d and f Block Elements

Among given, $\mathrm{Eu}^{2+}, \mathrm{Gd}^{3+}$ has same number of electron and are paramagnetic in nature.

$\mathrm{Eu}(\mathrm{Z}=63) \rightarrow \mathrm{Eu}^{2+}=63-2=61$ electron.

$\mathrm{Gd}(\mathrm{Z}=64) \rightarrow \mathrm{Gd}^{3+}=64-3=61$

electrons.

Both are paramagnetic (unpaired $f$-electrons)

Among the given ions, $\mathrm{Fe}^{2+}, \mathrm{Cu}^{2+}$ and $\mathrm{V}^{3+}$ are correctly matched with their colour.

The colour of $\mathrm{Fe}^{3+}$ (aquated) is yellow.

Atomic radius generally increases down a group. However, due to the lanthanoid contraction, elements of the $5 d$ series have nearly the same atomic radii as their counterparts in the $4 d$ series. Lanthanoid contraction is the poor shielding effect of the $4 f$ electrons, which increases the effective nuclear charge and pulls the valence electrons closer to the nucleus.

• $\operatorname{Zr}(4 d)$ and Hf ( $5 d$ ) are in the same group, and size of Hf is reduced by the lanthanoid contraction, making their radii nearly identical.

• Mo (4d) and W (5d) are also a pair in the same group where the size of $W$ is reduced by the lanthanoid contraction, resulting in almost identical radii.

• The other pairs, Y-La and Cr-Mo, do not exhibit this effect to the same extent, and their atomic radii differ significantly.

Therefore, the pairs with nearly the same atomic radius are $\mathrm{Zr}-\mathrm{Hf}$ and $\mathrm{Mo}-\mathrm{W}$.

$E_{M^{3+} / M^{2+}}^{\circ}$ is highest for Mn i.e., $=+1.57 \mathrm{~V}$

It is due to stability of $\mathrm{Mn}^{2+}$ ion which has a half filled $d$-orbitals.

Statements I and IV are correct, while II and III are incorrect. Their correct form is,

II $\mathrm{Eu}^{2+}$ and $\mathrm{Yb}^{2+}$ acts as reducing agent.

IV Misch metal is primarily composed of lanthanoids metal ( $\sim 95 \%$ ) and iron ( $\sim 5 \%$ ) and traces of S, C, Ca and Al.

The correct match is A-III, B-IV, C-I, D-II.

A. Ostwald's process is an industrial process to make nitric acid. Catalyst used is Pt gauge.

B. Haber's process is industrial process to make ammonia. Iron is used as catalyst in the process.

C. Deacon's process is used to make chlorine. $ \mathrm{CuCl}_{2} $ is used as catalyst.

D. Cracking of hydrocarbon is done in presence of zeolite catalyst.

Total 3 elements, $ \mathrm{Cu}, \mathrm{Gd} $ and Cm have half filled $ f $-orbitals.

The alloy formed by beryllium with copper ($X = \text{Cu}$) is commonly used in the preparation of high-strength springs. This alloy is known as beryllium copper and is prized for its excellent combination of strength, electrical conductivity, and resistance to fatigue and corrosion, making it ideal for various applications including spring manufacturing.

Therefore, the correct option is:

Option C: Cu

In the given options, statement (c) is correct, while the others contain inaccuracies. Here are the correct versions for the incorrect options:

For option (a), the correct order of standard electrode potential for the reaction $ M^{3+}/M^{2+} $ in volts is: $\mathrm{Cr} < \mathrm{Fe} < \mathrm{Mn}\, (-0.41\,V, +0.77\,V, +1.57\,V) $

For option (b), the correct order of magnetic moments is: $\mathrm{Cr}^{2+}(d^4) = \mathrm{Fe}^{2+}(d^6) < \mathrm{Mn}^{2+}(d^5) $

For option (d), the correct order of the first ionization enthalpy in $\mathrm{kJ/mol}$ is: $\mathrm{V} < \mathrm{Cr} < \mathrm{Ti} \, (650 < 653 < 656)$

Hs, Hg, W elements obey $ (n-1) d^{1-10} n s^{2} $ electronic configuration. Hs - $ 108-[\mathrm{Rn}] 5 f^{14} 6 d^{6} 7 s^{2} $

$ \mathrm{Hg}-80-[\mathrm{Xe}] 4 f^{14} 5 d^{10} 6 s^{2} $

$ \mathrm{W}-74-[\mathrm{Xe}] 4 f^{14} 5 d^{4} 6 s^{2} $

The complete reaction is as follows,

$ \begin{gathered} \mathrm{P}_4+3 \mathrm{NaOH}+3 \mathrm{H}_2 \mathrm{O} \xrightarrow{\mathrm{CO}_2} \underset{\text { (Product } X \text { ) }}{ \mathrm{PH}_3 }\begin{array}{c} \end{array}+3 \mathrm{NaH}_2 \mathrm{PO}_2 \\ 4 \mathrm{PH}_3+6 \mathrm{CuSO}_4 \longrightarrow 6 \mathrm{H}_2 \mathrm{SO}_4+\underset{\text { (Product } Y \text { ) }}{2 \mathrm{Cu}_3 \mathrm{P}_2} \end{gathered} $

Hence, product $Y$ is $\mathrm{Cu}_3 \mathrm{P}_2$.

Due to inert pair effect, $\mathrm{Tl}(+1)$ is more stable than $\mathrm{Tl}(+3)$. The valence ' $s$ ' electrons acts as inner core electrons due to poor shielding of ' $d$ ' and ' $f$ ' electrons.

This is the reason Tl is more stable in +1 oxidation state.

Let's examine each statement carefully:

(i) "Ti(IV) is more stable than Ti(III) and Ti(II)."

True. Among titanium’s common oxidation states (+2, +3, +4), the +4 state is generally the most stable due to the high lattice or coordination stabilization and the noble-gas-like configuration of Ti^(4+) (3d^0).

(ii) "Among 3d-series elements (Z = 22 to 29), only copper has a positive reduction potential (M^(2+)/M)."

True. Standard reduction potentials for most 3d metals (like Ti, V, Cr, Mn, Fe, Co, Ni, Zn) are negative. Copper is unique with a standard reduction potential of about +0.34 V, making it the only one in that range with a positive reduction potential for the M^(2+)/M couple.

(iii) "Both Sc and Zn exhibit +1 oxidation state."

False.

Sc typically shows +3 oxidation state (it is isoelectronic with Ar when Sc^(3+)). Instances of Sc in +1 or +2 are either extremely rare or not well-established.

Zn commonly shows +2. While there are organometallic complexes where Zn can appear in unusual oxidation states, Zn(+1) is not a stable or typical oxidation state in standard inorganic chemistry contexts.

Conclusion

(i) is correct

(ii) is correct

(iii) is incorrect

Hence, the correct statements are (i) and (ii) only, matching Option A.

The thermal decomposition of ammonium dichromate gives nitrogen gas, chromium (III) oxide and water vapours.

A good reducing agent oxidises itself readily. Eu and Yb act as reducing agent in +2 oxidation state by using two electron and forming $\mathrm{Eu}^{2+}$ and $\mathrm{Yb}^{2+}$, these will get stable half-filled [Xe]4f ${ }^7$ and fully-filled ( $[\mathrm{Xe}], 4 f^{14}$ ) configuration respectively.

$ \begin{aligned} & { }_{25} \mathrm{Mn}=[\mathrm{Ar}] 3 d^5 4 s^2 \\ & \mathrm{Mn}^{2+}=[\mathrm{Ar}] 3 d^5 4 s^0 \end{aligned} $

$ \mathrm{Mn}^{3+}=[\mathrm{Ar}] 3 d^{4+} 4 s^0 $

Hence, $\mathrm{Mn}^{2+}$ due to half-filled orbital is more stable than $\mathrm{Mn}^{3+}$.

Thus, $\mathrm{Zn}^{2+}, \mathrm{Sc}^{3+}, \mathrm{Ti}^{4+}$ and $\mathrm{Cu}^{+}$have zero magnetic moment.

The melting points of given elements are as follows :

$ \begin{array}{ll} \text { Element } & \text { Melting point } \\ \hline \mathrm{W} & 3422^{\circ} \mathrm{C} \\ \hline \mathrm{Cr} & 1907^{\circ} \mathrm{C} \\ \hline \text { Mo } & 2623^{\circ} \mathrm{C} \\ \hline \end{array} $

Hence, correct order of melting point is $\mathrm{W}>\mathrm{Mo}>\mathrm{Cr}$.

To determine the number of $f$-electrons in the +3 oxidation state of gadolinium (Gd) and +2 oxidation state of ytterbium (Yb), we can examine their electronic configurations:

Gadolinium (Gd):

Atomic number: 64

Electronic configuration: $[\mathrm{Xe}] \, 4f^7 \, 5d^1 \, 6s^2$

In the +3 oxidation state ($\text{Gd}^{3+}$), gadolinium loses three electrons. These electrons are typically removed from the outermost orbitals, i.e., $6s$ and $5d$, leaving the $f$-orbital unchanged.

Configuration of $\text{Gd}^{3+}$: $[\mathrm{Xe}] \, 4f^7$

Number of $f$-electrons ($x$): 7

Ytterbium (Yb):

Atomic number: 70

Electronic configuration: $[\mathrm{Xe}] \, 4f^{14} \, 6s^2$

In the +2 oxidation state ($\text{Yb}^{2+}$), ytterbium loses two electrons, typically from the $6s$ orbital, leaving the $f$-orbital full.

Configuration of $\text{Yb}^{2+}$: $[\mathrm{Xe}] \, 4f^{14}$

Number of $f$-electrons ($y$): 14

Finally, the sum of $x$ and $y$ is:

$ x + y = 7 + 14 = 21 $

Mo atomic number is 42 , so the ground state electronic configuration can be written as

Exchange energy is energy liberated when 2 or more electrons with similar spin exchange their positions in the degenerate orbitals.

More the number of unpaired electrons, more will be the exchange energy.

As both $4 d$ and $5 s$-orbitals in Mo are exactly half-filled and it has 5 unpaired electrons in $d$-orbitals which are degenerate.

∴ Will have highest exchange energy amongst the second row elements which might have either less than 5 electrons in $d$-orbitals or more than 5 electrons in $d$-orbitals.

Hence, both (A) and (R) is true but (R) is not the correct explanation for (A). The highest exchange energy is due to half-filled electronic configuration.

Atomic number of Fe is 26. Electronic configuration (Ground state)

$ \begin{aligned} {[\mathrm{Fe}] } & =[\mathrm{Ar}] 4 s^2 3 d^6 \\ {\left[\mathrm{Fe}^{2+}\right] } & =[\mathrm{Ar}] 4 s^0 3 d^6 \\ {\left[\mathrm{Fe}^{3+}\right] } & =[\mathrm{Ar}] 4 s^0 3 d^5 \end{aligned} $

$\mathrm{Fe}^{3+}$ has exactly half-filled $d$-orbital. $\therefore \mathrm{Fe}^{3+}$ is more stable than $\mathrm{Fe}^{2+}$.

$ \begin{aligned} & \mathrm{Zn}^{2+}=[\mathrm{Ar}] 3 d^{10} \\ & \mathrm{Cd}^{2+}=[\mathrm{Kr}] 4 d^{10} \\ & \mathrm{Hg}^{2+}=[\mathrm{Xe}] 4 f^{14} 5 d^{10} \end{aligned} $

All the above elements in their " +2 " oxidation state possess $d^{10}$ electronic configuration.

The increase in atomic radii of $5 d$-series of transition elements is very small due to filling of $4 f$-orbitals before $5 f$-orbitals. This is due to lanthanoid contraction. The $4 f$-shell has weak shielding effect and is highly attracted towards nucleus due to high nuclear charge.

Orbitals are filled according to the Aufbau principle, in order of their energy which is decided by the ( $n+l$ ) value. Thus, $4 s$ orbital is filled prior to $3 d$ orbital. Hence, the expected outer shell electronic configuration of Cr and Cu are $3 d^4 4 s^2$ and $3 d^9 4 s^2$ respectively. This is but their actual configurations are $3 d^5 4 s^1$ and $3 d^{10} 4 s^1$ respectively. This is because half-filled and fully-filled configurations impart extra stability.

(A) On moving from La to Lu , size of cation decreases due to poor shielding of $f$-orbitals. This contraction is called lanthanoide contraction. As size of cation decreases, its polarising power increases thus, covalent character or properties of lanthanoide metal hydroxides increases.

(B) The first few lanthanoids are quite reactive like calcium but with increasing atomic number, their reactivity becomes similar to Al, i.e. reactivity decreases. This is because of lanthanoid contraction. The ionisation becomes difficult as size of atom decreases. Hence, chemical reactivity decreases from La to Lu.

(C) From La to Lu , size of metal cation decreases, i.e. covalent character in metal hydroxides increases and basic strength decreases. Thus, $\mathrm{La}(\mathrm{OH})_3$ can easily give $\mathrm{OH}^{-}$ions. Hence, $\mathrm{La}(\mathrm{OH})_3$ is more basic than $\mathrm{Lu}(\mathrm{OH})_3$.

(D) On moving down the group from Zr to Hf, size of atom should increases. But due to lanthanoide contraction, Zr and Hf have almost same radius. This can explained on the basis of shielding effect. Due to presence of $f$-orbitals and their poor shielding, effective nuclear charge on the outer shell electron increases and as a result, size of Zr and Hf becomes equal.

(E) Due to lanthanoid contraction, the ionic radii of lanthanoides decreases which also affect their chemistry. Thus, the separation of lanthanoides from one another is not easy.

On treating $\mathrm{SO}_2$ with aqueous solution of $\mathrm{KMnO}_4$, Manganese in potassium permanganate is reduced to +2 oxidation state $\left(\mathrm{Mn}^{2+}\right)$ from +7 oxidation state $\left(\mathrm{Mn}^{7+}\right)$. Also, sulphur in $\mathrm{SO}_2$ is oxidised from +4 to +6 oxidation state $\left(\mathrm{SO}_4^{2-}\right)$.

On moving down the group, size of ion increases thus, $\mathrm{Y}^{3+}$ is smallest in size than $\mathrm{La}^{3+}, \mathrm{Lu}^{3+}$ and $\mathrm{Eu}^{3+}$.

On moving left to right in lanthanoids, size of cation decreases because of lanthanoid contraction. Hence, the correct order of ionic radii is $\mathrm{Y}^{3+}<\mathrm{Lu}^{3+}<\mathrm{Eu}^{3+}<\mathrm{La}^{3+}$.

$\mathrm{KMnO}_4$ decomposes on heating at 513 K to give $\mathrm{K}_2 \mathrm{MnO}_4$ and $\mathrm{MnO}_2$. Oxidation state of Mn is higher in $\mathrm{K}_2 \mathrm{MnO}_4$, i.e. +6 .

$ \mathrm{Mn}^{+6} \longrightarrow[\mathrm{Ar}] 3 d^1 4 s^0 $

So, $\mathrm{K}_2 \mathrm{MnO}_4$ is paramagnetic in nature and

also, its colour is green.

The melting points of transition metals are generally high because they have higher number of electrons (as compared to $s$-block metals) which form metal-metal bond. More the number of metal-metal bonds, higher is the melting point.

$\mathrm{La}^{3+}(Z=57)$ has 54 electrons.

$\mathrm{Ti}^{3+}(Z=22)$ has 19 electrons.

$\mathrm{Lu}^{3+}(Z=71)$ has 68 electrons.

and $\mathrm{Sc}^{3+}(Z=21)$ has 18 electrons.

The species with unpaired electrons show colour in aqueous solution.

Because, $\quad \mathrm{Ti}^{3+}=[\mathrm{Ar}] \cdot 4 s^1$

Thus, has one unpaired electron.

Hence, option (b) is the correct answer.

Trans uranium elements are elements after uranium, i.e. trans uranium elements have atomic number, $Z>92$.

The highest oxidation state observed in the first row transition metals is +7 .

In first row ' $\mathrm{Mn}^{\prime}$ with outer configuration : $3 d^5, 4 s^2$ show +7 oxidation state, i.e.

It has 7 electrons in outermost orbital, so has +7 oxidation state.

Among the given series of transition metal ions, one in which all metal ions have $3 d^2, 3 p^6$ electronic configuration is :

Option (d), i.e.

$\mathrm{Ti}(Z=22) \rightarrow \mathrm{Ti}^{2+}$ (has 20 electrons)

$\mathrm{V}(\mathrm{Z}=23) \rightarrow \mathrm{V}^{3+}$ (has 20 electrons)

$\mathrm{Cr}(Z=24) \rightarrow \mathrm{Cr}^{4+}$ (has 20 electrons)

and $\mathrm{Mn}(Z=25) \rightarrow \mathrm{Mn}^{5+}$ (has 20 electrons)

Thus, all can show $3 d^2 3 p^6$ electronic configuration i.e. option (d) is the correct answer.

Interstitial compounds are formed, when small atoms are trapped inside the crystal lattice of metals.

(i) They are very hard and rigid.

(ii) They have high melting point, which are higher than these of the pure metals.

(iii) They show conductivity like the of the pure metals.

(iv) They aquire chemical inertness.