between s and p block, indicating a transition from metallic to non-metallic character.

Question 2

Define transition /d block elements.

Answer 2

Transition metals are defined as metals which have incomplete d subshell either in neutral atom or in their ions.

Question 3

Why are zinc, cadmium, and mercury not regarded as transition metals?

Answer 3

They have a full d10 configuration in their ground state as well as in their common oxidation states.

Question 4

What is the difference between transition and non-transition elements?

Answer 4

The presence of partly filled d or f orbitals in their atoms makes transition elements different from non-transition elements.

Question 5

What is the general Electronic Configuration of the d-Block Elements?

Answer 5

(n-1)d¹–¹⁰ ns¹–²

Question 6

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

Answer 6

On the basis of incompletely filled 3d orbitals in case of scandium atom in its ground state (3d¹), it is regarded as a transition element. On the other hand, zinc atom has completely filled d orbitals (3d¹⁰) in its ground state as well as in its oxidized state.

Question 7

Why is silver (4d¹⁰ 5s¹) considered a transition element?

Answer 7

Silver in +2 state has (4d⁹), so a transition element.

Question 8

The transition metals (with the exception of Zn, Cd and Hg) are very hard and have low volatility. Why?

Answer 8

Strong interatomic interaction resulting in strong metallic bonding due to the presence of unpaired d-electrons.

Question 9

The transition metals have highmelting and boiling points. Why?

Answer 9

Strong interatomic interaction resulting in strong metallic bonding due to the presence of unpaired d-electrons.

Question 10

The transition metals have high enthalpy of atomization. Why?

Answer 10

Strong interatomic interaction resulting in strong metallic bonding due to the presence of unpaired d-electrons.

Question 11

Along the transition series,m.p/b.p reaches maxima near the middle.Why?

Answer 11

Greater the number of valence electrons, stronger is the resultant bonding. Along the series, the number of unpaired d electrons increases from 1 to 5 and then decreases

Question 12

Zn, Cd and Hg have low m.p.Why?

Answer 12

Weak interatomic interaction due to absence of unpaired d electrons as the configuration is d¹⁰

Question 13

Zn, Cd and Hg have low enthalpy of atomization. Why?

Answer 13

Weak interatomic interaction due to absence of unpaired d electrons as the configuration is d¹⁰

Question 14

Zn, Cd and Hg are volatile. Why?

Answer 14

Weak interatomic interaction due to absence of unpaired d electrons as the configuration is d¹⁰

Question 15

Why is there more frequent metal-metal bonding in compounds of the heavy transition metals?

Answer 15

The metals of the second and third series have greater enthalpies of atomisation than the corresponding elements of the first series.

the frequent occurrence of metal-metal bonding in the compounds of heavy transition elements is primarily due to the presence of unpaired d electrons that facilitate bonding interactions.

Question 16

What trend is observed in ions of the same charge in a given transition series?

Answer 16

Ions show a progressive decrease in radius with increasing atomic number.

The new electron enters a d orbital each time the nuclear charge increases by unity along the series.

Question 17

The radii of the third (5d) series are virtually the same as those of the corresponding members of the second series(4d).

Answer 17

Due to lanthanoid contraction the second and the third d series exhibit similar radii (e.g., Zr 160 pm,Hf 159 pm) and have very similar physical and chemical properties.

Question 18

Zr and Hf have similar ionic size

Answer 18

Due to lanthanoid contraction the second and the third d series exhibit similar radii (e.g., Zr 160 pm,Hf 159 pm) and have very similar physical and chemical properties.

Question 19

The second and the third d series have very similar physical and chemical properties. Why

Answer 19

Due to lanthanoid contraction the second and the third d series exhibit similar radii (e.g., Zr 160 pm,Hf 159 pm) and have very similar physical and chemical properties.

Question 20

Why is the enthalpy of atomisation of zinc the lowest in the series Sc (Z = 21) to Zn (Z = 30)?

Answer 20

Zn has a fully filled 3d10 configuration. Weak interatomic interaction due to absence of unpaired d electrons as the configuration is d10.

Question 21

Why is the second ionization energy (I.E) of Mn low?

Answer 21

Mn+ (3d5 4s1) changes to more stable exactly half filled Mn2+ (3d5 4s0).

Question 22

Why is the third ionization energy (I.E) of Fe low?

Answer 22

Fe2+ (3d6 4s0) changes to more stable exactly half filled Fe3+ (3d5 4s0).

Question 23

Why is the second ionization energy (I.E) of Zn low?

Answer 23

Zn+ (3d10 4s1) changes to more stable completely filled Zn2+ (3d10 4s0).

Question 24

Why is the third ionization energy (I.E) of Zn high?

Answer 24

Stable completely filled Zn2+ (3d10 4s0) configuration lost, Zn3+ (3d9 4s0).

Question 25

Why do transition metals exhibit a large number of oxidation states?

Answer 25

Due to the presence of unpaired d electrons. Less energy gap between ns and (n-1)d orbitals.

Question 26

Why do transition elements at the extreme ends exhibit a lesser number of oxidation states?

Answer 26

Due to less no of unpaired d electrons.

Question 27

Why do Scandium and Zinc exhibit only one oxidation state?

Answer 27

Due to less no of unpaired d electrons.

Question 28

Where do the elements that give the greatest number of oxidation states occur?

Answer 28

In or near the middle of the series due to max no of unpaired d electrons.(d5).

Question 29

Why does Mn exhibit the maximum number of oxidation states in the first transition series?

Answer 29

Due to max no of unpaired d electrons.(d5).

Question 30

Why do the oxidation states of transition metals differ from each other by unity?

Answer 30

Due to incomplete filling of d orbitals, less energy gap between ns & (n-1)d orbitals.

Question 31

Why is Cr(VI) a strong oxidizing agent in acidic medium while MoO3 and WO3 are not?

Answer 31

The higher oxidation state becomes more stable down the group among the transition elements unlike the p block (inert pair effect).

Question 32

Mention compounds where transition metals exhibit low/zero/negative oxidation states.

Answer 32

In Ni(CO)4 and Fe(CO)5, the oxidation state of nickel and iron is zero.

Question 33

Name a transition element that does not exhibit variable oxidation states.

Answer 33

Scandium (Z = 21) does not exhibit variable oxidation states.

Question 34

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

Answer 34

Mn due to max no of unpaired d electrons.(d5).

Question 35

Why is Cu unable to liberate H2 from acids despite being in the 3d series?

Answer 35

The high energy to transform Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy.

Question 36

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

Answer 36

Cr2+ is reducing as its configuration changes from d4 to d3 Cr3+, which has a half-filled t2g level. Mn3+ to Mn2+ results in a half-filled (d5) configuration with extra stability.

Question 37

What is the possible reason for the positive E0(M2+/M) value for copper?

Answer 37

High enthalpy of atomization of Cu and low hydration enthalpy of Cu2+.

Question 38

Why are many copper (I) compounds unstable in aqueous solution?

Answer 38

The stability of Cu2+(aq) rather than Cu+(aq) is due to the much more negative ∆hydH⁰ of Cu2+(aq) than Cu+, which compensates for the second ionisation enthalpy of Cu.

Question 39

The highest Mn fluoride is MnF4 whereas the highest oxide is Mn2O7. Why?

Answer 39

The ability of oxygen to form multiple bonds to metals while fluorine cannot.

Question 40

Transition metals exhibit highest oxidation states in their oxides and not fluorides. Why?

Answer 40

The ability of oxygen to form multiple bonds to metals while fluorine cannot.

Question 41

How would you account for the increasing oxidizing power in the series VO2+ < Cr2O7 2– < MnO4–

Answer 41

This is due to the increasing stability of the lower species to which they are reduced. VO2+ (V +5) < Cr2O7 2– (Cr +6)< MnO4– (Mn+7) oxidation no: increases.

Question 42

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

Answer 42

The irregular variation of ionization enthalpies can be attributed to the extra stability of configuration such as d⁰,d⁵ and d¹⁰. Since these states are exceptionally stable, their ionization enthalpies are very high.

Question 43

Explain the irregularity in the E⁰(M2+/M) along the 3d series.

Answer 43

The E⁰(M2+/M) values are not regular which can be explained from the irregular variation of ionization enthalpies (IE 1 and IE 2)and also the sublimation enthalpies which are relatively much less for manganese and vanadium.

Question 44

Why is the value E0 for the Mn3+/Mn2+ couple much more positive than that for Cr3+/Cr2+ orFe3+/Fe2+? Explain.

Answer 44

IE 3 of Mn (where the required change is d⁵ to d⁴) very high.

Question 45

Transition metals are paramagnetic. Why?

Answer 45

Due to the presence of unpaired d electrons.

Question 46

The ‘spin-only’ formula to calculate magnetic moment

Answer 46

µ =√ n(n+2 ), where n is the no of unpaired d electrons.

Question 47

Sc3+ and Zn2+ ions have 0 magnetic moment. Why?

Answer 47

Sc3 - d⁰ and Zn2+- d¹⁰ - absence of unpaired d electrons.

Question 48

Mn2+ has high magnetic moment.(5.98)

Answer 48

Mn2+(d⁵) max no of unpaired d electrons.

Question 49

Calculate the magnetic moment of a divalent ion in aqueous solution if its atomic number is 25.

Answer 49

M2+ is Mn2+ so magnetic moment µ =√5(5+2 ) = 5.92BM

Question 50

Calculate the ‘spin only’ magnetic moment of M2+(aq) ion (Z = 27)

Answer 50

M2+is Co2+(d⁷), 3 unpaired d electrons so magnetic moment µ =√ 3(3+2 ) = 3.87 BM.

Question 51

Transition metals form coloured ions. Why?

Answer 51

Due to the presence of unpaired d electrons,which are excited and de excited to emit radiation in the visible region. Due to transition of electrons from one d level to another (d-d transition)

Question 52

Sc3+,Ti4+,Zn2+ salts are colourless.

Answer 52

Due to the absence of unpaid d electrons.(Sc3+,Ti4+- d⁰, Zn2+ - d¹⁰)

Question 53

Transition metals form complexes. Why

Answer 53

Due to the comparatively smaller sizes of the metal ions, their high ionic charges and the availability of d orbitals for bond formation

Question 54

The transition metals and their compounds are known for their catalytic activity. Why?

Answer 54

This activity is ascribed to their ability to adopt multiple oxidation states and to form complexes. Being solid also provides surface area.

Question 55

Transition metals form interstitial compounds.Why?

Answer 55

Transition metals being solids, traps small atoms in their crystal lattice. ##Footnote The compounds formed when small atoms of H, C or N get trapped inside the crystal lattice of metals are known as interstitial compounds.

Question 56

Transition metals form alloys.Why?

Answer 56

Transition metals have similar sizes.

Question 57

Why are lanthanides called inner transition elements?

Answer 57

They have the (n-2)f orbitals filled and are to be placed in between transition elements.

Question 58

What is lanthanoid contraction?

Answer 58

The overall decrease in atomic and ionic radii from lanthanum to lutetium due to poor shielding of 4f orbitals.

Question 59

What are the consequences of lanthanide contraction?

Answer 59

~ The radii of the members of the third transition series to be very similar to those of the corresponding members of the second series. ~ [The almost identical radii of Zr(160 pm) and Hf (159 pm), a consequence of the lanthanide contraction, account for their occurrence together in nature It is difficult to separate elements of the second and third transition series.] ~ The basic character of Lanthanide hydroxide decreases from La(OH)3 to Lu(OH)3

Question 60

What is the most common oxidation state of lanthanides?

Answer 60

+3

Question 61

Few elements like Ce,Eu,Gd etc exhibit +2 or +4 oxidation state. Why?

Answer 61

Due to the stability of f⁰ ,f⁷ or f¹⁴ configuration