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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

Edited By Shivam Jadoun | Updated on Aug 16, 2022 12:37 PM IST | #CBSE Class 12th

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 in chapter 8 NCERT Class 12 Chemistry solutions. The NCERT Class 12 Chemistry Chapter 8 solutions 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. Also, check NCERT solutions for class 12 other subjects.

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 elements of the f -block are those in which the 4f and 5f orbitals are filled. In NCERT Solutions for d and f block elements Class 12 Chemistry Chapter 8, you will find all the topic-wise questions given in NCERT Class 12 Chemistry book. These exercise-wise NCERT solutions Class 12 Chemistry Chapter 8 will help you to learn better.

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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 NCERT solutions for class 12 chemistry chapter 8 in this article.

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Exercise-wise NCERT Solutions for Class 12 Chemistry Chapter 8 'The d and f Block Elements'

Below, we have provided the NCERT Class 12 Chemistry solutions for exercises 8.1 to 8.10 of Chapter 8:


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.

Question 8.2 In the series Sc (z=21) to Zn (z=30), the enthalpy of atomisation of zinc is the lowest, i.e., 126 \; kJ \; mol^{-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.

Question 8.3 Which of the 3d series of the 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- MnO_{2}(+4),\:MnO_{4}^{-}(+7),\:MnO(+2) etc.

Question 8.4 The E^{\ominus }(M^{2+}/M) value for copper is positive (+0.34V) . What is possible reason for this? (Hint: consider its high \Delta _{a}H^{\ominus } and low \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.

Question 8.5 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 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 already in a stable state. And we know that removal of an electron from a stable state requires a lot of energy.

Question 8.6 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.

Question 8.7 Which is a stronger reducing agent Cr^{2+} or Fe^{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 .

Question 8.8 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. 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

Question 8.9 Explain why Cu^{+} ion is not stable in aqueous solutions?

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.

Question 8.10 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 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}

Question 8.2 Why are Mn^{2+} compounds more stable than Fe^{2+} towards oxidation to their +3 state?

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.

Question 8.3 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 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)

Question 8.4 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 half-filled d-electron( d^{5} -configuration).

Question 8.5 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} \; and\; 3d^{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 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.

Question 8.6 Name the oxometal anions of the first series of the transition metals in which the metal exhibits the oxidation state equal to its group number.

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

Question 8.7 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 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} ).

Question 8.8 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 as transition elements due to the fully filled d-orbitals.

Question 8.9 In what way is the electronic configuration of the transition elements different from that of the non transition elements?

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}

Question 8.10 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 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.

Question 8.12 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 8.13 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 ( Fe^{3+}-Fe^{2+} ; Fe^{3+}-Fe^{2+} ). But in non-transition elements, the oxidation states are differed by two (+2 and +4 or +3 and +5 etc.)

Question 8.14 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 (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, 2Na_{2}CrO_{4}+2H^{+}\rightarrow Na_{2}Cr_{2}O_{7}+2Na^{+}+H_{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
15948971176621594897115588

Question 8.15(i) Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(i) iodide

Answer :

Potassium dichromate (K_{2}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

Question 8.15(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 (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+}

Question 8.15(iii) Describe the oxidising action of potassium dichromate and write the ionic equations for its reaction with:

(iii)\; H_{2}S

Answer :

Potassium dichromate 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

Question 8.16(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( 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\\

Question 8.16(ii) Describe the preparation of potassium permanganate. How does the acidified permanganate solution react with

(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^{+}

Question 8.16(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 ( 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

Question 8.17(i) For M^{2+}/M and M^{3+}/M^{2+} systems the E^{\ominus } values for some metals are as follows:

\\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+}<Fe^{3+}<Cr^{3+}

Question 8.17(ii) For M^{2+}/M and M^{3+}/M^{2+} systems the E^{\ominus } values for some metals are as follows:

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

Mn>Cr>Fe

Question 8.18(i) Predict which of the following, will be coloured in aqueous solution?

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.

Question 8.18(ii) Predict for the following, will be coloured in aqueous solution? Give reason.

(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.

Question 8.18(iii) Predict for the following, will be coloured in aqueous solution? Give reason.

(iii)\; Cu^{+}

Answer :

No, Cu^{+} aqueous solution has no color 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}

Question 8.18(iv) Predict for the following, will be coloured in aqueous solution? Give reason.

(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}

Question 8.18(v) Predict for the following, will be coloured in aqueous solution? Give reason.

(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}

Question 8.18(vi) Predict for the following, will be coloured in aqueous solution? Give reason.

(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}

Question 8.18(vii) Predict for the following, will be coloured in aqueous solution? Give reason.

(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}

Question 8.19 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 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





Question 8.20(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 8.20(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 shows actinoid contraction. But the contraction is greater in actinoids because of poor shielding effects of 5f orbitals

Question 8.20(iii) Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(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.

Question 8.20(iv) Compare the chemistry of actinoids with that of the lanthanoids with special reference to:

(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.

Question 8.22 What is meant by ‘disproportionation’? Give two examples of disproportionation reaction in aqueous solution.

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

Question 8.23 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 Cu^{+} has stable electronic configuration of [Ar]3d^{10} . the fully filled d-orbital makes it highly stable.

Question 8.24(i) Calculate the number of unpaired electrons in the following gaseous ions:

(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.

Question 8.24(ii) Calculate the number of unpaired electrons in the following gaseous ions:

(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

Question 8.24(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 electron, V has 2 electron left d-orbital

Question 8.24(iv) Calculate the number of unpaired electrons in the following gaseous ions:

(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

Question 8.24 Which one of these is the most stable in aqueous solution?

\\(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.
Electronic configuration of Cr^{3+} = [Ar]3d^{3}4s^{0}

Question 8.25(i) Give examples and suggest reasons for the following feature of the transition metal chemistry:

(i) The lowest oxide of transition metal is basic, 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.

Question 8.25(ii) Give examples and suggest reasons for the following feature of the transition metal chemistry:

(ii) A transition metal exhibits 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)

Question 8.25(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 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:

(i) K_{2}Cr_{2}O_{7} from chromite ore .

Answer :

(i) P otassium 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
15948970926161594897089960

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

Question 8.27 What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses .

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

Question 8.28 What are inner transition elements? Decide which of the following atomic numbers are the atomic numbers of the inner transition elements : 29, 59, 74, 95, 102, 104.

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.

Question 8.29 The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these 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.

Question 8.30 Which is the last element in the series of the actinoids? Write the electronic configuration of this element. Comment on the possible oxidation state of this element .

Answer :

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.

Question 8.31 Use Hund’s rule to derive the electronic configuration of Ce^{3+} ion, and calculate its magnetic moment on the basis of ‘spin-only’ formula.

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 \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

Question 8.32 Name the members of the lanthanoid series which exhibit +4 oxidation states and those which exhibit +2 oxidation states. Try to correlate this type of behaviour with the electronic configurations of these elements.

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}

Question 8.34 Write the electronic configurations of the elements with the atomic numbers 61, 91, 101, and 109.

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}

Question 8.35(i) Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(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) = 4d 5 5s 1 , Tc (43) = 4d 6 5s 1 , Ru (44) = 4d 7 5s 1 , Rh (45) = 4d 8 5s 1, Pd (46) = 4d 10 5s 0 , Ag (47) = 4d 10 5s 1

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

W (74) = 5d 4 6s 2 , Pt (78) = 5d 9 6s 1


Question 8.35(ii) Compare the general characteristics of the first series of the transition metalswith those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(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.

Question 8.35(iii) Compare the general characteristics of the first series of the transition metalswith those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(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.

Question 8.35(iv) Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

(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.

Question 8.36 Write down the number of 3d electrons in each of the following ions: Ti^{2+}, V^{2+}, Cr^{3+}, Mn^{2+}, Fe^{2+}, Fe^{3+}, Co^{2+}, Ni^{2+} and\; \; Cu^{2+} . Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).

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}


Question 8.37 Comment on the statement that elements of the first transition series possess many properties different from those of heavier transition elements.

Answer :

E lements 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.

Question 8.38 What can be inferred from the magnetic moment values of the following complex species ?

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.

More about NCERT Solutions for Class 12 Chemistry Chapter 8

It is an important chapter because it holds 5 marks out of 70 in the CBSE Board exam. The NCERT Solutions for Class 12 Chemistry Chapter 8 PDF are available for other classes and other subjects as well which are going to help you in competitive exams like JEE, NEET, BITS, VITEEE and KVPY etc.

In NCERT Class 12 Chemistry chapter 8 solutions, there are eight sub-topics that cover essential concepts of d and f-block elements. 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 NCERT questions. By referring to the Class 12 Chemistry Chapter 8 NCERT solutions, students can easily understand and solve questions well enough before their examination.

Topics and Sub-topics of NCERT Solutions for 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
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  • Revision is a must, and it will be easy with the help of detailed NCERT Class 12 Chemistry chapter 8 solutions.

  • With these solutions, students will be able to practice questions easily.

Also, check NCERT Exemplar Class 12 Solutions

Also Check NCERT Books and NCERT Syllabus here:

Frequently Asked Question (FAQs)

1. What are the important topics of this chapter?

Below are the some important topics of this chapter:

  • Variable oxidation states of transition elements 

  • Formation of coloured ions by transition metal ions 

  • Lanthanoid contraction 

  • Oxidation states of actinides

2. Where can I find complete solutions for NCERT Class 12 Chemistry?

Click on the link to get all NCERT Solutions. Solutions for Class 6 to 12 are available for Science And Mathematics.

3. What is the weightage of NCERT Class 12 Chemistry chapter 8 in JEE Mains?

1 or 2 questions can be expected for JEE Main from the NCERT syllabus Chemistry chapter d and f Block elements Class 12.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

hello mahima,

If you have uploaded screenshot of your 12th board result taken from CBSE official website,there won,t be a problem with that.If the screenshot that you have uploaded is clear and legible. It should display your name, roll number, marks obtained, and any other relevant details in a readable forma.ALSO, the screenshot clearly show it is from the official CBSE results portal.

hope this helps.

Hello Akash,

If you are looking for important questions of class 12th then I would like to suggest you to go with previous year questions of that particular board. You can go with last 5-10 years of PYQs so and after going through all the questions you will have a clear idea about the type and level of questions that are being asked and it will help you to boost your class 12th board preparation.

You can get the Previous Year Questions (PYQs) on the official website of the respective board.

I hope this answer helps you. If you have more queries then feel free to share your questions with us we will be happy to assist you.

Thank you and wishing you all the best for your bright future.

Hello student,

If you are planning to appear again for class 12th board exam with PCMB as a private candidate here is the right information you need:

  • No school admission needed! Register directly with CBSE. (But if you want to attend the school then you can take admission in any private school of your choice but it will be waste of money)
  • You have to appear for the 2025 12th board exams.
  • Registration for class 12th board exam starts around September 2024 (check CBSE website for exact dates).
  • Aim to register before late October to avoid extra fees.
  • Schools might not offer classes for private students, so focus on self-study or coaching.

Remember , these are tentative dates based on last year. Keep an eye on the CBSE website ( https://www.cbse.gov.in/ ) for the accurate and official announcement.

I hope this answer helps you. If you have more queries then feel free to share your questions with us, we will be happy to help you.

Good luck with your studies!

Hi there,

Yes you are obviously eligible to participate  in the jee mains and advance examination in 2025. Since it was your first attempt in the year 2024, you can certainly take a one year gap , that is a drop year and can give the exam in the next academic year.


For appearing in the exam in 2025, you only have to give a gap certificate to confirm your participation in the upcoming exam. You can further read more details regarding the gap year and the documents required to further participate in the examination.


Hope this resolves your query.

Hi there,

Hope you are doing fine

Yes you are certainly eligible for giving the jee exam in the year 2025. You must pass the maths exam with at least 75% criteria as required by jee and provide the marksheet and the passing certificate while registering for the exam.


Pursuing maths as an additional subject while taking biology as your main subject does not offer any hindrance in you appearing for the jee examination. It is indeed an privilege to pursue both maths and biology as the subjects and prepare for the same.

There will be no issue in filling the form while registering for the exam as it will only require your basic details and marksheet which you can provide by attaching the marksheet of maths also. Also, a detailed roadmap is also available on the official websites on how to fill the registration form. So you can fill the form easily.


Hope this resolves your query.

View All

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

Option 1)

0.34\; J

Option 2)

0.16\; J

Option 3)

1.00\; J

Option 4)

0.67\; J

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 :

Option 1)

2.45×10−3 kg

Option 2)

 6.45×10−3 kg

Option 3)

 9.89×10−3 kg

Option 4)

12.89×10−3 kg

 

An athlete in the olympic games covers a distance of 100 m in 10 s. His kinetic energy can be estimated to be in the range

Option 1)

2,000 \; J - 5,000\; J

Option 2)

200 \, \, J - 500 \, \, J

Option 3)

2\times 10^{5}J-3\times 10^{5}J

Option 4)

20,000 \, \, J - 50,000 \, \, J

A particle is projected at 600   to the horizontal with a kinetic energy K. The kinetic energy at the highest point

Option 1)

K/2\,

Option 2)

\; K\;

Option 3)

zero\;

Option 4)

K/4

In the reaction,

2Al_{(s)}+6HCL_{(aq)}\rightarrow 2Al^{3+}\, _{(aq)}+6Cl^{-}\, _{(aq)}+3H_{2(g)}

Option 1)

11.2\, L\, H_{2(g)}  at STP  is produced for every mole HCL_{(aq)}  consumed

Option 2)

6L\, HCl_{(aq)}  is consumed for ever 3L\, H_{2(g)}      produced

Option 3)

33.6 L\, H_{2(g)} is produced regardless of temperature and pressure for every mole Al that reacts

Option 4)

67.2\, L\, H_{2(g)} at STP is produced for every mole Al that reacts .

How many moles of magnesium phosphate, Mg_{3}(PO_{4})_{2} will contain 0.25 mole of oxygen atoms?

Option 1)

0.02

Option 2)

3.125 × 10-2

Option 3)

1.25 × 10-2

Option 4)

2.5 × 10-2

If we consider that 1/6, in place of 1/12, mass of carbon atom is taken to be the relative atomic mass unit, the mass of one mole of a substance will

Option 1)

decrease twice

Option 2)

increase two fold

Option 3)

remain unchanged

Option 4)

be a function of the molecular mass of the substance.

With increase of temperature, which of these changes?

Option 1)

Molality

Option 2)

Weight fraction of solute

Option 3)

Fraction of solute present in water

Option 4)

Mole fraction.

Number of atoms in 558.5 gram Fe (at. wt.of Fe = 55.85 g mol-1) is

Option 1)

twice that in 60 g carbon

Option 2)

6.023 × 1022

Option 3)

half that in 8 g He

Option 4)

558.5 × 6.023 × 1023

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

Option 1)

less than 3

Option 2)

more than 3 but less than 6

Option 3)

more than 6 but less than 9

Option 4)

more than 9

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