NCERT Solutions For Class 12 Chemistry Chapter 4 The d and f block elements

Page 95

Question 4.2 In the series $Sc(z=21)$ to $Zn(z=30),$ the enthalpy of atomisation of zinc is the lowest, i.e., $126kJmo{l}^{-1}$ . Why?

Answer :

The enthalpy of atomisation of zinc is lowest due to the absence of an unpaired electron, which is responsible for metallic bonding in the elements. Therefore, the inter-atomic bonding is weak in zinc( $Zn$ ). Hence it has a low enthalpy of atomisation.

Page 97

Question 4.3 Which of the 3d series of transition metals exhibits the largest number of oxidation states and why?

Answer :

In 3d series of transition metals Manganese shows largest number of oxidation states because it has highest number of unpaired electrons in its $d$ -orbitals.So, that by removing its all electrons we get different oxidation states.
Example- $Mn{O}_2(+4),Mn{O}_4^{-}(+7),MnO(+2)$ etc.

Page 98

Question 4.4 The $E^{\ominus }(M^{2+}/M)$ value for copper is positive $(+0.34V)$ . What is possible reason for this? (Hint: consider its high ${\mathrm{\Delta }}_a{H}^{\ominus }$ and low ${\mathrm{\Delta }}_{hyd}{H}^{\ominus }$ )

Answer :

The $E^{\ominus }(M^{2+}/M)$ value for metal depends on-

• Sublimation energy
• Ionisation energy
• Hydration energy

Copper has a high value of atomisation enthalpy and low hydration energy. Thus, as a result, the overall effect is $E^{\ominus }(M^{2+}/M)$ for copper is positive.

Page 100

Question 4.5 How would you account for the irregular variation of ionisation enthalpies (first and second) in the first series of the transition elements?

Answer :

The irregular variation in ionisation enthalpies is due to the extra stability of the configuration like $d^0{d}^5{d}^{10}$ because these states are extremely stable and have high ionisation enthalpies.
In the case of chromium ( $Cr$ ) has a low 1st IE because after losing one electron it attains stable configuration ( $d^5$ ). But in the case of Zinc ( $Zn$ ), the first IE is very high, because we remove an electron from a stable configuration(3 $d^{10},4s^2$ ).
The second IE is much higher than the 1st IE. This is because it becomes difficult to remove an electron when we already did that and it already has a stable configuration (such as $d^0{d}^5{d}^{10}$ ). For example elements such as $C{r}^{+}$ and $C{u}^{+}$ the second IE is extremely high because they are already in a stable state. We know that removal of an electron from a stable state requires a lot of energy.

Page 101

Question 4.6 Why is the highest oxidation state of a metal exhibited in its oxide or fluoride only?

Answer :

Oxygen and fluorine are strong oxidising agents and both of their oxides and fluorides are highly electronegative in nature and also small in size. Because of these properties, they can oxidise the metal to its highest oxidation states.

Page 101

Question 4.7 Which is a stronger reducing agent $C{r}^{2+}$ or $F{e}^{2+}$ and why?

Answer :

Cr ⁺² is a better reducing agent as compared to Fe ⁺² , as this can be explained on the basis of standard electrode potential of Cr ⁺² (-0.41) and Fe ⁺² (+0.77).

It can also be explained on the basis of their electronic configuration achieved. Cr ⁺² obtained d ³ configuration whereas Fe ⁺² gets d ⁵ configuration upon reduction. It is known that d ³ is more stable than d ⁵ . So Cr ⁺² is a better reducing agent as compared to Fe ⁺² .

Page 103

Question 4.8 Calculate the ‘spin only’ magnetic moment of $M^{2+}{}_{(aq)}$ ion $(Z=27)$.

Answer :

Atomic number (Z)= 27
So the electronic configuration cobalt ( $Co$ ) is $3d^7,4s^2$
$M^{2+}{}_{(aq)}$ ion means, it loses its two electrons and become $d^7$ configuration. It has 3 unpaired electrons
So, $\mu =\sqrt{n(n+2)}$ , where n = no. of unpaired electron
by putting the value of n= 3
we get, $\mu =\sqrt{15}$
$\approx 4BM$

Page 105

Question 4.9 Explain why $C{u}^{+}$ ion is not stable in aqueous solutions?

Answer :

$C{u}^{+}$ ion is unstable in aq. solution and disproportionate to give $C{u}^{2+}$ and $Cu$
$2C{u}^{+}(aq)\to C{u}^{2+}(aq)+Cu(s)$
The hydration energy released during the formation of $C{u}^{2+}$ compensates the energy required to remove an electron from $d^{10}$ -configuration.

Page 113

Question 4.10 Actinoid contraction is greater from element to element than lanthanoid contraction. Why?

Answer :

Actinoid contraction is greater from element to element than lanthanoid contraction. The reason behind it is the poor shielding effect of 5 $f$ (in actinoids) orbitals than 4 $f$ orbitals( in lanthanoids). As a result, the effective nuclear charge experienced by valence electrons is more in actinoids than lanthanoid elements.

NCERT Solutions for Class 12 The d and f block elements- (Exercise Questions )

Question 4.1(i) Write down the electronic configuration of:

$(i)C{r}^{3+}$

Answer :

Chromium has atomic number 24. So, nearest noble gas element is Argon ( $Ar$ )
So electronic configuration of $(i)C{r}^{3+}$ = $[Ar{]}^{18}3d^{3}4s^0$

Question 4.1(ii) Write down the electronic configuration of:

$(ii)P{m}^{3+}$

Answer :

The atomic number of promethium is 61 and the nearest noble gas is xenon( $Xe$ )
So, atomic configuration of $P{m}^{3+}=[Xe{]}^{54}4f^4$

Question 4.1(iii) Write down the electronic configuration of:

$(iii)C{u}^{+}$

Answer :

The atomic number of copper is 29 and the previous noble element is Argon ( $Ar$ )
the electronic configuration of $C{u}^{+}=[Ar{]}^{18}3d^{10}$

Question 4.1(iv) Write down the electronic configuration of:

$(iv)C{e}^{4+}$

Answer :

The atomic number of cerium ( $Ce$ ) is 58 and the previous noble element is Xenon ( $Xe$ )
The electronic configuration of $C{e}^{4+}=[Xe{]}^{54}4f^0$

Question 4.1(v) Write down the electronic configuration of:

$(v)C{o}^{2+}$

Answer :

The atomic number of cobalt (Co) is 27 and the previous noble element is Argon ( $Ar$ )
Thus electronic configuration of $C{o}^{2+}=[Ar{]}^{18}3d^{7}4s^0$

Question 4.1(vi) Write down the electronic configuration of:

$(vi)L{u}^{2+}$

Answer :

The atomic number of lutetium is 71 and the previous noble element is Xe (xenon)
The electronic configuration of $L{u}^{2+}=[Xe{]}^{54}4f^{14}5d^{1}6s^0$

Question 4.1(vii) Write down the electronic configuration of:

$(vii)M{n}^{2+}$

Answer :

The atomic number of Manganese is 25 and the previous noble element is Ar (argon)
So, the electronic configuration of $M{n}^{2+}=[Ar{]}^{18}3d^{5}4s^0$

Question 4.1(viii) Write down the electronic configuration of:

$(viii)T{h}^{4+}$

Answer :

The atomic number of thorium (Th) is 90 and the previous noble gas element is Xenon (Xe)
So, the elelctronic configuration of $T{h}^{4+}=[Rn{]}^{86}5f^0$

Question 4.2 Why are $M{n}^{2+}$ compounds more stable than $F{e}^{2+}$ towards oxidation to their $+3$ state?

Answer :

$M{n}^{2+}=1s^2,2s^2{p}^6,3s^2{p}^6{d}^5(Halffilledd-orbital)$

$F{e}^{2+}=1s^2,2s^2{p}^6,3s^2{p}^6{d}^6$

In +2 oxidation state of manganese has more stability than +2 oxidation state of iron, it is because half-filled and fully filled d-orbitals are more stable and $M{n}^{2+}$ has half-filled electron stability Manganese ( $M{n}^{2+}$ ) has $d^5$ configuration so it wants to remain in this configuration. On the other hand, $F{e}^{2+}$ has $d^6$ configuration and after losing one electron it becomes $d^5$ configuration and attains its stability. That's why $M{n}^{2+}$ compounds more stable than $F{e}^{2+}$ towards oxidation to their $+3$ state.

Question 4.3 Explain briefly how $+2$ state becomes more and more stable in the first half of the first row transition elements with increasing atomic number?

Answer :

According to our observation, except scandium, all other elements of the first row show +2 oxidation state. On moving from Sc to Mn the atomic number increases from 21 to 25 and also the increasing number of electrons in 3d orbitals from $d^1-d^5$. When metals lose two electrons from their 4s orbital then they achieve +2 oxidation state. Since the number of d electrons in (+2) state increases from $Ti(+2)-Mn(+2)$, the stability of the +2 oxidation state increases as d-orbitals become more and more half-filled.

Mn(+2) has $d^5$ configuration, which is half filled (it makes it highly stable)

Question 4.4 To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.

Answer :

Elements of the first half of the transition series exhibit many oxidation states. manganese shows the maximum number of oxidation states (+2 to +7). The stability of +2 oxidation states increases with the increase in atomic number (as more number of electrons are filled in d-orbital). However, the $Sc$ does not exhibit +2 oxidation states, its EC is $3d^{1}4s^2$. It loses all three electrons to attain stable $d^0$ -configuration (noble gas configuration). $Ti(IV)$ and $V(+5)$ are stable for the same reason. In the case of manganese, (+2) oxidation state is very stable because of the half-filled d-electron( $d^5$ -configuration).

Question 4.5 What may be the stable oxidation state of the transition element with the following d electron configurations in the ground state of their atoms : $3d^3,3d^5,3d^{8}and3d^4?$

Answer :

$3d^3$
Vanadium (atomic number- 23)
E.C = $[Ar{]}^{18}3d^{3}4s^2$ ,
So the stable oxidation states are (+2, +3, +4, +5)

$3d^5$
Manganese (atomic number = 25)
E.C = $[Ar{]}^{18}3d^{5}4s^2$ ,
So the stable oxidation states are (+2, +4, +6, +7)

$3d^5$
Chromium (atomic number = 24)
E.C = $[Ar{]}^{18}3d^{5}4s^1$ ,
So the stable oxidation states are (+3, +4, +6)

$3d^4$
No elements have $d^4$ electronic configuration in their ground state.

Question 4.6 Name the oxometal anions of the first series of transition metals in which the metal exhibits the oxidation state equal to its group number.

Answer :

Following oxometal anions of the first series that exhibit the oxidation state equal to its group number-

1. Vanadate $(V{O}_3^{-})$
   The group number of vanadium $(V)$ is 5 and here the oxidation state is also +5
2. Chromate ion $(Cr{O}_4^{2-})$
   The group number of chromium is (VIB) and the oxidation state is +6
3. Permanganate ion $(Mn{O}_4^{-})$
   The group number of $(Mn)$ is VIIB and here the oxidation number is also +7

Question 4.7 What is lanthanoid contraction? What are the consequences of lanthanoid contraction?

Answer :

On moving along the lanthanoid series, the atomic number is gradually increased by one. It means the no. of electrons and protons of the atom is also increases by one. And because of it the effective nuclear charge increases (electrons are added in the same shell, and the nuclear attraction overcomes the interelectronic repulsion due to the addition of a proton). Also, with the increase in atomic number, the number of electrons in orbital also increases. Due to the poor shielding effect of the electrons, the effective nuclear charge experienced by an outer electron is increased, and also the attraction of the nucleus for the outermost electron is increased. As a result, there is a gradual decrease in the atomic size as an increase in atomic number. This is known as lanthanoid contraction.

Consequences of Lanthanoid contraction-

• Similarities in the properties of the second and third transition series
• Separation of lanthanoids can be possible due to LC.
• Due to LC, there is variation in the basic strength of the hydroxide of lanthanoid. (basic strength decrease from $La(OH{)}_3-Lu(OH{)}_3$ ).

Question 4.8 What are the characteristics of the transition elements and why are they called transition elements? Which of the d-block elements may not be regarded as the transition elements?

Answer :

Transition elements are those which have partially filled $d$ or $f$ orbitals. These elements lie in the $d-block$ and show transition properties between s block and p-block. Thus these are called transition elements.

$Zn,Hg,Cd$ are not considered transition elements due to the fully filled d-orbitals.

Question 4.9 In what way is the electronic configuration of the transition elements different from that of the non-transition elements?

Answer :

Transition elements have partially filled $d$ -orbitals. Thus general electronic configuration of transition elements is
$(n-1)d^{1-10}n{s}^{0-2}$

Non-transition elements either have fully filled d-orbitals or do not have d-orbitals. Therefore general electronic configuration is
$n{s}^{1-2}$ or $n{s}^{2}n{p}^{1-6}$

Question 4.10 What are the different oxidation states exhibited by the lanthanoids?

Answer :

In lanthanoid +3 oxidation states are more common. $Ln(III)$ compounds are most predominant. However, +2 and +4 oxidation are also formed by them in the solution or solid compounds.

Question 4.11(i) Explain giving reasons:

(i) Transition metals and many of their compounds show paramagnetic behaviour.

Answer :

Paramagnetism is arising due to the presence of unpaired electrons. And we know that transition metals have unpaired electrons in their -orbitals. That's why they show paramagnetic behaviour.

Question 4.11(ii) Explain giving reasons:

(ii) The enthalpies of atomisation of the transition metals are high.

Answer :

Transition metals have high effective nuclear charge and also high outermost electrons. Thus they form a very strong metallic bond and due to these, transition elements have a very high enthalpy of atomisation.

Question 4.11(iii) Explain giving reasons:

(iii) The transition metals generally form coloured compounds.

Answer :

Most of the complex transition elements are coloured. This is due to the absorption of radiation from the visible light region to excite the electrons from its one position to another position in d-orbitals. In the presence of ligands, d-orbitals split into two sets of different orbital energies. Here transition of electrons takes place and emits radiation which falls on the visible light region.

Question 4.11(iv) Explain giving reasons:

(iv) Transition metals and their many compounds act as good catalyst.

Answer :

The catalytic activity of transition metals is because of two reasons-

1. They provide a suitable surface for the reaction to occur.
2. Ability to show variable oxidation states and form complexes, transition metals are also able to form intermediate compounds and thus they give the new path, which has lower activation energy for the reaction.

Question 4.12 What are interstitial compounds? Why are such compounds well-known for transition metals?

Answer :

Transition metals contain lots of interstitial sites. These elements trap the other elements which are small in sizes such as Carbon, Hydrogen and Nitrogen in their interstitial site of the crystal lattice as a result forms interstitial compounds.

Question 4.13 How is the variability in oxidation states of transition metals different from that of the non-transition metals? Illustrate with examples.

Answer :

In transition metals, the variation of oxidation states id from +1 to the highest oxidation number, by removing all its valence electrons. Also in transition metals, the oxidation number is differed by one unit like ( $F{e}^{3+}-F{e}^{2+}$; $F{e}^{3+}-F{e}^{2+}$ ). But in non-transition elements, the oxidation states are differed by two (+2 and +4 or +3 and +5 etc.)

Question 4.14 Describe the preparation of potassium dichromate from iron chromite ore. What is the effect of increasing pH on a solution of potassium dichromate?

Answer :

potassium dichromate is obtained from the fusion of chromite ore $(FeC{r}_2{O}_4)$ with sodium and potassium carbonate in the free supply of air.

$4FeC{r}_2{O}_4+8N{a}_{2}C{O}_3+7O_2\to 8N{a}_{2}Cr{O}_4+2F{e}_2{O}_3+8C{O}_2$

Sodium chromate is filtered and acidified with sulphuric acid ( $H_{2}S{O}_4$ ) to form sodium dichromate, $2{\mathrm{Na}}_2{\mathrm{CrO}}_4+2{\mathrm{H}}^{+}\to {\mathrm{Na}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+2{\mathrm{Na}}^{+}+{\mathrm{H}}_2\mathrm{O}$ can be crystallized

$2N{a}_{2}Cr{O}_4+2H^{+}\to N{a}_{2}C{r}_2{O}_7+2N{a}^{+}+H_{2}O$

Sodium dichromate is more soluble than potassium dichromate. So, treat the solution of dichromate with the potassium chloride( $KCl$ )

$N{a}_{2}C{r}_2{O}_7+KCl\to K_{2}C{r}_2{O}_7+2NaCl$

The chromate and dichromate are interconvertible in aqueous solution at pH 4

Structures of chromate and dichromate ion

Question 4.15(i) Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(i) iodide

Answer :

Potassium dichromate $(K_{2}C{r}_2{O}_7)$ acts as a strong oxidising agent in an acidic medium. It takes the electron to get reduced.
$(K_{2}C{r}_2{O}_7)$ oxidises iodide to iodine

$$\begin{gathered} C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O \\ 2I^{-}\to I_2+2e^{-}]\times 3 \\ ---------------------- \\ C{r}_2{O}_7^{2-}+14H^{+}+6I^{-}\to 2C{r}^{3+}+3I_2+7H_{2}O \end{gathered}$$
In the first reaction oxidation state of chromium reduced from +6 to +3

Question 4.15(ii) Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(ii) iron(II) solution

Answer :

Potassium dichromate react with $(F{e}^{2+})$ ion to produce solution of $(F{e}^{3+})$ ion and chromium reduced to +3 oxidation state.
$$\begin{gathered} C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O \\ F{e}^{2+}\to F{e}^{3+}+e^{-}]\times 6 \end{gathered}$$

----------------------------------------------------------------------------------
$C{r}_2{O}_7^{2-}+14H^{+}+F{e}^{2+}\to 2C{r}^{3+}+7H_{2}O+6F{e}^{3+}$

Question 4.15(iii) Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

$(iii)H_{2}S$

Answer :

Potassium dichromate oxidises $H_{2}S$ (hydrogen sulphide ) to sulphur (zero oxidation state)
The oxidizing action of dichromate ion is -
$$\begin{gathered} C{r}_2{O}_7^{2-}+14H^{+}+6e^{-}\to 2C{r}^{3+}+7H_{2}O \\ {H}_{2}S\to S+2H^{+}+2e^{-}]\times 3 \end{gathered}$$
---------------------------------------------------------------------------
$C{r}_2{O}_7^{2-}+14H^{+}+3H_{2}S\to 2C{r}^{3+}+3S+7H_{2}O$

Question 4.16(i) Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with

(i) iron

Write the ionic equations for the reactions.

Answer :

Potassium permanganate can be prepared from the fusion of pyrolusite ore( $Mn{O}_2$ ) with an alkali metal hydroxide and an oxidising agent (like $KN{O}_3$ ). This gives dark green $K_{2}Mn{O}_4$ . It disproportionates in an acidic or neutral medium to give permanganate.

$$\begin{gathered} 2Mn{O}_2+4KOH+O_2\to 2K_{2}Mn{O}_4+2H_{2}O \\ 3Mn{O}_4^{2-}+4H^{+}\to 2Mn{O}_4^{-}+Mn{O}_2+H_{2}O \end{gathered}$$

(i)Acidified permanganate ion reacts with iron-

$$\begin{gathered} Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O \\ F{e}^{2+}\to F{e}^{3+}+e^{-}]\times 5 \end{gathered}$$
-------------------------------------------------------------------------------
$Mn{O}_4^{-}+5F{e}^{2+}+8H^{+}\to M{n}^{2+}+5F{e}^{3+}+4H_{2}O$

Question 4.16(ii) Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with

$(ii)S{O}_2$

Write the ionic equations for the reaction.

Answer :

Reaction of acidified permanganate solution with sulphur dioxide ( $S{O}_2$ ). It oxidizes the $S{O}_2$ to sulphuric acid ( $H_{2}S{O}_4$ )
Here are the reactions-

$$\begin{gathered} Mn{O}_4^{2-}+6H^{+}+5e^{-}\to M{n}^{2+}+3H_{2}O]\times 2 \\ S{O}_2+2H_{2}O+O_2\to 4H^{+}+2S{O}_4^{2-}+2e^{-}]\times 5 \end{gathered}$$
------------------------------------------------------------------------------------------------
$2Mn{O}_4^{2-}+10S{O}_2+4H_{2}O+5O_2\to 2M{n}^{2+}+10S{O}_4^{2-}+8H^{+}$

Question 4.16(iii) Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with

(iii) oxalic acid

Write the ionic equations for the reactions.

Answer :

When acidified permanganate solution react with oxalic acid ( $H_2{C}_2{O}_4$ ) it converts oxalic acid into carbon dioxide ( $C{O}_2$ )
Here are the reactions-

$$\begin{gathered} Mn{O}_4^{-}+8H^{+}+5e^{-}\to M{n}^{2+}+4H_{2}O]\times 2 \\ {C}_2{O}_4^{2-}\to 2C{O}_2+2e^{-}]\times 5 \end{gathered}$$
--------------------------------------------------------------------------------------------------
$2Mn{O}_4^{2-}+5C_2{O}_4^{2-}+16H^{+}\to 2M{n}^{2+}+10C{O}_2+8H_{2}O$ overall reaction

Question 4.17(i) For $M^{2+}/M$ and $M^{3+}/M^{2+}$ systems the $E^{\ominus }$ values for some metals are as follows:

$$\begin{gathered} C{r}^{2+}/Cr-0.9V \\ M{n}^{2+}/Mn-1.2V \\ F{e}^{2+}/Fe-0.4V \end{gathered}$$

$$\begin{gathered} C{r}^{3+}/C{r}^{2+}-0.4V \\ M{n}^{3+}/M{n}^{2+}+1.5V \\ F{e}^{3+}/F{e}^{2+}+0.8V \end{gathered}$$

Use this data to comment upon:
(i) the stability of $F{e}^{3+}$ in acid solution as compared to that of $C{r}^{3+}$ or $M{n}^{3+}$

Answer :

The $E^{\mathrm{\Theta }}$ value of $F{e}^{3+}/F{e}^{2+}$ is higher than that of $C{r}^{3+}/C{r}^{2+}$ but less than that of $M{n}^{3+}/M{n}^{2+}$ . So, the reduction of ferric ion ( $F{e}^{3+}$ ) to ferrous ion( $F{e}^{2+}$ ) is easier than $M{n}^{3+}/M{n}^{2+}$ but as not easy as $C{r}^{3+}/C{r}^{2+}$ . Hence ferric ion is more stable than manganese ion( $M{n}^{3+}$ ), but less stable than chromium ion( $C{r}^{3+}$ ).

The order of relative stabilities of different ions is-

$M{n}^{3+}<F{e}^{3+}<C{r}^{3+}$

Question 4.17(ii) For $M^{2+}/M$ and $M^{3+}/M^{2+}$ systems the $E^{\ominus }$ values for some metals are as follows:

$$\begin{gathered} C{r}^{2+}/Cr-0.9V \\ M{n}^{2+}/Mn-1.2V \\ F{e}^{2+}/Fe-0.4V \end{gathered}$$

$$\begin{gathered} C{r}^3/C{r}^{2+}-0.4V \\ M{n}^{3+}/M{n}^{2+}+1.5V \\ F{e}^{3+}/F{e}^{2+}+0.8V \end{gathered}$$

Use this data to comment upon:

(ii) the ease with which iron can be oxidised as compared to a similar process for either chromium or manganese metal.

Answer :

From the values of $E^o$, the order of oxidation of the given metal to the divalent cation is-

$Mn>Cr>Fe$

Question 4.18(i) Predict which of the following, will be coloured in aqueous solution?

$T{i}^{3+},V^{3+},C{u}^{+},S{c}^{3+},M{n}^{3+},F{e}^{3+}andC{o}^{2+}$

Answer :

Ions which have incomplete d-orbital, they are able to do $d-d$ transition, which is responsible for colour. And those which has vacant d-orbitals or complete $d$ -orbitals are colourless

From the table, we notice that $S{c}^{3+}$ and $C{u}^{+}$ have $3d^0$ and $3d^{10}$ configuration, so their aqueous solutions are colourless. All others are coloured in an aqueous medium.

Question 4.18(ii) Predict for the following, will be coloured in aqueous solution? Give reason.

$(ii)V^{3+}$

Answer:

Yes, $V^{3+}$ (vanadium) ions have coloured aqueous solution because vanadium has two electrons in its $d$ -orbitals, as a result d-d transition will occur and which is responsible for the colour of the solution.

Question 4.18(iii) Predict for the following, will be coloured in aqueous solution? Give reason.

$(iii)C{u}^{+}$

Answer :

No, $C{u}^{+}$ aqueous solution has no colour because it has fully filled d-orbitals. So, that d-d transition will not happen, which is responsible for colour.
electronic configuration of $C{u}^{+}$ = $[Ar{]}^{18}3d^{10}4s^0$

Question 4.18(iv) Predict for the following, will be coloured in aqueous solution? Give reason.

$(iv)S{c}^{3+}$

Answer :

No, the aqueous solution of $S{c}^{3+}$ ion will have no colour because it has empty d-orbitals. Thus the d-d transition will ot happen (due absence of electrons), which is responsible for colour.
The electronic configuration of $S{c}^{3+}$ $=[Ar{]}^{18}3d^{0}4s^0$

Question 4.18(v) Predict for the following, will be coloured in aqueous solution? Give reason.

$(v)M{n}^{2+}$

Answer :

Yes, the aqueous solution of $M{n}^{2+}$ (manganese ion) will be coloured due to half-filled electron in its $d$ -orbitals( $d^5$ ) and because of that d-d transition will occurs, which is responsible for colour.
The electronic configuration of $M{n}^{2+}$ $=[Ar{]}^{18}{d}^{5}4s^0$

Question 4.18(vi) Predict for the following, will be coloured in aqueous solution? Give reason.

$(vi)F{e}^{3+}$

Answer :

Yes, the aqueous solution of $F{e}^{3+}$ (ferric ion) will be coloured due to half-filled electron in it $d$ -orbitals( $d^5$ ) and because of that d-d transition will occurs, which is responsible for the colour
electronic configuration of $F{e}^{3+}$ $=[Ar{]}^{18}3d^{5}4s^0$

Question 4.18(vii) Predict for the following, will be coloured in aqueous solution? Give reason.

$(vii)C{o}^{2+}$

Answer :

Yes, the aqueous solution of $C{o}^{2+}$ (ferric ion) will be coloured due to the presence of an electron in it $d$ -orbitals( $d^7$ ) and because of that d-d transition will occurs, which is responsible for colour
electronic configuration of $C{o}^{2+}$ $=[Ar{]}^{18}{d}^{7}4s^0$

Question 4.19 Compare the stability of +2 oxidation state for the elements of the first transition series.

Answer :

According to our observation, except scandium, all other elements of the first row shows +2 oxidation state. On moving from $Sc$ to $Mn$ the number of oxidation states increases but from $Mn$ to $Zn$ number of oxidation states decreases due to a decrease in unpaired electrons. The stability of +2 oxidation state increases on moving from $Sc$ to $Zn$ due to the increase in the difficulty level of removal of the third electron from d -orbital.

Question 4.20(i) Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(i) electronic configuration

Answer :

The general electronic configuration of actinoids series is $[Rn{]}^{86}5f^{1-14}6d^{0-1}7s^2$ and that for lanthanoids are $[Xe{]}^{54}4f^{1-14}5d^{0-1}6s^2$ . 5 $f$ orbitals do not deeply participate in bonding to a large extent.

Question 4.20(ii) Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(ii) atomic and ionic sizes

Answer :

Similar to lanthanoids, actinoids also show actinoid contraction. But the contraction is greater in actinoids because of poor shielding effects of 5f orbitals

Question 4.20(iii) Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(iii) oxidation state

Answer :

The principal oxidation state of lanthanoids is +3, but sometimes it also shows +2 and +4 oxidation states. This is due to the extra stability of fully-filled and half-filled orbitals.
Actinoids have a greater range of oxidation states due to comparable energies of and it also have a principal oxidation state is +3 but have more compounds in +3 oxidation states than lanthanoids.

Question 4.20(iv) Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(iv) chemical reactivity.

Answer :

In the lanthanoid series, an earlier member of the series is more reactive, and that is comparable to. With an increase in atomic number, lanthanoids start behaving similarly to aluminium.
Actinoids are highly reactive metals, especially when they are finally divided. When we add them into the water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoid metals

Question 4.21(i) How would you account for the following:

(i) Of the $d^4$ species, $C{r}^{2+}$ is strongly reducing while manganese(III) is strongly oxidising.

Answer :

$C{r}^{2+}$ is strongly reducing in nature. It has $d^4$ configuration. By losing one electron it gets oxidised to $C{r}^{3+}$ (electronic configuration $d^3$ ) which can be written as $t_{2g}^3$ and it is a more stable configuration. On the other hand, $M{n}^{3+}$ has also $d^4$ configuration by accepting one electron it gets reduced and acts as a strongly oxidising agent(electronic configuration $d^5$ ). Thus it is extra stable due to half-filled with d-orbital.

Question 4.21(ii) How would you account for the following:

(ii) Cobalt(II) is stable in aqueous solution but in the presence of complexing reagents, it is easily oxidised.

Answer :

Cobalt (II) is more stable in aq. solution but in the presence of strong field ligand complexing agents, it gets oxidised to Cobalt (III). Though the third ionisation energy of $Co$ is high but the CFSE ( crystal field stabilisation energy ) is very high in the presence of a strong field ligand which overcomes the ionisation energy.

Question 4.21(iii) How would you account for the following:

(iii) The $d^1$ configuration is very unstable in ions.

Answer :

The $d^1$ configuration is very unstable in ions because after losing one more electron it attains stable $d^0$ configuration.

Question 4.22 What is meant by ‘disproportionation’? Give two examples of disproportionation reactions in aqueous solution.

Answer :

In a chemical reaction a substance gets oxidised as well as reduced simultaneously is called a disproportionation reaction. For example-

• $3Cr{O}_4^{3-}(V)+8H^{+}\to 2Cr{O}_4^{2-}(VI)+C{r}^{3+}(III)+4H_{2}O$
• $3Mn{O}_4^{2-}(VI)+4H^{+}\to 2Mn{O}_4^{-}(VII)+Mn{O}_2(IV)+2H_{2}O$

Question 4.23 Which metal in the first series of transition metals exhibits +1 oxidation state most frequently and why?

Answer :

In the first transition series, Cu (copper) exhibits +1 oxidation states most frequently. This is because $C{u}^{+}$ has stable electronic configuration of $[Ar]3d^{10}$ . The fully filled d-orbital makes it highly stable.

Question 4.24(i) Calculate the number of unpaired electrons in the following gaseous ions:

$(i)M{n}^{3+}$

Answer :

The number of unpaired electron in $M{n}^{3+}$ is 4
$M{n}^{3+}(Z=25)=\left[Ar\right]3d^4$
After losing 3 electrons, Mn has 4 electrons left.

Question 4.24(ii) Calculate the number of unpaired electrons in the following gaseous ions:

$(ii)C{r}^{3+}$

Answer :

Electronic configuration of chromium is $Cr=3d^{5}4s^1$ . The number of unpaired electron in $C{r}^{3+}$ is 3
$C{r}^{3+}(Z=24)=\left[Ar\right]3d^3$
After losing 3 electrons, Cr has 3 electrons left d-orbital

Question 4.24(iii) Calculate the number of unpaired electrons in the following gaseous ions:

$(iii)V^{3+}$

Answer :

Electronic configuration of $V=3d^{3}4s^2$ . The number of unpaired electron in $V^{3+}$ is 2
$V^{3+}(Z=23)=\left[Ar\right]3d^2$
After losing 3 electrons, V has 2 electrons left d-orbital

Question 4.24(iv) Calculate the number of unpaired electrons in the following gaseous ions:

$(iv)T{i}^{3+}$

Answer :

Electronic configuration of $Ti=3d^{2}4s^2$ . The number of unpaired electron in $T{i}^{3+}$ is 1
$T{i}^{3+}(Z=22)=\left[Ar\right]3d^1$
After losing 3 electrons, Ti has 1 electron left d-orbital

Question 4.24 Which one of these is the most stable in aqueous solution?

$$\begin{gathered} (i)M{n}^{3+} \\ (ii)C{r}^{3+} \\ (iii)V^{3+} \\ (iv)T{i}^{3+} \end{gathered}$$

Answer :

$C{r}^{3+}$ is the most stable in the aqueous solution because it attains the $t_{2g}^3$ configuration, which is a stable $d-$ configuration.
Electronic configuration of $C{r}^{3+}$ = $[Ar]3d^{3}4s^0$

Question 4.25(i) Give examples and suggest reasons for the following feature of the transition metal chemistry:

(i) The lowest oxide of transition metal is basic, and the highest is amphoteric/acidic.

Answer :

The lowest oxidation states of transition metals are basic because some of their valence electrons do not participate in bonding. Thus they have free electrons, which they can donate and act as a base. In the higher oxide of transition metals, valence electrons of their participate in bonding, so they are unavailable. But they can accept electrons and behave as an acid. For example $MnO$ (+2)- behave as a base and $M{n}_2{O}_7$ (+7)behave as an acid.

Question 4.25(ii) Give examples and suggest reasons for the following feature of the transition metal chemistry:

(ii) A transition metal exhibits the highest oxidation state in oxides and fluorides.

Answer :

Oxygen and fluorine are a strong oxidizing agent because of their small in size and high electronegativity. So, they help transition metals to exhibit the highest oxidation states. Examples of oxides and fluorides of transition metals are $Os{F}_6(+6)$ and $V_2{O}_5(+5)$

Question 4.25(iii) Give examples and suggest reasons for the following feature of the transition metal chemistry:

(iii) The highest oxidation state is exhibited in oxoanions of a metal.

Answer :

Oxygen is a strong oxidizing agent because of its small in size and high electronegativity. Thus oxo-anions of metals show the highest oxidation state.
For example- $KMn{O}_4$, here manganese shows +4 oxidation state.

Question 4.26(i) Indicate the steps in the preparation of:

(i) $K_{2}C{r}_2{O}_7$ from chromite ore .

Answer :

(i) Potassium dichromate is obtained from the fusion of chromite ore $(FeC{r}_2{O}_4)$ with sodium and potassium carbonate in the free supply of air.

$4FeC{r}_2{O}_4+8N{a}_{2}C{O}_3+7O_2\to 8N{a}_{2}Cr{O}_4+2F{e}_2{O}_3+8C{O}_2$

(ii) Sodium chromate is filtered and acidified with sulphuric acid ( $H_{2}S{O}_4$ ) to form sodium dichromate, $(N{a}_{2}C{r}_2{O}_7.2H_{2}O)$ can be crystallized