# NCERT Solutions For Class 12 Chemistry Chapter 8 The d and f block elements

NCERT solutions for class 12 chemistry chapter 8 The d and f block elements -  The general and physical properties of d-block and f-block elements have been discussed here in this chapter and alongwith that solutions of NCERT class 12 chemistry chapter 8 The d and f block elements has all the answers pertaining to questions based on such properties. The d-block elements are elements of groups 3 to 12 of the periodic table. The elements of the f -block are those in which the 4f and 5f orbitals are filled. Also, you will study that these elements are formal members of group 3 from which they have been taken out to form a separate f-block of the periodic table. In NCERT solutions for class 12 chemistry chapter 8 The d and f block elements, you will find all the topic wise questions as well as the exercise questions which will make your learning easier. The d-block elements are often referred to as transition metals and f-block elements as inner transition metals. There are three series of the d block elements, 3d series (Sc to Zn), 4d series (Y to Cd) and 5d series (La to Hg, except Ce to Lu). The fourth 6d series which begins with Ac is still incomplete. The two series of the f-block elements, (4f and 5f) are known as lanthanoids and actinoids respectively. You will find all the CBSE NCERT solutions for class 12 chemistry chapter 8 The d and f block elements here in this article once you scroll down.

This chapter is not lengthy like chapter 7 the p-block elements but it is an important chapter because it holds 5 marks out of 70 in the CBSE Board exam. The NCERT solutions are available for other classes and other subjects as well which are going to help you in competitive exams like JEE, NEET, BITS, VITEE and KVPY etc. In NCERT solutions for class 12 chemistry chapter 8 The d and f block elements, there are eight sub-topics which cover essential concepts of d and f-block elements and to check your knowledge, 38 questions are given in the exercise at the end of the chapter. The NCERT solutions for class 12 chemistry chapter 8 The d and f block elements are created by subject experts to give a clear understanding of the concept used to solve the question.

In chapter 8 The d and f block elements you will first read about the occurrence, electronic configuration and general characteristics of transition elements and discuss important trends in the properties of the first row (3d) transition metals, preparation and properties of some important compounds. After this, the chapter will talk about oxidation states, electronic configurations and chemical reactivity of the inner transition metals.

## Topics and Sub-topics of NCERT Class 12 Chemistry Chapter 8 The d and f block elements-

8.1 Position in the Periodic Table

8.2 Electronic Configurations of the d-Block Elements

8.3 General Properties of the Transition Elements (d-Block)

8.4 Some Important Compounds of Transition Elements

8.5 The Lanthanoids

8.6 The Actinoids

8.7 Some Applications of d- and f-Block Elements

## Question 8.1     Silver atom has completely filled d-orbitals $(4d^{10})$ in its ground state. How can you say that it is a transition element?

Answer:

Silver atom(atomic no. = 47) has completely filled d-orbital in its ground state(4$d^{10}$). However, in +2 oxidation state, the electron of d-orbitals get removed. As a result, the d-orbital become incomplete($d^{9}$) . Hence it is a transition element.

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.

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- $\dpi{80} MnO_{2}(+4),\:MnO_{4}^{-}(+7),\:MnO(+2)$ etc.

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.

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 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$\dpi{100} d^{10},4s^{2}$).
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 $\dpi{100} d^{0}\ ,d^{5}\ ,d^{10}$). For example elements such as $Cr^{+}$and $Cu^{+}$ the second IE is extremely high because they already in a stable state. And we know that removal of an electron from a stable state requires a lot of energy.

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.

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 d3 configuration whereas Fe+2 gets d5 configuration upon reduction. It is known that d3 is more stable than d5. So Cr+2 is a better reducing agent as compared to Fe+2.

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. And 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, $\mu = \sqrt{15}$
$\approx 4\ BM$

Answer:

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

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 lanthanoids elements.

## NCERT Solutions for Class 12 Chemistry Chapter 8 The d and f block elements- Exercise Questions

Question 8.1(i)     Write down the electronic configuration of:

$(i)\; Cr^{3+}$

Answer:

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

Question 8.1(ii)     Write down the electronic configuration of:

$(ii)Pm^{3+}$

Answer:

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

Question 8.1(iii)     Write down the electronic configuration of:

$(iii)Cu^{+}$

Answer:

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

Question 8.1(iv)     Write down the electronic configuration of:

$(iv)\; Ce^{4+}$

Answer:

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

Question  8.1(v)     Write down the electronic configuration of:

$(v)\; Co^{2+}$

Answer:

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

Question 8.1(vi)     Write down the electronic configuration of:

$(vi)\; Lu^{2+}$

Answer:

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

Question 8.1(vii)     Write down the electronic configuration of:

$(vii)\; Mn^{2+}$

Answer:

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

Question 8.1(viii)     Write down the electronic configuration of:

$(viii)\; Th^{4+}$

Answer:

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

Answer:

$Mn^{2+}= 1s^2,2s^2p^6,3s^2p^6d^5\:\:\:(Half\:filled \:d-orbital)$

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

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 $Mn^{2+}$has half filled electron stability Manganese ($Mn^{2+}$) has $d^{5}$ configuration so it wants to remain in this configuration. On the other hand, $Fe^{2+}$has $d^{6}$ configuration and after losing one electron it becomes$d^{5}$ configuration and attains its stability. That's why $Mn^{2+}$ compounds more stable than $Fe^{2+}$ towards oxidation to their $+3$ state.

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 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 its 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 is becoming more and more half filled.

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

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 half-filled d-electron($d^{5}$-configuration).

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 state are (+2, +4, +6, +7)

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

$3d^{4}$
No elements has $d^{4}$  electronic configuration in their ground state.

Answer:

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

1. Vanadate $(VO_{3}^{-})$
Group number of vanadium$(V)$ is 5 and here the oxidation state is also +5

2. Chromate ion $(CrO_{4}^{2-})$
Group number of chromium is (VIB) and the oxidation state is +6

3. Permanganate ion $(MnO_{4}^{-})$
Group number of $(Mn)$ is VIIB and here the oxidation number is also +7

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 adding in the same shell, and the nuclear attraction overcomes the interelectronic repulsion due to adding 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 second and third transition series

• Separation of lanthanoid can be possible due to LC.

• Due to LC, there is variation in basic strength of hydroxide of lanthanoid. (basic strength decrease from $La(OH)_{3}-Lu(OH)_{3}$).

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 as transition elements due to the fully filled d-orbitals.

Answer:

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

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

Answer:

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

Question 8.11(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 electron. And we know that transition metals have unpaired electrons in their -orbitals. That's why they show paramagnetic behaviour.

Question 8.11(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 outer most electrons. Thus they form a very strong metallic bond and due to these, transition elements have a very high enthalpy of atomisation.

Question 8.11(iii)     Explain giving reasons:

(iii) The transition metals generally form coloured compounds.

Answer:

Most of the complex of transition elements are coloured. This is due to the absorption of radiation from 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 electron takes place and emit radiation which falls on the visible light region.

Question 8.11(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 occurs.

2. Ability to show variable oxidation states and form complexes, transition metals also able to form intermediate compounds and thus they give the new path, which has lower activation energy for the reaction.

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.

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 ($Fe^{3+}-Fe^{2+}$;$Cu^{+}-Cu^{2+}$). But in non-transition elements, the oxidation states are differed by two (+2 and +4 or +3 and +5 etc.)

Answer:

potassium dichromate is obtained from the fusion of chromite ore $(FeCr_{2}O_{4})$ with sodium and potassium carbonate in the free supply of air.

$4FeCr_{2}O_{4}+8Na_{2}CO_{3}+7O_{2}\rightarrow 8Na_{2}CrO_{4}+2Fe_{2}O_{3}+8CO_{2}$

Sodium chromate is filtered and acidified with sulphuric acid ($H_{2}SO_{4}$) to form sodium dichromate, $(Na_{2}Cr_{2}O_{7}.2H_{2}O)$ can be crystallised

$2Na_{2}CrO_{4}+2H^{+}\rightarrow Na_{2}Cr_{2}O_{7}+2Na^{+}+H_{2}O$

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

$Na_{2}Cr_{2}O_{7}+KCl\rightarrow K_{2}Cr_{2}O_{7}+2NaCl$

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

Structures of chromate and dichromate ion

(i) iodide

Answer:

Potassium dichromate $(K_{2}Cr_{2}O_{7})$ act as a strong oxidising agent in acidic medium. It takes the electron to get reduced.
$(K_{2}Cr_{2}O_{7})$ oxidises iodide to iodine

$\\Cr_{2}O^{2-}_{7}+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_{2}O\\ 2I^{-}\rightarrow I_{2}+2e^{-}]\times 3\\ ----------------------\\ Cr_{2}O^{2-}_{7}+14H^{+}+6I^{-}\rightarrow 2Cr^{3+}+3I_{2}+7H_{2}O$
In first reaction oxidation state of chromium reduced from +6 to +3

(ii) iron(II) solution

Answer:

Potassium dichromate react with $(Fe^{2+})$ ion to produce solution of  $(Fe^{3+})$ ion and chromium reduced to +3 oxidation state.
$\\Cr_{2}O_{7}^{2-}+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_{2}O\\ Fe^{2+}\rightarrow Fe^{3+}+e^{-}]\times 6$

----------------------------------------------------------------------------------
$\\Cr_{2}O_{7}^{2-}+14H^{+}+Fe^{2+}\rightarrow 2Cr^{3+}+7H_{2}O+ 6Fe^{3+}$

$(iii)\; H_{2}S$

Answer:

Potassium dichromate oxidise $H_{2}S$ (hydrogen sulphide ) to sulphur (zero oxidation state)
The oxidising action of dichromate ion is -
$\\Cr_{2}O_{7}^{2-}+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_{2}O\\ H_{2}S\rightarrow S + 2H^{+}+2e^{-}]\times 3$
---------------------------------------------------------------------------
$Cr_{2}O_{7}^{2-}+14H^{+}+3H_{2}S\rightarrow 2Cr^{3+}+3S+7H_{2}O$

(i) iron

Answer:

Potassium permanganate can be prepared from the fusion of pyrolusite ore($MnO_{2}$) with an alkali metal hydroxide and an oxidising agent (like $KNO_{3}$). This gives dark green $K_{2}MnO_{4}$. It disproportionates in acidic or neutral medium to give permanganate.

$\\2MnO_{2}+4KOH+O_{2}\rightarrow 2K_{2}MnO_{4}+2H_{2}O\\ 3MnO_{4}^{2-}+4H^{+}\rightarrow 2MnO_{4}^{-}+MnO_{2}+H_{2}O$

(i)Acidified permanganate ion reacts with iron-

$\\MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O\\ Fe^{2+}\rightarrow Fe^{3+}+e^{-}]\times 5$
-------------------------------------------------------------------------------
$MnO_{4}^{-}+5Fe^{2+}+8H^{+}\rightarrow Mn^{2+}+ 5Fe^{3+}+4H_{2}O\\$

$(ii)\; SO_{2}$

Write the ionic equations for the reaction.

Answer:

Reaction of acidified permanganate solution with sulphur dioxide ($SO_{2}$). It oxidises the $SO_{2}$ to sulphuric acid ($H_{2}SO_{4}$)
Here are the reactions-

$\\MnO_{4}^{2-}+6H^{+}+5e^{-}\rightarrow Mn^{2+}+3H_{2}O]\times 2\\ SO_{2}+2H_{2}O+O_{2}\rightarrow 4H^{+}+2SO_{4}^{2-}+2e^{-}]\times 5$
------------------------------------------------------------------------------------------------
$2MnO_{4}^{2-}+10SO_{2}+4H_{2}O+5O_{2}\rightarrow 2Mn^{2+}+10SO_{4}^{2-}+8H^{+}$

(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 ($CO_{2}$)
Here are the reacions-

$\\MnO_{4}^{-}+8H^{+}+5e^{-}\rightarrow Mn^{2+}+4H_{2}O]\times 2\\ C_{2}O_{4}^{2-}\rightarrow 2CO_{2}+2e^{-}]\times 5$
--------------------------------------------------------------------------------------------------
$2MnO_{4}^{2-}+5C_{2}O_{4}^{2-}+16H^{+}\rightarrow 2Mn^{2+}+10CO_{2}+8H_{2}O$      overall reaction

$\\Cr^{2+}/Cr\; \; \; \; \; \; \; \; \; \; \; \; -0.9V\\Mn^{2+}/Mn\; \; \; \; \; \; \; \;\; \; -1.2V\\Fe^{2+}/Fe\; \; \; \; \; \; \; \; \; \; \; \; \; -0.4V$

$\\Cr^{3+}/Cr^{2+}\; \; \; \; \; \; \; \; \; \; \; \; -0.4V\\Mn^{3+}/Mn^{2+}\; \; \; \; \; \; \; \; +1.5V\\Fe^{3+}/Fe^{2+}\; \; \; \; \; \; \; \; \; \; \; +0.8V$

Use this data to comment upon:
(i) the stability of $Fe^{3+}$ in acid solution as compared to that of $Cr^{3+}$ or $Mn^{3+}$

Answer:

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

Order of relative stablities of different ions is-

$Mn^{3+}

$\\Cr^{2+}/Cr\; \; \; \; \; \; \; \; \; \; \; \; -0.9V\\Mn^{2+}/Mn\; \; \; \; \; \; \; \;\; \; -1.2V\\Fe^{2+}/Fe\; \; \; \; \; \; \; \; \; \; \; \; \; -0.4V$

$\\Cr^{3}/Cr^{2+}\; \; \; \; \; \; \; \; \; \; \; \; -0.4V\\Mn^{3+}/Mn^{2+}\; \; \; \; \; \; \; \; +1.5V\\Fe^{3+}/Fe^{2+}\; \; \; \; \; \; \; \; \; \; \; +0.8V$

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 $\dpi{100} E^{o}$, the order of oxidation of the given metal to the divalent cation is-

$\dpi{100} Mn>Cr>Fe$

$Ti^{3+}, \:V^{3+}\:,\:Cu^+,\:Sc^{3+},\:Mn^{3+},\:Fe^{3+}\: and \:Co^{2+}$

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

 $Ti^{3+}$$=[Ar]3d^{1}$ Purple $V^{3+}$$=[Ar]3d^{1}$ green $Sc^{3+}$$=[Ar]3d^{0}$ colourless $Mn^{2+}$$=[Ar]^{18}d^{5}4s^{0}$ pink $Fe^{3+}$   $=[Ar]^{18}3d^{5}4s^{0}$ yellow $Co^{2+}$$=[Ar]^{18}d^{7}4s^{0}$ blue pink $Cu^{+}$ $=[Ar]^{18}3d^{10}4s^{0}$ colourless

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

$(ii)\; V^{3+}$

Answer:

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

$(iii)\; Cu^{+}$

Answer:

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

$(iv)\; Sc^{3+}$

Answer:

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

$(v)\; Mn^{2+}$

Answer:

Yes, the aqueous solution of $Mn^{2+}$ (manganese 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 colour.
the electronic configuration of $Mn^{2+}$$=[Ar]^{18}d^{5}4s^{0}$

$(vi)\; Fe^{3+}$

Answer:

Yes, the aqueous solution of $Fe^{3+}$ (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 colour
electronic configuration of
$Fe^{3+}$$=[Ar]^{18}3d^{5}4s^{0}$

$(vii)\; Co^{2+}$

Answer:

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

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 increase on moving from $Sc$ to $Zn$ due to increase in difficulty level of removal of the third electron from d -orbital.

 Sc Ti V Cr Mn Fe Co Ni Cu Zn +3 +2 +3 +4 +2 +3 +4 +5 +2 +3 +4 +5 +6 +2 +3 +4 +5 +6 +7 +2 +3 +4   +6 +2 +3 +4 +2 +3 +4 +1 +2 +2

(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.

(ii) atomic and ionic sizes

Answer:

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

(iii) oxidation state

Answer:

The principle oxidation state of lanthanoids are +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 has principle oxidation state is +3 but have more compounds in +3 oxidation states than lanthanoids.

(iv) chemical reactivity.

Answer:

In lanthanoid series, an earlier member of the series is more reactive, and that is comparable to . with an increase in atomic number, lanthanoids starts behaving similar 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 temperature. Alkalies have no action on these actinoids metals

Question 8.21(i) How would you account for the following:

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

Answer:

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

Question 8.21(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 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 presence of strong field ligand which overcomes the ionisation energy.

Question 8.21(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.

Answer:

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

• $3CrO_{4}^{3-}(V)+8H^{+}\rightarrow 2CrO_{4}^{2-}(VI)+Cr^{3+}(III)+4H_{2}O$

• $3MnO_{4}^{2-}(VI)+4H^{+}\rightarrow 2MnO_{4}^{-}(VII)+MnO_{2}(IV)+2H_{2}O$

Answer:

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

$(i) Mn^{3+}$

Answer:

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

$(ii)\; Cr^{3+}$

Answer:

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

$(iii)\; V^{3+}$

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 electron, V has 2 electron left d-orbital

$(iv)\; Ti^{3+}$

Answer:

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

$\\(i)Mn^{3+}\\(ii)Cr^{3+}\\(iii)V^{3+}\\(iv)Ti^{3+}$

Answer:

$Cr^{3+}$ is the most stable in the aqueous solution solution because it attains the $t^{3}_{2g}$ configuration, which is stable $d-$configurtaion.
Eelectronic configuration of $Cr^{3+}$$[Ar]3d^{3}4s^{0}$

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

Answer:

The lowest oxidation states of transition metals are basic because some of their valence electrons are not participating in bonding. Thus they have free electrons, which they can donate and act as a base. In the higher oxide of transition metals, valence electron 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 $Mn_{2}O_{7}$ (+7)behave as an acid.

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 $OsF_{6}(+6)$ and $V_{2}O_{5}(+5)$

(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 shows the highest oxidation state.
For example- $KMnO_{4}$, here manganese shows +4 oxidation state.

Question 8.26(i) Indicate the steps in the preparation of:

Answer:

(i) Potassium dichromate is obtained from the fusion of chromite ore $(FeCr_{2}O_{4})$ with sodium and potassium carbonate in the free supply of air.

$4FeCr_{2}O_{4}+8Na_{2}CO_{3}+7O_{2}\rightarrow 8Na_{2}CrO_{4}+2Fe_{2}O_{3}+8CO_{2}$

(ii) Sodium chromate is filtered and acidified with sulphuric acid ($H_{2}SO_{4}$) to form sodium dichromate, $(Na_{2}Cr_{2}O_{7}.2H_{2}O)$ can be crystallised

$2Na_{2}CrO_{4}+2H^{+}\rightarrow Na_{2}Cr_{2}O_{7}+2Na^{+}+H_{2}O$

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

$Na_{2}Cr_{2}O_{7}+KCl\rightarrow K_{2}Cr_{2}O_{7}+2NaCl$

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

Structures of chromate and dichromate ion

Question 8.26(ii) Indicate the steps in the preparation of:

(ii) $KMnO_{4}$ from pyrolusite ore.

Answer:

Potassium permanganate can be prepared from the fusion of pyrolusite ore($MnO_{2}$) with an alkali metal hydroxide and an oxidising agent (like $KNO_{3}$).

This gives dark green $K_{2}MnO_{4}$. It disproportionates in acidic or neutral medium to give permanganate.

$\\2MnO_{2}+4KOH+O_{2}\rightarrow 2K_{2}MnO_{4}+2H_{2}O\\ 3MnO_{4}^{2-}+4H^{+}\rightarrow 2MnO_{4}^{-}+MnO_{2}+H_{2}O$

Answer:

It is a solid solution of two or more elements in a metallic matrix. Alloys possess different physical properties than component materials.
An important alloy of lanthanoid is mischmetal.

uses-

• mischmetal is used in cigarettes and gas lighters

• Used in flame-throwing tanks

• It is used in tracer bullets and shells

Answer:

Inner transition metals are those in which the last electrons are filled in f-orbitals. The elements in which 4f and 5f are filled are called f block elements. 59, 95 and 102 are the inner transition elements.

Answer:

Lanthanoid primarily shows three oxidation states +2, +3, and +4 and out of these +3 is most common in lanthanoids. they show limited no. of oxidation states due to the large difference in energies of 4$f$, 5$d$ and 6$s$ orbitals. But, actinoids shows large no. of oxidation state because they have comparable energy difference in 5$f$,6$d$ and 7$s$ orbitals. For example $U$ and $Pu$ exhibits +3, +4, +5 and +6 oxidation states.

Answer:

The last element of the actinoid series is Lawrencium ($Lr$). Its atomic number is 103. The electronic configuration of $Lr$ is $\dpi{100} [Rn]^{86}5f^{14}6d^{1}7s^{2}$.
The possible oxidation state of lawrencium is +3 because after losing 3 electrons it becomes a stable molecule.

Answer:

Electronic configuration of $Ce= 1s^{2}2s^{2}2p^{6}3s^{2}3p^{6}3d^{10}4s^{2}4p^{6}4d^{10}5s^{2}5p^{6}4f^{1}5d^{1}6s^{2}$
Magnetic moment can be calculated as $\dpi{100} \mu = \sqrt{n(n+2)}$, where n= no. of unpaired electrons

in Cerium n = 2
So, by putting the value of n we get $\dpi{100} \mu = \sqrt{2(2+2)}= \sqrt{8}=2.828BM$

Answer:

Members of the lanthanoids which exhibits +4 oxidation states are- $Ce, Pr,Nd,Tb,Dy$
members who exhibit +2 oxidation states = $Nd,Sm,Eu,Tm,Yb$

After losing 4 electrons $Ce^{4+}$attains stable configuration $[Xe]$ and also the same thing happen to $Tb = [Xe]4f^{7}$

In the case of $Eu$ and $Yb$ ,  after losing two electrons they also get their stable electronic configuration.
$\\Eu^{2+}=[Xe]4f^{7}\\ Yb^{2+}= [Xe]4f^{14}$

Answer:

Atomic number = 61, Promethium
the electronic configuration is $[Xe]^{54}4f^{5}5d^{0}6s^{2}$

atomic number = 91, protactinium
the electronic configuration is $[Rn]^{86}5f^{2}6d^{1}7s^{2}$

Atomic number = 101, Mendelevium
the electronic configuration is $[Rn]^{86}5f^{13}6d^{0}7s^{2}$

Atomic number = 109, Meitnerium
the electronic configuration is $[Rn]^{86}5f^{14}6d^{7}7s^{2}$

(i) electronic configurations

Answer:

Electronic configurations-
In 1st, 2nd and 3rd transition metal series 3$d$, 4$d$ and 5$d$ orbitals are used respectively. In first series copper and zinc show unusual electronic configuration.

$\\Cr = 3d^54s^1\\ Cu = 3d^{10}4s^9$

In the second transition series different electron configuration shown by following metals,

$Mo$(42) = 4d5 5s1$Tc$(43) = 4d6 5s1$Ru$(44) = 4d7 5s1$Rh$(45) = 4d8 5s1, $Pd$(46) = 4d10 5s0$Ag$(47) = 4d10 5s1

In 3rd series there are also some metals which show this types of behaviour such as;

$W$(74) = 5d4 6s2$Pt$(78) = 5d9 6s1

(ii) oxidation states

Answer:

In each of the three transition series, the no. of oxidation state is minimum at the extremes and the highest at the middle of the row. In the first transition series, the +2 and +3 oxidation state are quite stable. Elements of first transition series metals form stable compounds of +2 and +3 oxidation state. But the stability of +2 and +3 oxidation state decreases in second and third series.
Second and third transition series metals formed complexes in which their oxidation state is high ($WCl_{6},ReF_{7}$) and in first transition series ($[Co(NH_{3})_{6}]^{3+}, [Ti(H_{2}O)_{6}]^{3+}$) are stable complexes.

(iii) ionisation enthalpies

Answer:

In all of the three transition series, the 1st ionisation energy increases from the left side to right side. But, there are some exceptions like the first ionisation enthalpies of the third transition series are more significant than those of the first and second transition series. This is happening due to the weak shielding effect of 4 electrons in the third series.
Some elements in the second series have higher first IE than elements of the same column in the first transition series. There are also elements in the 2nd transition series whose first IE are lower than those of the elements corresponding to the same vertical column in the 1st transition series.

(iv) atomic sizes.

Answer:

Generally, atomic sizes decrease from left to right across the period. In among the three transition series, the size of the second series element is bigger than that of the first transition element of the same vertical group. But the atomic size of the third transition element is nearly the same as the element of the second transition series element. This is because of Lanthanoid contraction.

Answer:

For $Ti^{2+}$ d-orvbital has two electron. So, filling of d-orbitals can be $t^{2}_{2g}$

In $V^{2+}$  d-orbital has three electron. So, the filling of d-orbital can be $t^{3}_{2g}$
Similarily

 $Cr^{3+}$(Ions) $d^{3}$ (No. of d electrons) $t^{3}_{2g}$ (Filling of d-orbitals) $Mn^{2+}$ $d^{5}$ $t^{3}_{2g}, e^{2}_{g}$ $Fe^{2+}$ $d^{6}$ $t^{4}_{2g}, e^{2}_{g}$ $Fe^{3+}$ $d^{5}$ $t^{3}_{2g}, e^{2}_{g}$ $Co^{2+}$ $d^{7}$ $t^{5}_{2g}, e^{2}_{g}$ $Ni^{2+}$ $d^{8}$ $t^{6}_{2g}, e^{2}_{g}$ $Cu^{2+}$ $d^{9}$ $t^{6}_{2g}, e^{3}_{g}$

Answer:

Elements of the first transition series possess many properties different from those of heavier transition elements in the following ways-

1. The atomic size of the 1st transition series is smaller than those of 2nd and 3rd series elements. But due to lanthanoid contraction, atomic size of the 2nd series elements are nearly the same as 3rd series element of the corresponding same vertical group.

2. In 1st transition series +2 and +3 oxidation states are more common but in the 2nd and 3rd series higher oxidation states are more common.

3. The enthalpy of atomisation of first series elements is lower than 2nd and 3rd series elements.

4. The melting and boiling point of the 1st transition series is less than that of heavier metals. This is because of strong metallic bonding in heavier metals.

Example                                    Magnetic Moment (BM)

$K_{4}[Mn(CN)_{6})$                                2.2

$[Fe(H_{2}O)_{6}]^{2+}$                                          5.3

$K_{2}[MnCl_{4}]$                                            5.9

Answer:

Magnetic moment is given as - $\mu = \sqrt{n(n+2)}$
Putting the value on n = 1, 2, 3, 4, 5  (number of unpaired electrons in d-orbital)
we get the value of $\mu$ are 1.732, 2.83, 3.87, 4.899, 5.92 respectively.

$K_{4}[Mn(CN)_{6})$
By comparing with our calculation we get the values n nearest to 1. It means, in above compound d-orbital has one unpaired electron($Mn^{2+} = [d^{5}]$), which means $CN$ is astrong field ligand that cause force pairing of the electron.

$[Fe(H_{2}O)_{6}]^{2+}$
After comparing with our calculation the nearest value of n = 4. Here iron is in +2 oxidation state ($d^{6}$ configuration). So, we can say that $H_{2}O$ is a weak field ligand, which not cause any force pairing.

$K_{2}[MnCl_{4}]$
By observing we get the nearest value of n is 5. So, in this complex Manganese has $d^{5}$ configuration. So, we conclude that $Cl$  ligand does not cause any force pairing and hence it is a weak ligand.

## NCERT Solutions Class 12 Chemistry

 Chapter 1 CBSE NCERT solutions for class 12 chapter 1 The Solid State Chapter 2 NCERT solutions for class 12 chemistry chapter 2 Solutions Chapter 3 Solutions of NCERT class 12 chemistry chapter 3 Electrochemistry Chapter 4 CBSE NCERT solutions for class 12 chemistry chapter 4 Chemical Kinetics Chapter 5 Solutions of NCERT class 12 chemistry chapter 5 Surface chemistry Chapter 6 NCERT solutions for class 12 chemistry General Principles and Processes of isolation of elements Chapter 7 CBSE NCERT solutions for class 12 chemistry chapter 7 The P-block elements Chapter 8 NCERT solutions for class 12 chemistry chapter 8 The d and f block elements Chapter 9 NCERT solutions for class 12 chemistry chapter 9 Coordination compounds Chapter 10 Solutions of NCERT class 12 chemistry chapter 10 Haloalkanes and Haloarenes Chapter 11 CBSE NCERT solutions for class 12 chemistry Alcohols, Phenols, and Ethers Chapter 12 Solutions of NCERT class 12 chemistry chapter 12 Aldehydes, Ketones and Carboxylic Acids Chapter 13 NCERT solutions for class 12 chemistry chapter 13 Amines Chapter 14 CBSE NCERT solutions for class 12 chemistry chapter 14 Biomolecules Chapter 15 Solutions of NCERT class 12 chemistry chapter 15 Polymers Chapter 16 NCERT solutions for class 12 chemistry chapter 16 Chemistry in Everyday life

## NCERT Solutions for Class 12 Subject wise

 Solutions of NCERT class 12 biology NCERT solutions for class 12 maths CBSE NCERT solutions for class 12 chemistry Solutions of NCERT class 12 physics

## Benefits of NCERT solutions for class 12 chemistry chapter 8 The d and f block elements

• The comprehensive answers given in the NCERT solutions for class 12 chemistry chapter 8 The d and f block elements will help you to understand chapter easily.

• Revision will be easy because with the help of the detailed solutions you will always remember the concepts and get very good marks in your class.

• Homework problems won't bother you anymore, all you need to do is check the detailed NCERT solutions for class 12 chemistry and you are stress-free.

If you have a doubt or question that is not available here or in any of the chapters, contact us. You will get all the answers that will help you score well in your exams.

Exams
Articles
Questions