Did you know the middle of the periodic table has special metals known as d and f-block elements that are used in industries, colorful fireworks and important chemical reactions? This chapter features transition metals known as d-block elements and the inner transition metals known as f-block elements, as well as the principles and theories that govern their behaviour.The important topics like electronegativity, ionic size and colour in transition metal complexes are all discussed in this chapter.
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NCERT Solutions for Class 12 Chemistry Chapter 4: Download PDF
NCERT Solutions of Class 12 Chemistry Chapter 4 ( Intext Questions )
NCERT Solutions for Class 12 Chemistry Chapter 4 (Exercise Questions )
NCERT Class 12 Chemistry Chapter 4: Higher Order Thinking Skill (HOTS) Questions
Approach to Solve Questions of Class 12 Chemistry Chapter 4
Topics and subtopics covered in the NCERT Textbook Class 12 Chemistry Chapter 4
What Extra Should Students Study Beyond the NCERT for JEE?
What Students Learn from NCERT Solutions for Class 12 Chemistry Chapter 4 The d- and f-Block Elements
NCERT Solutions Class 12 Chemistry Chapter-Wise
NCERT Solutions for Class 12 Subject-wise
NCERT Exemplar Solutions Class 12 Subject-Wise
NCERT Books and NCERT Syllabus
NCERT Solutions For Class 12 Chemistry Chapter 8
The NCERT Solutions for Class 12 Chemistry are designed to offer a systematic approach to help students develop a clear understanding of critical concepts through a series of solved examples and conceptual explanations. These solutions also provide a valuable resource to enhance performance in competitive exams like NEET, JEE Mains, etc. The higher-order thinking skills questions are added in this article to improve your analytical thinking. We have also added some important points that will help you build a good approach for NCERT Solutions.
NCERT Solutions for Class 12 Chemistry Chapter 4: Download PDF
Students can download the d and f block elements ncert solutions pdf for free from the download icon given below. These solutions of NCERT are designed to help you understand the fundamental concepts and solve textbook questions with ease.
NCERT Solutions of Class 12 Chemistry Chapter 4 ( Intext Questions )
These class 12 chemistry chapter 4 d and f block elements solutions provide step by step explanations for all the in text questions, helping students grasp the key concepts with clarity and confidence.
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 gets removed. As a result, the d-orbital becomes incomplete( $d^{9}$ ). Hence it is a transition element.
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.
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- $MnO_{2}(+4),\:MnO_{4}^{-}(+7),\:MnO(+2)$ etc.
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.
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 $Cr^{+}$ and $Cu^{+}$ 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.
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.
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.
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, $\mu = \sqrt{15}$
$\approx 4\ BM$
$Cu^{+}$ 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 released during the formation of $Cu^{2+}$ compensates the energy required to remove an electron from $d^{10}$ -configuration.
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 in lanthanoid elements.
NCERT Solutions for Class 12 Chemistry Chapter 4 (Exercise Questions )
These d and f block elements class 12 question answer explain all exercise questions in a simple and clear way. They help students understand key concepts of Chemical Kinetics and solve numerical problems effectively. Practising with these solutions improves accuracy and exam confidence.
Chromium has an atomic number 24. So, the nearest noble gas element is Argon ( $Ar$ )
So electronic configuration of $(i)\; Cr^{3+}$ = $[Ar]^{18}3d^{3}4s^{0}$
The atomic number of cobalt (Co) is 27 and the previous noble element is Argon ( $Ar$ )
Thus electronic configuration of $Co^{2+}=[Ar]^{18}3d^{7}4s^{0}$
The atomic number of lutetium is 71 and the previous noble element is Xe (xenon)
The electronic configuration of $Lu^{2+}=[Xe]^{54}4f^{14}5d^{1}6s^{0}$
The atomic number of thorium (Th) is 90 and the previous noble gas element is Xenon (Xe)
So, the elelctronic configuration of $Th^{4+}= [Rn]^{86}5f^{0}$
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.
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 theyachieve +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)
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).
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 aslanthanoid 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 $La(OH)_{3}-Lu(OH)_{3}$ ).
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.
Transition elementshave partially filled $d$ -orbitals. Thus general electronic configuration of transition elements is
$(n-1)d^{1-10}ns^{0-2}$
Non-transition elementseither have fully filledd-orbitalsor do not have d-orbitals. Therefore general electronic configuration is
$ns^{1-2}$ or $ns^{2}np^{1-6}$
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.
(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.
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.
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.
The catalytic activity of transition metals is because of two reasons-
They provide a suitable surface for the reaction to occur.
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.
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.
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 ( $Fe^{3+}-Fe^{2+}$; $F e^{3+}-F e^{2+}$ ). But in non-transition elements, the oxidation states are differed by two (+2 and +4 or +3 and +5 etc.)
Sodium chromate is filtered and acidified with sulphuric acid ( $H_{2}SO_{4}$ ) to form sodium dichromate, $2 \mathrm{Na}_2 \mathrm{CrO}_4+2 \mathrm{H}^{+} \rightarrow \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{Na}^{+}+\mathrm{H}_2 \mathrm{O}$ can be crystallized
Potassium dichromate $(K_{2}Cr_{2}O_{7})$ acts as a strong oxidising agent in an acidic medium. It takes the electron to get reduced.
$(K_{2}Cr_{2}O_{7})$ oxidises iodide to iodine
$-------\mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{I}^{-} \rightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O}$ In the first reaction oxidation state of chromium reduced from +6 to +3
Potassium dichromate react with $(Fe^{2+})$ ion to produce solution of $(Fe^{3+})$ ion and chromium reduced to +3 oxidation state.
$Cr_{2}O_{7}^{2-}+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_{2}O$
Potassium dichromate oxidises $H_{2}S$ (hydrogen sulphide ) to sulphur (zero oxidation state)
The oxidizing action of dichromate ion is -
$Cr_{2}O_{7}^{2-}+14H^{+}+6e^{-}\rightarrow 2Cr^{3+}+7H_{2}O$
$H_{2}S\rightarrow S + 2H^{+}+2e^{-}]\times 3$
---------------------------------------------------------------------------
$Cr_{2}O_{7}^{2-}+14H^{+}+3H_{2}S\rightarrow 2Cr^{3+}+3S+7H_{2}O$
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 an acidic or neutral medium to give permanganate.
Reaction of acidified permanganate solution with sulphur dioxide ( $SO_{2}$ ). It oxidizes the $SO_{2}$ to sulphuric acid ( $H_{2}SO_{4}$ )
Here are the reactions-
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 reactions-
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( $Cr^{3+}$ ).
The order of relative stabilities of different ions is-
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 an aqueous medium.
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.
No, $Cu^{+}$ 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 $Cu^{+}$ = $[Ar]^{18}3d^{10}4s^{0}$
No, the 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 electrons), which is responsible for colour.
The electronic configuration of $Sc^{3+}$ $=[Ar]^{18}3d^{0}4s^{0}$
Yes, the aqueous solution of $Mn^{2+}$ (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 $Mn^{2+}$ $=[Ar]^{18}d^{5}4s^{0}$
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 the colour
electronic configuration of $Fe^{3+}$ $=[Ar]^{18}3d^{5}4s^{0}$
Yes, the aqueous solution of $Co^{2+}$ (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 $Co^{2+}$ $=[Ar]^{18}d^{7}4s^{0}$
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.
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.
Similar to lanthanoids, actinoids also show actinoid contraction. But the contraction is greater in actinoids because of poor shielding effects of 5f orbitals
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.
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
(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 acts as a strongly oxidising agent(electronic configuration $d^{5}$ ). Thus it is extra stable due to half-filled with d-orbital.
(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.
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.
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 electrons, Cr has 3 electrons left d-orbital
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
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 electrons, Ti has 1 electron left d-orbital
$Cr^{3+}$ is the most stable in the aqueous solution because it attains the $t^{3}_{2g}$ configuration, which is a stable $d-$ configuration.
Electronic configuration of $Cr^{3+}$ = $[Ar]3d^{3}4s^{0}$
(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 $Mn_{2}O_{7}$ (+7)behave as an acid.
(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 $OsF_{6}(+6)$ 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 show the highest oxidation state.
For example- $KMnO_{4}$, here manganese shows +4 oxidation state.
(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 crystallized
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 an acidic or neutral medium to give permanganate.
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 lanthanoids is mischmetal.
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.
Lanthanoid primarily shows three oxidation states +2, +3, and +4 and out of these +3 is most common in lanthanoids. they show a 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.
The last element of the actinoid series is Lawrencium ( $Lr$ ). Its atomic number is 103. The electronic configuration of $Lr$ is $[Rn]^{86}5f^{14}6d^{1}7s^{2}$ .
The possible oxidation state of lawrencium is +3 because after losing 3 electrons, it becomes a stable molecule.
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 $\mu = \sqrt{n(n+2)}$ , where n= no. of unpaired electrons
in Cerium n = 2
So, by putting the value of n we get $\mu = \sqrt{2(2+2)}= \sqrt{8}=2.828BM$
Members of the lanthanoids which exhibit +4 oxidation states are- $Ce, Pr, Nd, Tb, D y$
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.
$\\Eu^{2+}=[Xe]4f^{7}\\ Yb^{2+}= [Xe]4f^{14}$
Electronic configurations-
In 1st, 2nd and 3rd transition metal series 3 $d$ , 4 $d$ and 5 $d$ orbitals are used respectively. In the first series copper and zinc show unusual electronic configurations.
$\\Cr = 3d^54s^1\\ Cu = 3d^{10}4s^9$
In the second transition series, different electron configurations are shown by the following metals,
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 states are quite stable. Elements of first transition series metals form stable compounds of +2 and +3 oxidation states. But the stability of +2 and +3 oxidation state decreases in the 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 the first transition series ( $[Co(NH_{3})_{6}]^{3+}, [Ti(H_{2}O)_{6}]^{3+}$ ) are stable complexes.
In all of the three transition series, the 1st ionisation energy increases from the left side to the 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.
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 : Elements of the first transition series possess many properties different from those of heavier transition elements in the following ways-
The atomic size of the 1st transition series is smaller than those of the 2nd and 3rd series elements. But due to lanthanoid contraction, the atomic size of the 2nd series elements is nearly the same as 3rd series element of the corresponding same vertical group.
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.
The enthalpy of atomisation of first series elements is lower than 2nd and 3rd series elements.
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.
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 the above compound d-orbital has one unpaired electron( $Mn^{2+} = [d^{5}]$ ), which means $CN$ is a strong 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 does 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 the $Cl$ ligand does not cause any force pairing and hence it is a weak ligand.
NCERT Class 12 Chemistry Chapter 4: Higher Order Thinking Skill (HOTS) Questions
These HOTS questions are designed to help students understand class 12 chemistry chapter 4 d and f block elements solutions beyond the basics. These NCERT Solutions for Class 12 helps develop analytical skills, critical thinking, and problem-solving abilities required for board exams and competitive tests.
Question 1. $KMnO_4$ acts as an oxidising agent in acidic medium. ‘X’ is the difference between the oxidation states of Mn in reactant and product. ‘Y’ is the number of ‘d’ electrons present in the brown red precipitate formed at the end of the acetate ion test with neutral ferric chloride. The value of X + Y is _________.
Answer:
X is the difference in oxidation state.
7 – 2 = 5
So X = 5
Question 3. Among, $\mathrm{Sc}, \mathrm{Mn}, \mathrm{Co}$ and Cu , identify the element with highest enthalpy of atomisation. The spin only magnetic moment value of that element in its +2 oxidation state is ___________ BM (in nearest integer).
Approach to Solve Questions of Class 12 Chemistry Chapter 4
The elements present in the middle of the periodic table from Group 3 to 12 are called d-block elements. The name d-blocks because the last electron enters the d-orbital of the penultimate shell. This chapter includes both theory-based understanding and application-based questions. The step by step approach for class 12 chemistry chapter 4 d and f block elements solutions are given below:
1. Understand the basics
Start by clearly understanding the electronic configuration, oxidation states and general properties of d- and f-block elements. These concepts form the foundation for most questions. These question answer will help you understand these concepts better.
2. Memorize important trends
Make your focus on periodic trends such as atomic size, ionization enthalpy, melting/boiling points and magnetic properties within the transition and inner transition elements.
3. Learn colour and complex formation
Try to remember common colored compounds and their oxidation states, especially for transition metals. You can make flowcharts or flashcards to revise it. You can learn these concepts in d and f block elements ncert notes.
4. Practice reactions and uses
Give proper attention to the important reactions like the preparation and properties of potassium dichromate and potassium permanganate, and their oxidizing behaviour in acidic, basic, and neutral media.
5. Use NCERT examples and exercises
You can refer to the NCERT solved examples and try solving in-text questions for a better understanding of the concept. Also, solve the textbook exercise questions as they are often directly asked in board exams. You can also refer to the NCERT exemplar for better learning. Practice previous year questions and solve mock tests.
Topics and subtopics covered in the NCERT Textbook Class 12 Chemistry Chapter 4
These d and f block elements ncert solutions focuses on fundamental concepts, reaction rates, and factors affecting reactions. The chapter is divided into several topics and subtopics to help students understand the subject systematically
4.1 Position in the Periodic Table
4.2 Electronic Configurations of the d-Block Elements
4.3 General Properties of the Transition Elements (d-Block)
4.3.1. Physical Properties
4.3.2. Variation in Atomic and Ionic Sizes of Transition Metals
4.3.3. Ionisation Enthalpies
4.3.4. Oxidation States
4.3.5. Trends in the M2+/M Standard ElectrodePotentials
4.3.6. Trends in the M3+/M2+ Standard Electrode Potentials
4.3.7. Trends in Stability of Higher Oxidation States
4.3.8. Chemical Reactivity and Eo Values
4.3.9. Magnetic Properties
4.3.10. Formation of Coloured Ions
4.3.11 Formation of Complex Compounds
4.3.12 Catalytic Properties
4.3.13 Formation of Interstitial Compounds
4.3.14 Alloy Formation
4.4 Some Important Compounds of Transition Elements
4.4.1 Oxides and Oxoanions of Metals
4.5 The Lanthanoids
4.5.1 Electronic Configurations
4.5.2 Atomic and Ionic Sizes
4.5.3 Oxidation States
4.5.4 General Characteristics
4.6 The Actinoids
4.6.1 Electronic Configurations
4.6.2 Ionic Sizes
4.6.3 Oxidation States
4.6.4 General Characteristics and Comparison with Lanthanoids
4.7 Some Applications of d- and f-Block Elements
What Extra Should Students Study Beyond the NCERT for JEE?
Along with class 12 chemistry chapter 4 d and f block elements question answer, students preparing for JEE should explore advanced problem solving books and practice higher level numerical questions. Here's a comparison table highlighting what to study beyond the NCERT for JEE:
What Students Learn from NCERT Solutions for Class 12 Chemistry Chapter 4 The d- and f-Block Elements
These class 12 chemistry chapter 4 d and f block elements solutions help students understand the unique properties and trends of transition and inner transition elements. Given key some points on key learning of this chapter:
Using these solutions students can understand where the d-block and f-block elements are placed and their general characteristics.
They help in learning the electron filling order, variable oxidation states, and reasons for electronic anomalies.
Trends in metallic character, atomic and ionic sizes, ionisation enthalpies, and catalytic properties are well explained in these d and f block elements ncert solutions.
Here students will learn about crystal field theory and how unpaired electrons cause colour.
These d and f block elements class 12 question answers cover topics like how transition metals form coordination complexes, decrease in ionic radii of lanthanides, similarities and differences between actinides and lanthanides, and their variable oxidation states.
NCERT Solutions Class 12 Chemistry Chapter-Wise
Along with NCERT Solutions for Class 12 Chemistry Chapter 4 The d and f block elements, here are the links to chapter-wise NCERT Class 12 solutions:
To support students in their board exam preparation as well as competitive exams like JEE and NEET, NCERT solutions for Class 12 are available subject-wise.
In NCERT Class 12 Chemistry, d-block elements are transition elements (Groups 3–12) where the last electron enters the d-orbital. f-block elements are inner transition elements where the last electron enters the f-orbital, and they are placed separately at the bottom of the periodic table.
Q: What is NCERT Solutions for Class 12 Chemistry Chapter 4 The d and f block elements?
A:
Class 12 Chemistry Chapter 4 The d and f block elements solutions are step by step answers to all the textbook questions from this chapter. They help students understand the concepts, properties, trends, and reactions of transition d-block and inner transition f-block elements clearly, making it easier to revise.
Q: Why are compounds of d-block elements often colored?
A:
The colour of d-block elements is due to the d-d transitions. When white light falls on a transition metal ion, electrons in the partially filled d-orbitals absorb specific wavelengths of light to jump to higher energy d-orbitals. The remaining unabsorbed wavelengths constitute the observed colour.
Q: What are d-block elements, and where are they located on the periodic table?
A:
D-block elements are those in which the last electron enters the d-orbital of the penultimate (n-1) shell. They are located in the middle of the periodic table, between the s-block and p-block elements, in Groups 3 to 12. They are also known as transition elements or transition metals.
Q: What are f-block elements, and where are they located on the periodic table?
A:
f-block elements are those in which the last electron enters the f-orbital of the antepenultimate (n-2) shell. They are located at the bottom of the periodic table as two separate series: the lanthanides and the actinides. They are also known as inner transition elements.
The date of 12 exam is depends on which board you belongs to . You should check the exact date of your exam by visiting the official website of your respective board.
Class 12 biology questions papers 2023-2025 are available on cbseacademic.nic.in , and other educational website. You can download PDFs of questions papers with solution for practice. For state boards, visit the official board site or trusted education portal.
Taking a drop year to reappear for the Karnataka Common Entrance Test (KCET) is a well-defined process. As a repeater, you are fully eligible to take the exam again to improve your score and secure a better rank for admissions.
The main procedure involves submitting a new application for the KCET through the official Karnataka Examinations Authority (KEA) website when registrations open for the next academic session. You must pay the required application fee and complete all formalities just like any other candidate. A significant advantage for you is that you do not need to retake your 12th board exams. Your previously secured board marks in the qualifying subjects will be used again. Your new KCET rank will be calculated by combining these existing board marks with your new score from the KCET exam. Therefore, your entire focus during this year should be on preparing thoroughly for the KCET to achieve a higher score.
For more details about the KCET Exam preparation,
CLICK HERE.
I hope this answer helps you. If you have more queries, feel free to share your questions with us, and we will be happy to assist you.
Thank you, and I wish you all the best in your bright future.
Yes, you can switch from Science in Karnataka State Board to Commerce in CBSE for 12th. You will need a Transfer Certificate from your current school and meet the CBSE school’s admission requirements. Since you haven’t studied Commerce subjects like Accountancy, Economics, and Business Studies, you may need to catch up before or during 12th. Not all CBSE schools accept direct admission to 12th from another board, so some may ask you to join Class 11 first. Make sure to check the school’s rules and plan your subject preparation.
For the 12th CBSE Hindi Medium board exam, important questions usually come from core chapters like “Madhushala”, “Jhansi ki Rani”, and “Bharat ki Khoj”.
Questions often include essay writing, letter writing, and comprehension passages. Grammar topics like Tenses, Voice Change, and Direct-Indirect Speech are frequently asked.
Students should practice poetry questions on themes and meanings. Important questions also cover summary writing and translation from Hindi to English or vice versa.
Previous years’ question papers help identify commonly asked questions.
Focus on writing practice to improve handwriting and presentation. Time management during exams is key to answering all questions effectively.
A block of mass 0.50 kg is moving with a speed of 2.00 ms-1 on a smooth surface. It strikes another mass of 1.00 kg and then they move together as a single body. The energy loss during the collision is
A person trying to lose weight by burning fat lifts a mass of 10 kg upto a height of 1 m 1000 times. Assume that the potential energy lost each time he lowers the mass is dissipated. How much fat will he use up considering the work done only when the weight is lifted up ? Fat supplies 3.8×107 J of energy per kg which is converted to mechanical energy with a 20% efficiency rate. Take g = 9.8 ms−2 :
A pulley of radius 2 m is rotated about its axis by a force F = (20t - 5t2) newton (where t is measured in seconds) applied tangentially. If the moment of inertia of the pulley about its axis of rotation is 10 kg m2 , the number of rotations made by the pulley before its direction of motion if reversed, is