The d- and f- Block Elements

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Questions:

Generic 5

8 x 5 = 40

(a) Account for the following:
(i) Mn shows the highest oxidation state of +7 with oxygen but with fluorine, it shows the highest oxidation state of +4.
(ii) Zirconium and Hafnium exhibit similar properties.
(iii) Transition metals act as catalysts.
(b) Complete the following equations:
(i) \(2MnO_{ 2 }+4KOH+O_{ 2 }\overset { \Delta }{ \longrightarrow } \)
(ii) \(Cr_{ 2 }O_{ 7 }^{ 2- }+14H^{ + }+6I^{ - }\longrightarrow \)

(i) With reference to structural variability and chemical reactivity, write the differences between lanthanoids and actinoids.
(ii) Name a member of the lanthanoid series which is well known to exhibit +4 oxidation state.
(iii) Complete the following equation:
\(MnO_{ 4 }^{ - }8H^{ + }5e^{ - }\longrightarrow \)
(iv) Out of \(Mn^{ 3+ }\) and \(Cr^{ 3+ }\) , which is more paramagnetic and why?
(Atomic nos.: Mn = 25, Cr = 24)

Complete the following chemical reaction equations:
(i) \(CrO_{ 7 }^{ 2\quad - }(aq)+H_{ 2 }S(g)+H^{ + }(aq)\longrightarrow \)
(ii) \(CrO_{ 7 }^{ 2\quad - }(aq)+H_{ 2 }S(g)+H^{ + }(aq)\longrightarrow \)
(b) Explain the following observations:
(i) Transition metals form compounds which are usually coloured.
(ii) Transition metals exhibit variable oxidation states.
(iii) The actinoids exhibit a greater range of oxidation states than the lanthanoids.

(a) Complete the following chemical equations :
\((i)\ { Cr }_{ 2 }{ O }_{ 7 }^{ 2- }(aq)+{ H }_{ 2 }S(g)+{ H }^{ + }(aq)\longrightarrow \)
\((ii)\ { Cu }^{ 2+ }(aq)+{ I }^{ - }(aq)\longrightarrow\)
(b) How would you account for the following ?
(i) The oxidizing power of oxoanions is the order : \({ VO }_{ 2 }^{ + }<{ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }<{ MnO }_{ 4 }^{ - }\)
(ii) The third ionization enthalpy of manganese (Z = 25) is exceptionally high.
(iii) Cr²⁺ is stronger reducing agent than Fe²⁺

(a) Answer the following questions:
(i) Which element of the first transition series has highest second ionisation enthalpy?
(ii) Which element of the first transition series has highest third ionisation enthalpy?
(iii) Which element of the first transition series has lowest enthalpy of atomisation?
(b) Identify the metal and justify your answer.
(i) Carbonyl M(CO)₅
(ii) MO₃F

On the basis of Lanthanoid contraction, explain the following:
(i) Nature of bonding in La₂O₃ and Lu₂O₃.
(ii) Trends in the stability of oxo salts of lanthanoids from LA to Lu.
(iii) Stability of the comlexes of lanthanoids.
(iv) Radii of 4d and 5d block elements.
(v) trends in acidic character of lanthanoid oxides.

Write down the number of 3d electrons in each of the following ions: Ti²⁺, V²⁺, Cr³⁺, Mn²⁺, Fe²⁺, Fe³⁺, Co²⁺, Ni²⁺ and Cu²⁺. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

(a) Given below are the electrode potential values, E^o for the some of the first row of transition elements:

Element

E^o_M²⁺_/M (V)

V(23)

Cr(24)

Mn(25)

Fe(26)

Co(27)

Ni(28)

Cu(29)

-1.18

-0.91

-1.18

-0.44

-0.28

-0.25

+0.34

Explain the irregularities in these values on the basis of electronic structures of atoms.
(b) Complete the following reaction equations:
(i) Cr₂O₇^2-+Sn²⁺+H⁺\(\longrightarrow \)
(ii) MnO₄^-+Fe²⁺+H⁺\(\longrightarrow \)

Generic 2

1 x 2 = 2

Calculate the number of unpaired electrons in following gaseous ions: Mn³⁺, Cr³⁺, V³⁺ and Ti³⁺. Which one of these is the most stable in aqueous solution?

Generic 3

1 x 3 = 3

10.

Give examples and suggest reasons for the following features of the transition metal chemistry:
(i) The lowest oxide of transition metal is basic, the highest is amphoteric/acidic.
(ii) A transition metal exhibits highest oxidation state in oxides and fluorides.
(iii) The highest oxidation state is exhibited in oxoanions of a metal.

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Answers Key:

Generic 5

8 x 5 = 40

(i) Manganese shows the highest oxidation state of +7 with oxygen but +4 with fluorine. This
is because oxygen has a tendency to form multiple bonds and hence stabilize the high oxidation state.
(ii) Due to lanthanoid contraction, Zr and Hf show similar properties.
(iii) The transition metals act as catalysts. This is due to their ability to show multiple oxidation states
\((i)2{ MnO }_{ 2 }+4KOH+{ O }_{ 2 }\rightarrow 2{ K }_{ 2 }{ MnO }_{ 4 }+2{ H }_{ 2 }O\)
\(\\ (ii){ Cr }_{ 2 }{ O }_{ 7 }^{ 2- }+14{ H }^{ + }+6{ I }^{ - }\rightarrow 2{ Cr }^{ 3+ }+7{ H }_{ 2 }O+3{ I }_{ 2 }\)

(i) Zinc (Z= 30, 1822822p6 3823p6 3d10 482)does not have partially filled d-subshell in its elementary state or in its commonly occurring oxidation state (Zn²⁺: 3dlO). Therefore, it is not regarded as a transition element.
(ii) Transition elements or the d-block elements from a large number of coordination complexes. The transition metal ions bind to a number of anions or neutral molecules in these complexes. The common examples are [Ni(NHJJ²⁺, [Co(NHJJ^3-, [Fe(CN)J³⁺,
[Fe(CN)J^4-, [Cu(NH3)4]²⁺e,etc. The high tendency of transition metal ions to form complexes is due to
(a) small size of the atoms and ions of transition metals
(b) high nuclear charge
(c) availability of vacant d-orbitals of suitable energy to accept lone pairs of electrons donated by other groups (called ligands).
(iii) Mn²⁺ has 3^d5 electronic configuration. It is stable because of the half-filled configuration of d-subshell. Therefore, Mn has very high third ionization enthalpy for the change from d5 to d4 and it is responsible for much more positive E^ovalue for Mn³⁺/Mn²⁺couple in comparison to Cr3⁺/Cr²⁺couple

(a) (i) This is due to increase stability of lower species to which they are reduced.
(ii) The electronic configuration of manganese is 3^d548z. After the loss of the outer 48 electrons; its electronic configuration becomes stable because of half filled configuration. Therefore, it becomes difficult to remove the third electrons and hence its third ionization enthalpy is exceptionally high.
(iii) Cr²⁺ is reducing because its configuration changes from 3^d4 to 3d³. The 3d³ configuration of Cr²⁺in its compounds (expressed as t2 3) is stable because it has half filled t²g subshell. In the other hand, Fe²⁺on changing to Fe³⁺becomes 3^d5 which is not as stable as 3d3 in its compounds. Therefore, Cr²⁺ is stronger reducing than Fe²⁺.

(a) (i) \( \mathrm{Cr}_{2} \mathrm{O}_{7}^{2-}(a q)+3 \mathrm{H}_{2} \mathrm{~S}(g)+8 \mathrm{H}^{+}(a q) \) \(\longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{~S}+7 \mathrm{H}_{2} \mathrm{O}\)
(ii) \( 2 \mathrm{Cu}^{2+}(a q)+2 \mathrm{I}^{-}(a q) \longrightarrow 2 \mathrm{Cu}^{+}(a q)+\mathrm{I}_{2}(s)\)
(b) (i) It is because V in lower oxidation the state is less stable than Cr which is less stable than Mn. That is why MnO₄ is a best oxidizing agent and VO²⁺ is least.
(ii) M n (2 5) has electronic configuration [Ar]4s23ds, electronic configuration of Mn²⁺ is [Ar]4so3ds. After losing 2 electrons, it has half filled d-orbital, which is more stable that is why Mn²⁺ has exceptionally high third ionization energy, i.e. the energy required to remove
the third electron is very high.
(iii) It is because in Cr³⁺, d3 (half filled t2g orbitals) is more stable in aqueous solution than Fe³⁺ i.e. Cr³⁺ is more stable than Fe³⁺.

(i) Cu. This is because the electronic configuration of Cu is 3d10 4s1. After the loss of a 1st electron, it acquires stable configuration of 3dlO. Hence, removal of the 2nd electron is very difficult.
(ii) Zn. This is because Zn = 3dlO 4; and Zn2+ = 3d10 which is again fully filled and hence is very stable. Removal of 3rd electron requires very high energy.
(iii) Zn. This is because it has completely filled 3d subshell and no unpaired electron is available for metallic bonding.
(b) (i) Following EAN rule (page 9/102), M(CO)s is Fe(CO)s x -2
(ii) M03 p-l, X - 6 - 1 = 0, x = + 7, i.e., M is in oxidation state + 7. Mn shows an oxidation state of + 7. Hence, the compound is Mn03F.

(i) As the size decreases from La to Lu, covalent character increases (according to Fajan's rule). Hence, La203 is more ionic and LU203 is more covalent.
(ii) As the size decreases from La to Lu, stability of oxo salts decreases.
(iii) As the size of lanthanoids decreases, charge/size ratio increases and hence the stability of the complexes increases.
(iv) Due to lanthanoid contraction, radii of 4d and 5d block elements are nearly equal.
(v) As explained in
(a) covalent character of oxides increases from La to Lu, therefore, their basic character decreases or acidic character increases

Metal ion	Number of d-electrons	Filling of d-orbitals
Ti²⁺	2	\(t_{2 g}^{2}\)
V²⁺	3	\(t_{2 g}^{3}\)
Cr³⁺	3	\(t_{2 g}^{3}\)
Mn²⁺	5	\(t_{g}^{3} e_{g}^{2}\)
Fe²⁺	6	\(t_{2 g}^{4} e_{g}^{2}\)
Fe³⁺	5	\(t_{2 g}^{3} e_{g}^{2}\)
CO²⁺	7	\(t_{2 g}^{5} e_{g}^{2}\)
Ni²⁺	8	\(t_{2 g}^{6} e_{g}^{2}\)
Cu²⁺	9	\(t_{2 g}^{6} e_{g}^{3}\)

(a) It is due to irregular variations in sum of first and second ionisation energies and sublimation energies. It is also due to stability of electronic configuration.
(b) (i) Cr₂O₇^2- + 3Sn²⁺ + 14H⁺➝ 2Cr³⁺+ 3Sn⁴⁺ + 7H₂O
(ii) MnO₄^- + 5Fe²⁺ + 8H⁺ ➝ SFe³⁺ + Mn²⁺ + 4H₂O

Generic 2

1 x 2 = 2

Mn³⁺ = 3d⁴ = 4 unpaired electron, Cr³⁺ = 3d³ = 3 unpaired electrons, V³⁺ = 3d² = 2 unpaired electrons, Ti³⁺= 3d¹ = 1 unpaired electron. Cr³⁺ is most stable out of these in aqueous solution because it has half filled t_2g level (i.e., t³_2g) .

Generic 3

1 x 3 = 3

10.

(i) The lower oxide of transition metal is basic because the metal atom has low oxidation state whereas n in highest is acidic due to highest oxidation state. For example, MnO is basic whereas Mn₂O₇is acidic. In the low oxidation state of the metal, some of the valence electrons of the metal atom are not involved in bonding. Hence, it can donate electrons and behave as a base. In the higher oxidation state, valence electrons are involved in the bonding and are not available. Instead, the effective nuclear charge is high. Hence, it can accept electrons and hence behave as an acid.
(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 oxidizing agents). For example, osmium shows an oxidation state of + 6 in O₃F₆ and vanadium shows an oxidation state of + 5 in V₂O₅
(iii) Oxometal unions have highest oxidation state, e.g., Cr in Cr₂O has an oxidation state of +6 whereas Mn 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 oxidizing element.