Answer

(i) $\mathrm{Cr}^{3+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{3}$

Or, $[\mathrm{Ar}]^{18} 3 d^{3}$

(ii) $\mathrm{Pm}^{3+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6} 4 f^{4}$

Or, $[\mathrm{Xe}]^{54} 3 d^{3}$

(iii) $\mathrm{Cu}^{+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10}$

Or, $[\operatorname{Ar}]^{18} 3 d^{10}$

(iv) $\mathrm{Ce}^{4+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6}$

Or, $[\mathrm{Xe}]^{54}$

(v) $\mathrm{Co}^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{7}$

Or, $[\mathrm{Ar}]^{18} 3 d^{7}$

(vi) $\mathrm{Lu}^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6} 4 f^{14} 5 d^{1}$

Or, $[X e]^{54} 2 f^{14} 3 d^{3}$

(vii) $\mathrm{Mn}^{2+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{5}$

Answer

Electronic configuration of $\mathrm{Mn}^{2+}$ is $[\mathrm{Ar}]^{18} 3 d^{5}$.

Electronic configuration of $\mathrm{Fe}^{2+}$ is $[\mathrm{Ar}]^{18} 3 d^{6}$.

It is known that half-filled and fully-filled orbitals are more stable. Therefore, $\mathrm{Mn}$ in (+2) state has a stabled ${ }^{5}$ configuration. This is the reason $\mathrm{Mn}^{2+}$ shows resistance to oxidation to $\mathrm{Mn}^{3+}$. Also, $\mathrm{Fe}^{2+}$ has $3 d^{6}$ configuration and by losing one electron, its configuration changes to a more stable $3 d^{5}$ configuration. Therefore, $\mathrm{Fe}^{2+}$ easily gets oxidized to $\mathrm{Fe}^{+3}$ oxidation state.

Answer

Element (+2 state) $ _{21} \mathrm{Sc}^{2+} , _{22} \mathrm{Ti}^{2+} , _{23} \mathrm{~V}^{2+} , _{24} \mathrm{Cr}^{2+} , _{25} \mathrm{Mn}^{2+}$

Electronic configuration $3 d^1 3 d^2 3 d^3 3 d^4 3 d^5$

In all the elements listed, the removal of two $4 \mathrm{~s}$ electrons (in $\mathrm{Cr}^{2+}$ i.e., from $4 \mathrm{~s}$ and $1 e^{-}$from $3 d$ ), the $3 d$-orbitals get gradually occupied. Since, the number of empty d-orbitals decreases or the number of unpaired electrons in 3d-orbitals increases with increase in atomic number of cations, so the stability of the cations $\left(\mathrm{M}^{2+}\right)$ increases from $\mathrm{Sc}^{2+}$ to $\mathrm{Mn}^{2+}$

Answer

The elements in the first-half of the transition series exhibit many oxidation states with Mn exhibiting maximum number of oxidation states ( +2 to +7 ). The stability of +2 oxidation state increases with the increase in atomic number. This happens as more electrons are getting filled in the $d$-orbital. However, $\mathrm{Sc}$ does not show +2 oxidation state. Its electronic configuration is $4 s^{2} 3 d^{1}$. It loses all the three electrons to form $\mathrm{Sc}^{3+} .+3$ oxidation state of $\mathrm{Sc}$ is very stable as by losing all three electrons, it attains stable noble gas configuration, $[\mathrm{Ar}]$. $\mathrm{Ti}(+4)$ and $\mathrm{V}(+5)$ are very stable for the same reason. For Mn, +2 oxidation state is very stable as after losing two electrons, its $d$-orbital is exactly halffilled, $[\mathrm{Ar}] 3 d^{5}$.

Answer

Answer

$\mathrm{MnO}_4^{-}: \mathrm{Mn}$ has +7 oxidation number and group number is 7 .

$\mathrm{CrO}_4^{2-}: \mathrm{Cr}$ has +6 oxidation state and group number is 6 .

$\mathrm{VO}_3^{-}: \mathrm{V}$ has +5 oxidation state and the group number is 5 .

Answer

As we move along the lanthanoid series, the atomic number increases gradually by one. This means that the number of electrons and protons present in an atom also increases by one. As electrons are being added to the same shell, the effective nuclear charge increases. This happens because the increase in nuclear attraction due to the addition of proton is more pronounced than the increase in the interelectronic repulsions due to the addition of electron. Also, with the increase in atomic number, the number of electrons in the 4 forbital also increases. The 4 felectrons have poor shielding effect. Therefore, the effective nuclear charge experienced by the outer electrons increases. Consequently, the attraction of the nucleus for the outermost electrons increases. This results in a steady decrease in the size of lanthanoids with the increase in the atomic number. This is termed as lanthanoid contraction.

Consequences of lanthanoid contraction

(i) There is similarity in the properties of second and third transition series.

ii. Separation of lanthanoids is possible due to lanthanide contraction.

(iii) It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides. (Basic strength decreases from $\mathrm{La}\mathrm{(OH) _3}$ to $\mathrm{Lu}(\mathrm{OH}) _{3}$.)

Answer

Transition elements are those elements in which the atoms or ions (in stable oxidation state) contain partially filled $d$ orbital. These elements lie in the $d$-block and show a transition of properties between $s$-block and $p$-block. Therefore, these are called transition elements.

Elements such as $\mathrm{Zn}, \mathrm{Cd}$, and $\mathrm{Hg}$ cannot be classified as transition elements because these have completely filled $d$ subshell.

Answer

Transition metals have a partially filled $d$-orbital. Therefore, the electronic configuration of transition elements is ( $n$ 1) $d^{1-10} \mathrm{~ns}^{0.2}$.

The non-transition elements either do not have a $d$-orbital or have a fully filled $d$-orbital. Therefore, the electronic configuration of non-transition elements is $n s^{1-2}$ or $n s^{2} n p^{1-6}$.

Answer

In the lanthanide series, +3 oxidation state is most common i.e., Ln(III) compounds are predominant. However, +2 and +4 oxidation states can also be found in the solution or in solid compounds.

Answer

(i) Transition metals show paramagnetic behaviour. Paramagnetism arises due to the presence of unpaired electrons with each electron having a magnetic moment associated with its spin angular momentum and orbital angular momentum. However, in the first transition series, the orbital angular momentum is quenched. Therefore, the resulting paramagnetism is only because of the unpaired electron.

(ii) Transition elements have high effective nuclear charge and a large number of valence electrons. Therefore, they form very strong metallic bonds. As a result, the enthalpy of atomization of transition metals is high.

(iii) Most of the complexes of transition metals are coloured. This is because of the absorption of radiation from visible light region to promote an electron from one of the $d$-orbitals to another. In the presence of ligands, the $d$ orbitals split up into two sets of orbitals having different energies. Therefore, the transition of electrons can take place from one set toanother. The energy required for these transitions is quite small and falls in the visible region of radiation. The ions of transition metals absorb the radiation of a particular wavelength and the rest is reflected, imparting colour to the solution.

(iv) The catalytic activity of the transition elements can be explained by two basic facts.

(a) Owing to their ability to show variable oxidation states and form complexes, transition metals form unstable intermediate compounds. Thus, they provide a new path with lower activation energy, $E_{\mathrm{a}}$, for the reaction.

(b) Transition metals also provide a suitable surface for the reactions to occur.

Answer

Transition metals are large in size and contain lots of interstitial sites. Transition elements can trap atoms of other elements (that have small atomic size), such as H, C, N, in the interstitial sites of their crystal lattices. The resulting compounds are called interstitial compounds.

Answer

In transition elements, the oxidation state can vary from +1 to the highest oxidation state by removing all its valence electrons. Also, in transition elements, the oxidation states differ by $1\left(\mathrm{Fe}^{2+}\right.$ and $\mathrm{Fe}^{3+} ; \mathrm{Cu}^{+}$ and $\left.\mathrm{Cu}^{2+}\right)$. In non-transition elements, the oxidation states differ by 2 , for example, +2 and +4 or +3 and +5 , etc.

Answer

(i) Preparation of $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ from chromite ore

(a) Potassium dichromate is obtained by the fusion of chromite ore $\left(\mathrm{FeCr}_2 \mathrm{O}_4\right)$ with sodium (or potassium) carbonate in the presence of excess air.

$ 4 \mathrm{FeCr}_2 \mathrm{O}_4+8 \mathrm{Na}_2 \mathrm{CO}_3+7 \mathrm{O}_2 \rightarrow \underset{\left(\substack{ \text{Yellow} \\ \text{solution} }\right)}{8 \mathrm{Na}_2 \mathrm{CrO}_4}+2 \mathrm{Fe}_2 \mathrm{O}_3+8 \mathrm{CO}_2 $

(b) The solution is filtered and treated with sulphuric acid.

$ 2 \mathrm{Na}_2 \mathrm{CrO}_4+\underset{\left(\mathrm{From} \mathrm{H}_2 \mathrm{SO}_4\right)}{2 \mathrm{H}^{+}} \rightarrow \underset{\substack{\text{Sodium}\\ \substack{\text{dichromate}\\ \text{(orange)}}}}{\mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7(\mathrm{~s})}+2 \mathrm{Na}^{+}+\mathrm{H}_2 \mathrm{O} $

(c) Now sodium dichromate is treated with potassium chloride.

As a result, potassium dichromate is produced.

$ \mathrm{Na}_2 \mathrm{Cr}_2 \mathrm{O}_7+2 \mathrm{KCl} \rightarrow \underset{\substack{\text { Potassium } \\ \text { dichromate } \\ \text { crystals }}}{\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7} +2 \mathrm{NaCl} $

(ii) Effect of increasing $\mathrm{pH}$ on $\mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ solution

$ \underset{\text { Orange }}{\mathrm{Cr}_2 \mathrm{O}_7^{2-}}+2 \mathrm{OH}^{-} \rightarrow \underset{\text { Yellow }}{2 \mathrm{CrO}_4^{2-}}+\mathrm{H}_2 \mathrm{O} $

On increasing $\mathrm{pH} ; \mathrm{K}_2 \mathrm{Cr}_2 \mathrm{O}_7$ changes into $\mathrm{K}_2 \mathrm{CrO}_4$ (orange to yellow).

Answer

$\mathrm{K_2} \mathrm{Cr_2} \mathrm{O_7}$ acts as a very strong oxidising agent in the acidic medium.

$\mathrm{K_2} \mathrm{Cr_2} \mathrm{O_7}+4 \mathrm{H_2} \mathrm{SO_4} \longrightarrow \mathrm{K_2} \mathrm{SO_4}+\mathrm{Cr_2}\left(\mathrm{SO_4}\right)_{3}+4 \mathrm{H_2} \mathrm{O}+3[\mathrm{O}]$

$\mathrm{K_2} \mathrm{Cr_2} \mathrm{O_7}$ takes up electrons to get reduced and acts as an oxidising agent. The reaction of $\mathrm{K_2} \mathrm{Cr_2} \mathrm{O_7}$ with other iodide, iron (II) solution, and $\mathrm{H_2} \mathrm{~S}$ are given below.

(i) $\mathrm{K_2} \mathrm{Cr_2} \mathrm{O_7}$ oxidizes iodide to iodine.

$ \begin{aligned} \mathrm{Cr}_2 \mathrm{O}_7^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H}_2 \mathrm{O} \\ 2 \mathrm{I}^{-} & \left.\longrightarrow \mathrm{I}_2+2 \mathrm{e}^{-}\right] \times 3 \\ \hline \mathrm{Cr}_2 \mathrm{O}_7^{2-}+6 \mathrm{I}^{-}+14 \mathrm{H}^{+} & \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{I}_2+7 \mathrm{H}_2 \mathrm{O} \\ \hline \end{aligned} $

(ii) $\mathrm{K_2} \mathrm{Cr_2} \mathrm{O_7}$ oxidizes iron (II) solution to iron (III) solution i.e., ferrous ions to ferric ions.

$ \begin{aligned} \mathrm{Cr_2} \mathrm{O_7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H_2} \mathrm{O} \\ \mathrm{Fe}^{2+} & \left.\longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-}\right] \times 6 \\ \hline \mathrm{Cr_2} \mathrm{O_7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{Fe}^{2+} & \longrightarrow 2 \mathrm{Cr}^{3+}+6 \mathrm{Fe}^{3+}+7 \mathrm{H_2} \mathrm{O} \\ \hline \end{aligned} $

(iii) $\mathrm{K_2} \mathrm{Cr_2} \mathrm{O_7}$ oxidizes $\mathrm{H_2} \mathrm{~S}$ to sulphur.

$ \begin{aligned} \mathrm{Cr_2} \mathrm{O_7}^{2-}+14 \mathrm{H}^{+}+6 \mathrm{e}^{-} & \longrightarrow 2 \mathrm{Cr}^{3+}+7 \mathrm{H_2} \mathrm{O} \\ \mathrm{H_2} \mathrm{~S} & \left.\longrightarrow \mathrm{S}+2 \mathrm{H}^{+}+2 \mathrm{e}^{-}\right] \times 3 \\ \hline \mathrm{Cr_2} \mathrm{O_7}^{2-}+3 \mathrm{H_2} \mathrm{~S}+8 \mathrm{H}^{+} & \longrightarrow 2 \mathrm{Cr}^{3+}+3 \mathrm{~S}+7 \mathrm{H_2} \mathrm{O} \\ \hline \end{aligned} $

Answer

Potassium permanganate can be prepared from pyrolusite $\left(\mathrm{MnO_2}\right)$. The ore is fused with $\mathrm{KOH}$ in the presence of either atmospheric oxygen or an oxidising agent, such as $\mathrm{KNO_3}$ or $\mathrm{KClO_4}$, to give $\mathrm{K_2} \mathrm{MnO_4}$.

$ 2 \mathrm{MnO_2}+4 \mathrm{KOH}+\mathrm{O_2} \xrightarrow{\text { heat }} {\text{(Green)}}{2 \mathrm{~K_2} \mathrm{MnO_4}}+2 \mathrm{H_2} \mathrm{O} $

The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.

Electrolytic oxidation

$ \begin{aligned} & \mathrm{K_2} \mathrm{MnO_4} \longleftrightarrow 2 \mathrm{~K}^{+}+\mathrm{MnO_4}^{2-} \\ & \mathrm{H_2} \mathrm{O} \longleftrightarrow \mathrm{H}^{+}+\mathrm{OH}^{-} \end{aligned} $

At anode, manganate ions are oxidized to permanganate ions.

$ \underset{\text{Green}}{\mathrm{MnO_4}^{2-}} \longleftrightarrow \underset{\text{Green}}{\mathrm{MnO_4}^{-}}+\mathrm{e}^{-} $

Oxidation by chlorine

$ \begin{aligned} & 2 \mathrm{~K_2} \mathrm{MnO_4}+\mathrm{Cl_2} \longrightarrow 2 \mathrm{KMnO_4}+2 \mathrm{KCl} \\ & 2 \mathrm{MnO_4}^{2-}+\mathrm{Cl_2} \longrightarrow 2 \mathrm{MnO_4}^{-}+2 \mathrm{Cl}^{-} \end{aligned} $

Oxidation by ozone

$ \begin{aligned} 2 \mathrm{~K_2} \mathrm{MnO_4}+\mathrm{O_3}+\mathrm{H_2} \mathrm{O} \longrightarrow 2 \mathrm{KMnO_4}+2 \mathrm{KOH}+\mathrm{O_2} \\ 2 \mathrm{MnO_4}^{2-}+\mathrm{O_3}+\mathrm{H_2} \mathrm{O} \longrightarrow 2 \mathrm{MnO_4}^{2-}+2 \mathrm{OH}^{-}+\mathrm{O_2} \end{aligned} $

(i) Acidified $\mathrm{KMnO_4}$ solution oxidizes $\mathrm{Fe}$ (II) ions to $\mathrm{Fe}$ (III) ions i.e., ferrous ions to ferric ions.

$ \begin{aligned} \mathrm{MnO_4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} & \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H_2} \mathrm{O} \\ \mathrm{Fe}^{2+} & \longrightarrow \mathrm{Fe}^{3+}+\mathrm{e}^{-} ] \times 5 \\ \hline \mathrm{MnO_4}^{-}+5 \mathrm{Fe}^{2+}+8 \mathrm{H}^{+} & \longrightarrow \mathrm{Mn}^{2+}+5 \mathrm{Fe}^{3+}+4 \mathrm{H_2} \mathrm{O} \\ \hline \end{aligned} $

(ii) Acidified potassium permanganate oxidizes $\mathrm{SO_2}$ to sulphuric acid.

$ \begin{aligned} & \mathrm{MnO_4}^{-}+6 \mathrm{H}^{+}+5 \mathrm{e}^{-}\left.\longrightarrow \mathrm{Mn}^{2+}+3 \mathrm{H_2} \mathrm{O}\right] \times 2 \\ & 2 \mathrm{H_2} \mathrm{O}+2 \mathrm{SO_2}+\mathrm{O_2}\left.\longrightarrow 4 \mathrm{H}^{+}+2 \mathrm{SO_4}^{2-}+2 \mathrm{e}^{-}\right] \times 5 \\ \hline & 2 \mathrm{MnO_4}^{-}+10 \mathrm{SO_2}+5 \mathrm{O_2}+4 \mathrm{H_2} \mathrm{O} \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{SO_4}^{2-}+8 \mathrm{H}^{+}\\ \hline \end{aligned} $

(iii) Acidified potassium permanganate oxidizes oxalic acid to carbon dioxide.

$ \begin{aligned} \mathrm{MnO _4}^{-}+8 \mathrm{H}^{+}+5 \mathrm{e}^{-} & \longrightarrow \mathrm{Mn}^{2+}+4 \mathrm{H_2} \mathrm{O}] \times 2 \\ \mathrm{C _2} \mathrm{O_4}^{2-} & \longrightarrow 2 \mathrm{CO _2}+2 \mathrm{e}^{-}] \times 5 \\ \hline 2 \mathrm{MnO _4}^{-}+5 \mathrm{C _2} \mathrm{O_4}^{2-}+16 \mathrm{H}^{+} & \longrightarrow 2 \mathrm{Mn}^{2+}+10 \mathrm{CO _2}+8 \mathrm{H _2} \mathrm{O} \\ \hline \end{aligned} $

Answer

(i) The $\mathrm{E}^{\ominus}$ value for $\mathrm{Fe}^{3+} / \mathrm{Fe}^{2+}$ is higher than that for $\mathrm{Cr}^{3+} / \mathrm{Cr}^{2+}$ and lower than that for $\mathrm{Mn}^{3+} / \mathrm{Mn}^{2+}$. So, the reduction of $\mathrm{Fe}^{3+}$ to $\mathrm{Fe}^{2+}$ is easier than the reduction of $\mathrm{Mn}^{3+}$ to $\mathrm{Mn}^{2+}$, but not as easy as the reduction of $\mathrm{Cr}^{3+}$ to $\mathrm{Cr}^{2+}$. Hence, $\mathrm{Fe}^{3+i}$ is more stable than $\mathrm{Mn}^{3+}$, but less stable than $\mathrm{Cr}^{3+}$. These metal ions can be arranged in the increasing order of their stability as: $\mathrm{Mn}^{3+}<\mathrm{Fe}^{3+}<\mathrm{Cr}^{3+}$

(ii) The reduction potentials for the given pairs increase in the following order.

$\mathrm{Mn}^{2+} / \mathrm{Mn}<\mathrm{Cr}^{2+} / \mathrm{Cr}<\mathrm{Fe}^{2+} / \mathrm{Fe}$

So, the oxidation of $\mathrm{Fe}$ to $\mathrm{Fe}^{2+}$ is not as easy as the oxidation of $\mathrm{Cr}$ to $\mathrm{Cr}^{2+}$ and the oxidation of $\mathrm{Mn}$ to $\mathrm{Mn}^{2+}$. Thus, these metals can be arranged in the increasing order of their ability to get oxidised as: $\mathrm{Fe}<\mathrm{Cr}<\mathrm{Mn}$

Answer

Only the ions that have electrons in $d$-orbital and in which $d$-d transition is possible will be coloured. The ions in which $d$-orbitals are empty or completely filled will be colourless as no $d$-d transition is possible in those configurations.

From the above table, it can be easily observed that only $\mathrm{Sc}^{3+}$ has an empty $d$-orbital and $\mathrm{Cu}^{+}$has completely filled $\mathrm{d}$ orbitals. All other ions, except $\mathrm{Sc}^{3+}$ and $\mathrm{Cu}^{+}$, will be coloured in aqueous solution because of $d$ - $d$ transitions.

Answer

$ \begin{array}{|c|c|c|c|c|c|c|c|} \hline \text{Sc} & & & +3 & & & \\ \hline \mathrm{Ti} & +1 & +2 & +3 & +4 & & \\ \hline \mathrm{V} & +1 & +2 & +3 & +4 & +5 & \\ \hline \mathrm{Cr} & +1 & +2 & +3 & +4 & +5 & +6 \\ \hline \mathrm{Mn} & +1 & +2 & +3 & +4 & +5 & +6 & +7 \\ \hline \mathrm{Fe} & +1 & +2 & +3 & +4 & +5 & +6 & \\ \hline \mathrm{Co} & +1 & +2 & +3 & +4 & +5 & & \\ \hline \mathrm{Ni} & +1 & +2 & +3 & +4 & & & \\ \hline \mathrm{Cu} & +1 & +2 & +3 & & & & \\ \hline \mathrm{Zn} & & +2 & & & & & \\ \hline \end{array} $

From the above table, it is evident that the maximum number of oxidation states is shown by $\mathrm{Mn}$, varying from +2 to +7. The number of oxidation states increases on moving from Sc to Mn. On moving from Mn to Zn, the number of oxidation states decreases due to a decrease in the number of available unpaired electrons. The relative stability of the +2 oxidation state increases on moving from top to bottom. This is because on moving from top to bottom, it becomes more and more difficult to remove the third electron from the $d$-orbital.

Answer

(i) Electronic configuration

The general electronic configuration for lanthanoids is $[\mathrm{Xe}]^{54} 4 f^{0-14} 5 d^{0-1} 6 s^{2}$ and that for actinoids is $[\mathrm{Rn}]^{86} 5 f^{-14} 6 d^{0-1} 7 s^{2}$. Unlike 4 forbitals, $5 f$ orbitals are not deeply buried and participate in bonding to a greater extent.

(ii) Oxidation states

The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4 . This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the $5 f, 6 d$, and 7 slevels are of comparable energies. Again, (+3) is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.

(iii) Atomic and lonic sizes

Similar to lanthanoids, actinoids also exhibit actinoid contraction (overall decrease in atomic and ionic radii). The contraction is greater due to the poor shielding effect of 5 forbitals.

(iv) Chemical reactivity

In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to $\mathrm{Ca}$. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

Answer

(i) $\mathrm{Cr}^{2+}$ is strongly reducing in nature. It has a $d^{4} c$ configuration. While acting as a reducing agent, it gets oxidized to

$\mathrm{Cr}^{3+}\left(\right.$ electronic configuration, $\left.d^{\beta}\right)$. This $d^{3}$ configuration can be written as $t_{2 \mathrm{~g}}^{3}$ configuration, which is a more stable configuration. In the case of $\mathrm{Mn}^{3+}\left(d^{4}\right)$, it acts as an oxidizing agent and gets reduced to $\mathrm{Mn}^{2+}\left(d^{5}\right)$. This has an exactly half-filled $d$-orbital and is highly stable.

(ii) $\mathrm{Co}(\mathrm{II})$ is stable in aqueous solutions. However, in the presence of strong field complexing reagents, it is oxidized to Co (III). Although the 3rionization energy for Co is high, but the higher amount of crystal field stabilization energy (CFSE) released in the presence of strong field ligands overcomes this ionization energy.

(iii) The ions in $d^{11}$ configuration tend to lose one more electron to get into stable $d^{0}$ configuration. Also, the hydration or lattice energy is more than sufficient to remove the only electron present in the $d$-orbital of these ions. Therefore, they act as reducing agents.

Answer

It is found that sometimes a relatively less stable oxidation state undergoes an oxidation - reduction reaction in which it is simultaneously oxidised and reduced. This is called disproportionation.

Forexample,

(i) $\underset{\mathrm{Cr(V)}}{\mathrm{3CrO_{4}^{3-}}} + \mathrm{8H^+} \longrightarrow \underset{\mathrm{Cr(VI)}}{\mathrm{3CrO_{4}^{2-}}} + \underset{\mathrm{Cr(III)}}{\mathrm{CrO^{3+}}} + \mathrm{4H_2O} $

$\mathrm{Cr}(\mathrm{V})$ is oxidized to $\mathrm{Cr}(\mathrm{VI})$ and reduced to $\mathrm{Cr}(\mathrm{III})$.

(ii) $\underset{\mathrm{Mn(VI)}}{3 \mathrm{MnO_{4}^{2-}}}+4 \mathrm{H}^{+} \longrightarrow \underset{\mathrm{Mn(VII)}}{2 \mathrm{MnO_{4}^{-}}}+\underset{\mathrm{Mn(IV)}}{\mathrm{MnO_2}}+2 \mathrm{H_2} \mathrm{O}$

$\mathrm{Mn}(\mathrm{VI})$ is oxidized to $\mathrm{Mn}(\mathrm{VII})$ and reduced to Mn (IV).

Answer

In the first transition series, Cu exhibits +1 oxidation state very frequently. It is because $\mathrm{Cu}(+1)$ has an electronic configuration of $[\mathrm{Ar}] 3 d^{10}$. The completely filled $d$-orbital makes it highly stable.

Answer

$\mathrm{Cr}^{3+}$ is the most stable in aqueous solutions owing to a $t_{2 \mathrm{~g}}^{3}$ configuration.

Answer

(i) In the case of a lower oxide of a transition metal, the metal atom has a low oxidation state. This means that some of the valence electrons of the metal atom are not involved in bonding. As a result, it can donate electrons and behave as a base.

On the other hand, in the case of a higher oxide of a transition metal, the metal atom has a high oxidation state. This means that the valence electrons are involved in bonding and so, they are unavailable. There is also a high effective nuclear charge.

As a result, it can accept electrons and behave as an acid.

For example, $\mathrm{Mn}^{\mathrm{II}} \mathrm{O_\text {is basic and }} \mathrm{Mn_2} \mathrm{O_7}$ is acidic.

(ii) Oxygen and fluorine act as strong oxidising agents because of their high electronegativities and small sizes. Hence, they bring out the highest oxidation states from the transition metals. In other words, a transition metal exhibits higher oxidation states in oxides and fluorides. For example, in $\mathrm{OsF_6}$ and $\mathrm{V_2} \mathrm{O_5}$, the oxidation states of $\mathrm{Os}$ and V are +6 and +5 respectively.

(iii) Oxygen is a strong oxidising agent due to its high electronegativity and small size. So, oxo-anions of a metal have the highest oxidation state. For example, in $\mathrm{MnO_4}^{-}$, the oxidation state of $\mathrm{Mn}$ is +7 .

Answer

(i)

Potassium dichromate $\left(\mathrm{K_2} \mathrm{Cr_2} \mathrm{O_7}\right)$ is prepared from chromite ore $\left(\mathrm{FeCr_2} \mathrm{O_4}\right)$ in the following steps.

Step (1):Preparation of sodium chromate

$ 4 \mathrm{FeCr_2} \mathrm{O_4}+16 \mathrm{NaOH}+7 \mathrm{O_2} \longrightarrow 8 \mathrm{Na_2} \mathrm{CrO_4}+2 \mathrm{Fe_2} \mathrm{O_3}+8 \mathrm{H_2} \mathrm{O} $

Step (2):Conversion of sodium chromate into sodium dichromate

$ 2 \mathrm{Na_2} \mathrm{CrO_4}+\text { conc. } \mathrm{H_2} \mathrm{SO_4} \longrightarrow \mathrm{Na_2} \mathrm{Cr_2} \mathrm{O_7}+\mathrm{Na_2} \mathrm{SO_4}+\mathrm{H_2} \mathrm{O} $

Step(3):Conversion of sodium dichromate to potassium dichromate

$ \mathrm{Na_2} \mathrm{Cr_2} \mathrm{O_7}+2 \mathrm{KCl} \longrightarrow \mathrm{K_2} \mathrm{Cr_2} \mathrm{O_7}+2 \mathrm{NaCl} $

Potassium chloride being less soluble than sodium chloride is obtained in the form of orange coloured crystals and can be removed by filtration.

$ \underset{\substack{\text{Chromate}\\ \text{(Yellow)}}}{2 \mathrm{CrO_{4}^{2-}}} \stackrel{\text{Acid}}{\underset{\text{Alkali}}{\leftarrow\rightarrow}} \underset{\substack{\text{Hydrogen} \\ \text{(Chromate)}}}{2 \mathrm{HCrO_{4}^{-}}}\stackrel{\text{Acid}}{\underset{\text{Alkali}}{\leftarrow\rightarrow}} \underset{\substack{\text{Dichromate}\\ \text{(Orange)}}}{\mathrm{Cr_2} \mathrm{O_7}^{2-}} $

The dichromate ion $\left(\mathrm{Cr_2} \mathrm{O_7}^{2-}\right)$ exists in equilibrium with chromate $\left(\mathrm{CrO_4}^{2-}\right)$ ion at $\mathrm{pH} 4$. However, by changing the $\mathrm{pH}$, they can be interconverted. (ii)

Potassium permanganate $\left(\mathrm{KMnO_4}\right.$ ) can be prepared from pyrolusite $\left(\mathrm{MnO_2}\right)$. The ore is fused with $\mathrm{KOH}$ in the presence of either atmospheric oxygen or an oxidising agent, such as $\mathrm{KNO_3} \mathrm{Or} \mathrm{KClO_4}$, to give $\mathrm{K_2} \mathrm{MnO_4}$.

$ 2 \mathrm{MnO_2}+4 \mathrm{KOH}+\mathrm{O_2} \xrightarrow{\text { heat }} \underset{\text{(Green)}}{2 \mathrm{~K_2} \mathrm{MnO_4}}+2 \mathrm{H_2} \mathrm{O} $

The green mass can be extracted with water and then oxidized either electrolytically or by passing chlorine/ozone into the solution.

Electrolytic oxidation

$ \begin{aligned} & \mathrm{K_2} \mathrm{MnO_4} \longleftrightarrow 2 \mathrm{~K}^{+}+\mathrm{MnO_4}^{2-} \\ & \mathrm{H_2} \mathrm{O} \longleftrightarrow \mathrm{H}^{+}+\mathrm{OH}^{-} \end{aligned} $

At anode, manganate ions are oxidized to permanganate ions.

$ \underset{\text{Green}}{\mathrm{MnO_4}^{2-}} \longleftrightarrow \underset{\text{Purple}}{\mathrm{MnO_4}^{-}}+\mathrm{e}^{-} $

Green Purple

Oxidation by chlorine

$ \begin{aligned} & 2 \mathrm{~K_2} \mathrm{MnO_4}+\mathrm{Cl_2} \longrightarrow 2 \mathrm{KMnO_4}+2 \mathrm{KCl} \\ & 2 \mathrm{MnO_4}^{2-}+\mathrm{Cl_2} \longrightarrow 2 \mathrm{MnO_4}^{-}+2 \mathrm{Cl}^{-} \end{aligned} $

Oxidation by ozone

$ \begin{aligned} & 2 \mathrm{~K_2} \mathrm{MnO_4}+\mathrm{O_3}+\mathrm{H_2} \mathrm{O} \longrightarrow 2 \mathrm{KMnO_4}+2 \mathrm{KOH}+\mathrm{O_2} \\ & 2 \mathrm{MnO_4}^{2-}+\mathrm{O_3}+\mathrm{H_2} \mathrm{O} \longrightarrow 2 \mathrm{MnO_4}^{2-}+2 \mathrm{OH}^{-}+\mathrm{O_2} \end{aligned} $

Answer

An alloy is a solid solution of two or more elements in a metallic matrix. It can either be a partial solid solution or a complete solid solution. Alloys are usually found to possess different physical properties than those of the component elements.

An important alloy of lanthanoids is Mischmetal. It contains lanthanoids (94-95 $\%$), iron (5 $\%$), and traces of S, C, Si, $\mathrm{Ca}$, and $\mathrm{Al}$.

Uses

(1) Mischmetal is used in cigarettes and gas lighters.

(2) It is used in flame throwing tanks.

(3) It is used in tracer bullets and shells.

Answer

Inner transition metals are those elements in which the last electron enters the $f$-orbital. The elements in which the $4 f$ and the $5 f$ orbitals are progressively filled are called $f$-block elements. Among the given atomic numbers, the atomic numbers of the inner transition elements are 59, 95, and 102.

Answer

Lanthanoids primarily show three oxidation states (+2,+3,+4). Among these oxidation states, +3 state is the most common. Lanthanoids display a limited number of oxidation states because the energy difference between $4 f, 5 d$, and 6 sorbitals is quite large. On the other hand, the energy difference between $5 f, 6 d$, and $7 s$ orbitals is very less. Hence, actinoids display a large number of oxidation states. For example, uranium and plutonium display +3,+4,+5, and +6 oxidation states while neptunium displays +3,+4,+5, and +7 . The most common oxidation state in case of actinoids is also +3 .

Answer

The last element in the actinoid series is lawrencium, Lr. Its atomic number is 103 and its electronic configuration is $[\operatorname{Rn}] 5 f^{14} 6 d^{1} 7 s^{2}$. The most common oxidation state displayed by it is +3 ; because after losing 3 electrons it attains stablef ${ }^{14}$ configuration.

Answer

Ce : $1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6} 4 f^{1} 5 d^{1} 6 s^{2}$

Magnetic moment can be calculated as:

$\mu=\sqrt{n(n+2)}$

Where,

$n=$ number of unpaired electrons

The electronic configuration of $\mathrm{Ce}^{3+}: 1 s^{2} 2 s^{2} 2 p^{6} 3 s^{2} 3 p^{6} 3 d^{10} 4 s^{2} 4 p^{6} 4 d^{10} 5 s^{2} 5 p^{6} 4 f^{1}$

In $\mathrm{Ce}^{3+}, n=1$

$\therefore$ Magnetic moment, $\mu=\sqrt{\mathrm{n}(\mathrm{n}+2)}$ B.M

$ \begin{aligned} & \Longrightarrow \mu=\sqrt{1(1+2)} \text { B.M } \\ & \Longrightarrow \mu=\sqrt{3} \text { B.M } \\ & \Longrightarrow \mu=1.73 \text { B.M } \end{aligned} $

Answer

The lanthanides that exhibit +2 and +4 states are shown in the given table. The atomic numbers of the elements are given in the parenthesis.

Ce after forming $\mathrm{Ce}^{4+}$ attains a stable electronic configuration of $[\mathrm{Xe}]$.

$\mathrm{Tb}$ after forming $\mathrm{Tb}^{4+}$ attains a stable electronic configuration of $[\mathrm{Xe}] 4 f$.

Eu after forming $\mathrm{Eu}^{2+}$ attains a stable electronic configuration of $[\mathrm{Xe}] 4 f^{\prime}$.

$\mathrm{Yb}$ after forming $\mathrm{Yb}^{2+}$ attains a stable electronic configuration of $[\mathrm{Xe}] 4 f^{44}$

Answer

Electronic configuration

The general electronic configuration for lanthanoids is $[\mathrm{Xe}]^{54} 4 f^{0-14} 5 d^{0-1} 6 s^{2}$ and that for actinoids is $[\mathrm{Rn}]^{86} 5 f^{-14} 6 d^{0-1} 7 s^{2}$. Unlike 4 forbitals, $5 f$ orbitals are not deeply buried and participate in bonding to a greater extent.

Oxidation states

The principal oxidation state of lanthanoids is (+3). However, sometimes we also encounter oxidation states of +2 and +4 . This is because of extra stability of fully-filled and half-filled orbitals. Actinoids exhibit a greater range of oxidation states. This is because the $5 f, 6 d$, and 7 slevels are of comparable energies. Again, $(+3)$ is the principal oxidation state for actinoids. Actinoids such as lanthanoids have more compounds in +3 state than in +4 state.

Chemical reactivity

In the lanthanide series, the earlier members of the series are more reactive. They have reactivity that is comparable to Ca. With an increase in the atomic number, the lanthanides start behaving similar to Al. Actinoids, on the other hand, are highly reactive metals, especially when they are finely divided. When they are added to boiling water, they give a mixture of oxide and hydride. Actinoids combine with most of the non-metals at moderate temperatures. Alkalies have no action on these actinoids. In case of acids, they are slightly affected by nitric acid (because of the formation of a protective oxide layer).

Answer

Answer

(i) In the $1^{\text {st }}, 2^{\text {nd }}$ and $3^{\text {rd }}$ transition series, the $3 d, 4 d$ and $5 d$ orbitals are respectively filled.

We know that elements in the same vertical column generally have similar electronic configurations.

In the first transition series, two elements show unusual electronic configurations:

$ \begin{aligned} & \operatorname{Cr}(24)=3 d^{5} 4 s^{1} \\ & \mathrm{Cu}(29)=3 d^{10} 4 s^{1} \end{aligned} $

Similarly, there are exceptions in the second transition series. These are:

$ \begin{aligned} & \mathrm{Mo}(42)=4 d^{5} 5 s^{1} \\ & \mathrm{Tc}(43)=4 d^{6} 5 s^{1} \\ & \mathrm{Ru}(44)=4 d^{7} 5 s^{1} \\ & \mathrm{Rh}(45)=4 d^{8} 5 s^{1} \\ & \mathrm{Pd}(46)=4 d^{10} 5 s^{0} \\ & \mathrm{Ag}(47)=4 d^{10} 5 s^{1} \end{aligned} $

There are some exceptions in the third transition series as well. These are:

$ \begin{aligned} & \mathrm{W}(74)=5 d^{4} 6 s^{2} \\ & \mathrm{Pt}(78)=5 d^{9} 6 s^{1} \\ & \mathrm{Au}(79)=5 d^{10} 6 s^{1} \end{aligned} $

As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.

(ii) In each of the three transition series the number of oxidation states shown by the elements is the maximum in the middle and the minimum at the extreme ends.

However, +2 and +3 oxidation states are quite stable for all elements present in the first transition series. All metals present in the first transition series form stable compounds in the +2 and +3 oxidation states. The stability of the +2 and +3 oxidation states decreases in the second and the third transition series, wherein higher oxidation states are more important.

For example $\left[\stackrel{\text{II}}{\mathrm{Fe}}(\mathrm{Cn}) _{6}\right]^{4-},\left[\stackrel{\text{III}}{\mathrm{Co}}\left(\mathrm{NH _3}\right) _{6}\right]^{3+},\left[\mathrm{Ti} \left(\mathrm{H _2} \mathrm{O}\right) _{6}\right]^{3+}$ are stable complexes, but no such complexes are known for the second and third transition series such as Mo, W, Rh, In. They form complexes in which their oxidation states are high. For example: $\mathrm{WCl _6}, \mathrm{ReF _7}, \mathrm{RuO _4}$, etc.

(iii) In each of the three transition series, the first ionisation enthalpy increases from left to right. However, there are some exceptions. The first ionisation enthalpies of the third transition series are higher than those of the first and second transition series. This occurs due to the poor shielding effect of 4 felectrons in the third transition series.

Certain elements in the second transition series have higher first ionisation enthalpies than elements corresponding to the same vertical column in the first transition series. There are also elements in the $2^{\text {nd }}$ transition series whose first ionisation enthalpies are lower than those of the elements corresponding to the same vertical column in the $1^{\text {st }}$ transition series.

(iv) Atomic size generally decreases from left to right across a period. Now, among the three transition series, atomic sizes of the elements in the second transition series are greater than those of the elements corresponding to the same vertical column in the first transition series. However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

Answer

$ \begin{array}{|c|c|c|} \hline \text{Metal ion} & \text{Number of } \mathbf{d}\text{-electrons} & \text{Filling of } \mathbf{d}\text{-orbitals} \\ \hline \mathrm{Ti}^{2+} & 2 & t_{2 g}^{2} \\ \hline \mathrm{V}^{2+} & 3 & t_{2 g}^{3} \\ \hline \mathrm{Cr}^{3+} & 3 & t_{2 g}^{3} \\ \hline \mathrm{Mn}^{2+} & 5 & t_{2 g}^{3} e_{g}^{2} \\ \hline \mathrm{Fe}^{2+} & 6 & t_{2 g}^{4} e_{g}^{2} \\ \hline \mathrm{Fe}^{3+} & 5 & t_{2 g}^{3} e_{g}^{2} \\ \hline \mathrm{CO}^{2+} & 7 & t_{2 g}^{5} e_{g}^{2} \\ \hline \mathrm{Ni}^{2+} & 8 & t_{2 g}^{6} e_{g}^{2} \\ \hline \mathrm{Cu}^{2+} & 9 & t_{2 g}^{6} e_{g}^{3} \\ \hline \end{array} $

Answer

The properties of the elements of the first transition series differ from those of the heavier transition elements in many ways.

(i) The atomic sizes of the elements of the first transition series are smaller than those of the heavier elements (elements of $2^{\text {nd }}$ and $3^{\text {rd }}$ transition series).

However, the atomic sizes of the elements in the third transition series are virtually the same as those of the corresponding members in the second transition series. This is due to lanthanoid contraction.

(ii) +2 and +3 oxidation states are more common for elements in the first transition series, while higher oxidation states are more common for the heavier elements.

(iii) The enthalpies of atomisation of the elements in the first transition series are lower than those of the corresponding elements in the second and third transition series.

(iv) The melting and boiling points of the first transition series are lower than those of the heavier transition elements. This is because of the occurrence of stronger metallic bonding ( $M-M$ bonding).

(v) The elements of the first transition series form low-spin or high-spin complexes depending upon the strength of the ligand field. However, the heavier transition elements form only low-spin complexes, irrespective of the strength of the ligand field.

Answer

Magnetic moment $\left(\mu\right)$ is given as $\mu=\sqrt{n(n+2)}$.

For value $n=1, \mu=\sqrt{1(1+2)}=\sqrt{3}=1.732$.

For value $n=2, \quad \mu=\sqrt{2(2+2)}=\sqrt{8}=2.83$.

For value $n=3, \mu=\sqrt{3(3+2)}=\sqrt{15}=3.87$.

For value $n=4, \mu=\sqrt{4(4+2)}=\sqrt{24}=4.899$.

For value $n=5, \mu=\sqrt{5(5+2)}=\sqrt{35}=5.92$.

(i) $\mathrm{K_4}\left[\mathrm{Mn}(\mathrm{CN})_{6}\right]$

For in transition metals, the magnetic moment is calculated from the spin-only formula. Therefore,

$ \sqrt{n(n+2)}=2.2 $

We can see from the above calculation that the given value is closest to $n=1$. Also, in this complex, $\mathrm{Mn}$ is in the +2 oxidation state. This means that Mn has 5 electrons in the $d$-orbital.

Hence, we can say that $\mathrm{CN}$ - is a strong field ligand that causes the pairing of electrons.

(ii) $\left[\mathrm{Fe}\left(\mathrm{H_2} \mathrm{O}\right)_{6}\right]^{2+}$

$ \sqrt{n(n+2)}=5.3 $

We can see from the above calculation that the given value is closest to $n=4$. Also, in this complex, $\mathrm{Fe}$ is in the +2 oxidation state. This means that $\mathrm{Fe}$ has 6 electrons in the $d$-orbital.

Hence, we can say that $\mathrm{H_2} \mathrm{O}$ is a weak field ligand and does not cause the pairing of electrons.

(iii) $\mathrm{K_2}\left[\mathrm{MnCl_4}\right]$

$ \sqrt{n(n+2)}=5.9 $

We can see from the above calculation that the given value is closest to $n=5$. Also, in this complex, $\mathrm{Mn}$ is in the +2 oxidation state. This means that Mn has 5 electrons in the $d$-orbital.

Hence, we can say that $\mathrm{Cl}$ is a weak field ligand and does not cause the pairing of electrons.