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.

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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-

Vanadate $(VO_{3}^{-})$
Group number of vanadium $(V)$ is 5 and here the oxidation state is also +5
Chromate ion $(CrO_{4}^{2-})$
Group number of chromium is (VIB) and the oxidation state is +6
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-

They provide a suitable surface for the reaction to occurs.
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
1594897117662

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
1594897092616

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-

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.
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.

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.

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Questions related to CBSE Class 12th

Have a question related to CBSE Class 12th ?

I am interested in graphic designing and coding. how do I make a career .which stream should I go.what engineering subject to take

Hello aspirant,

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Sajal Trivedi 29 January,2024

i couldn't clear 1st compartment (which held in August 2022) cbse class 12th so can I give improvement in all subjects which to be held on March 2023. tell me a solution i need more than 75% in class 12th

hello,

Yes you can appear for the compartment paper again since CBSE gives three chances to a candidate to clear his/her exams so you still have two more attempts. However, you can appear for your improvement paper for all subjects but you cannot appear for the ones in which you have failed.

I hope this was helpful!

Good Luck

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Upasana Barman 24 August,2022

i was not able to clear 1st compartment and now I have to give 2nd compartment so can I give improvement exam next year? plz tell me very much tensed(cbse class 12th)

Hello dear,

If you was not able to clear 1st compartment and now you giving second compartment so YES, you can go for your improvement exam next year but if a student receives an improvement, they are given the opportunity to retake the boards as a private candidate the following year, but there are some requirements. First, the student must pass all of their subjects; if they received a compartment in any subject, they must then pass the compartment exam before being eligible for the improvement.

As you can registered yourself as private candidate for giving your improvement exam of 12 standard CBSE(Central Board of Secondary Education).For that you have to wait for a whole year which is bit difficult for you.

Positive side of waiting for whole year is you have a whole year to preparing yourself for your examination. You have no distraction or something which may causes your failure in the exams. In whole year you have to stay focused on your 12 standard examination for doing well in it. By this you get a highest marks as a comparison of others.

Believe in Yourself! You can make anything happen

All the very best.

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Mayankjha.student2007 23 August,2022

if i am giving improvement in one subject this year, will i be able to give improvement in more subjects next year?

Hello Student,

I appreciate your Interest in education. See the improvement is not restricted to one subject or multiple subjects and we cannot say if improvement in one subject in one year leads to improvement in more subjects in coming year.

You just need to have a revision of all subjects what you have completed in the school. have a revision and practice of subjects and concepts helps you better.

All the best.

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Nuthalapati Vyshnavi 21 August,2022

I got an RT for 2 subjects, so do I have to give exams for all the subjects in the coming year or just the 2. pls help me

Hi,

You just need to give the exams for the concerned two subjects in which you have got RT. There is no need to give exam for all of your subjects, you can just fill the form for the two subjects only.

Read Complete Answer

Muskan Dhalwal 17 August,2022

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Popular CBSE Class 12th Questions

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 3.8×10⁷J 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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A career as ethical hacker involves various challenges and provides lucrative opportunities in the digital era where every giant business and startup owns its cyberspace on the world wide web. Individuals in the ethical hacker career path try to find the vulnerabilities in the cyber system to get its authority. If he or she succeeds in it then he or she gets its illegal authority. Individuals in the ethical hacker career path then steal information or delete the file that could affect the business, functioning, or services of the organization.

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

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

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