d and f Block Elements

$\mathrm{PbS+dil.\,HNO_3\to Pb(NO_3)_2+S+NO+H_2O}$

NO is paramagnetic due to the presence of unpaired electron. It is a neutral oxide. It is colourless.

Hence, (A) and (D) are correct statements.

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.

Isoelectronic species have same number of electrons in them.

(a) $\mathrm{Pr}^{3+}$ number of electron $=59-3=56$

$\mathrm{Nd}^{3+}$ number of electron $=60-3=57$

$\therefore \mathrm{Pr}^{3+}$ and $\mathrm{Nd}^{3+}$ are not isoelectronic.

(b) $\mathrm{Tb}^{3+}$ : Total number of electron $=65-3=62$

$\mathrm{Dy}^{2+}$ : Total number of electron $=66-2=64$

$\therefore \mathrm{Tb}^{3+}$ and $\mathrm{Dy}^{2+}$ are not isoelectronic.

(c) $\mathrm{Eu}^{2+}$ number of electron $=63-2=61$ $\mathrm{Gd}^{3+}$ number of electron $=64-3=61$

$\therefore \mathrm{Eu}^{2+}$ and $\mathrm{Gd}^{3+}$ are isoelectronic.

(d) $\mathrm{Pr}^{3+}$ number of electron $=59-3=56$ $\mathrm{Ce}^{4+}$ number of electron $=58-4=54$

$\therefore \mathrm{Pr}^{3+}$ and $\mathrm{Ce}^{4+}$ are not isoelectronic.

Transition metals and their complexes show catalytic activity because of

I. presence of vacant $d$-orbitals.

II. ability to show variable oxidation valencies.

III. tendency to form complex compounds.

A catalyst is a chemical species which increases the rate of the reaction by lowering the activation energy of the reaction.

Thus, Assertion and Reason are correct but Reason is not correct explanation of Assertion.

The element with atomic number 24 is Chromium (Cr). Chromium commonly exhibits several positive oxidation states, which are due to the loss of electrons from both the 4s and 3d orbitals.

Here are the common oxidation states for Chromium :

1. +2 (as in ${Cr}^{2+} $) : This is observed where the electron configuration is $[{Ar}] 3d^4$


2. +3 (as in ${Cr}^{3+} )$ : This is the most stable state, with the electron configuration $[ {Ar} ] 3d^3$


3. +6 (as in ${CrO}_4^{2-} $ and ${Cr}_2\text{O}_7^{2-} )$ : In these cases, chromium exhibits an oxidation state of +6.

Hence, the common positive oxidation states for Chromium are +2, +3, and +6.

Therefore, the correct option is :

Option B : +2 to +6.

(A) ${}_{58}Ce \to [Xe]4{f^2}5{d^0}6{s^2}$

In complex $C{e^{ + 4}} \to [Xe]4{f^0}5{d^0}6{s^0}$

There is no unpaired electron, so, $\mu$_m = 0 BM

(B) ${}_{64}Gd \to [Xe]4{f^7}5{d^1}6{s^2}$

In complex, ${}_{64}G{d^{2 + }} \to [Xe]4{f^7}5{d^1}$

There are 8 unpaired electrons.

$\therefore$ ${\mu _m} = \sqrt {8(8 + 2)} = \sqrt {80} = 8.94BM$

(C) ${}_{63}Eu \to [Xe]4{f^9}5{d^0}6{s^0}$

In complex ${}_{63}E{u^{ + 3}} \to [Xe]4{f^6}5{d^0}6{s^0}$

contain six unpaired electrons.

So, ${\mu _m} = \sqrt {6(6 + 2)} = \sqrt {48} BM = 6.93BM$

Hence, order of spin only magnetic moment

$B > C > A$.

Generally, due to decrease in metallic radius and increase in atomic mass density increase across the period from left to right.

Correct order is Cu > Co > Fe > Cr > Zn.