Answer

(b) Electronic configuration of $X^{3+}$ is $[Ar] 3 d^{5}$

It repersents the total number of $e^{s}$ and oxidation state.

Therefore, atomic number of $X=18+5+3=26$

Hence, option (b) is correct.

Answer

(a) $Cu(II)$ is more stable than $Cu(I)$. As it is known that, $Cu(I)$ has $3 d^{10}$ stable configuration while $Cu(II)$ has $3 d^{9}$ configuration. But $Cu(II)$ is more stable due to greater effective nuclear charge of $Cu(II)$ i.e., it hold 17 electrons instead of 18 in $Cu(I)$.

Answer

(d) On moving left to right along period, metallic radius decreases while mass increases. Decrease in metallic radius coupled with increase in atomic mass results in increase in density of metal.

Hence, among the given four choices $Cu$ belongs to right side of Periodic Table in transition metal, and it has highest density $(89 g / cm^{3})$.

Answer

(b) Transition elements form coloured salt due to the presence of unpaired electrons. In $CuF_2, Cu(II)$ contain one unpaired electron hence, $CuF_2$ is coloured in solid state.

Answer

(a) On addition of $KMnO_4$ to concentrated $H_2 SO_4$, a green oily compound $Mn_2 O_7$ is obtained which is highly explosive in nature.

$$ 2 KMnO_4+2 H_2 SO_4 \text { (Conc.) } \longrightarrow Mn_2 O_7+2 KHSO_4+H_2 O $$

Thinking Process

This problem is based on calculation of magnetic moment can be done as Magnetic moment $(\mu)=\sqrt{n(n+2)} B M$.

Answer

(b) Greater the number of unpaired electron, higher will be its value of magnetic moment. Since, $3 d^{5}$ has 5 unpaired electrons hence highest magnetic moment.

$$ \begin{aligned} \mu & =\sqrt{5(5+2)} \\ & =\sqrt{35} \\ & =5.95 BM \end{aligned} $$

Answer

(b) Lanthanoids show common oxidation state of +3 . Some of which also show +2 and +4 stable oxidation state alongwith +3 oxidation state. These are shown by those elements which by losing 2 or 4 electrons acquire a stable configuration of $f^{0}, f^{7}$ or $f^{14}$, e.g., $Eu^{2+}$ is $[Xe] 4 f^{7}, Yb^{2+}$ is $[Xe] 4 f^{14}, Ce^{4+}$ is $[Xe] 4 f^{0}$ and $Tb^{4 f}$ is $[Xe] 4 f^{7}$.

Answer

(a) The reaction in which oxidation as well as reduction, occur upon same atom simultaneously is known as disproportionation reaction.

Answer

(d) When $KMnO_4$ solution is added to oxalic acid solution, the decolourisation is slow in the beginning but becomes instantaneous after some time because $Mn^{2+}$ acts as autocatalyst.

Reduction half $.MnO_4^{-}+8 H^{+}+5 e^{-} \longrightarrow Mn^{2+}+4 H_2 O] \times 2$

Oxidation half $.\quad C_2 O_4^{2-} \longrightarrow 2 CO_2+2 e^{-}] \times 5$

Overall equation

$$ 2 MnO_4^{-}+16 H^{+}+5 C_2 O_4^{2-} \longrightarrow 2 Mn^{2+}+10 CO_2+8 H_2 O $$

End point of this reaction Colourless to light pink

Answer

(c) $Tm(Z=69)$ do not belong to actinoid series. The actinoid series is with atomic numbers 90 to 103. Thulium (Tm) has atomic number 69 belongs to lanthanoids (4f series).

Answer

The reaction of $KMnO_4$ in which it acts as an oxidising agent in acidic medium is

$2 KMnO_4+3 H_2 SO_4 \longrightarrow K_2 SO_4+2 MnSO_4+3 H_2 O+5[O]$

$$ \frac{H_2 S+[O] \longrightarrow H_2 O+S] \times 5}{2 KMnO_4+3 H_2 SO_4+5 H_2 S \longrightarrow K_2 SO_4+2 MnSO_4+8 H_2 O+5 S} $$

5 moles of $S^{2-}$ ions react with 2 moles of $KMnO_4$. So, 1 mole of $S^{2-}$ ion will react with $\frac{2}{5}$ moles of $KMnO_4$

Answer

(a) $V_2 O_5$ and $Cr_2 O_3$ are amphoteric oxide because both react with alkalies as well as acids.

Note In lower oxides, the basic character is predominant while in higher oxides, the acidic character is predominant.

Answer

(a) Gadolinium belongs to $4 f$ series and has atomic number 64 . The correct electronic configuration of gadolinium is

$$ _{64} Gd= _{54}[Xe] 4 f^{7} 5 d^{1} 6 s^{2} $$

It has extra stability due to half-filled $4 f$ subshell.

Answer

(d) Interstitial compounds are formed when small atoms are trapped inside the crystal lattice of metals. Some of their important characteristics are as follows

(i) They are very hard and rigid.

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

(iii) They show conductivity like that of the pure metal.

(iv) They acquire chemical inertness.

Answer

(b) The magnetic moment is associated with its spin angular momentum and orbital angular momentum.

Spin only magnetic moment value of $Cr^{3+}$ ion is $3 d^{3}$

Hence, magnetic moment

$$ \begin{aligned} (\mu) & =\sqrt{n(n+2)} BM \\ & =\sqrt{3(3+2)}=\sqrt{15} \\ & =3.87 BM \end{aligned} $$

Answer

(c) $KMnO_4$ acts as an oxidising agent in alkaline medium. When alkaline $KMnO_4$ is treated with $KI$, iodide ion is oxidised to $IO_3^{-}$.

Reaction $ \quad \quad \quad \quad 2 \mathrm{KMnO}_4+\mathrm{H}_2 \mathrm{O}+\mathrm{KI} \longrightarrow 2 \mathrm{MnO}_4+2 \mathrm{KOH}+\mathrm{KIO}_3$

or, $\quad \quad v\quad \mathrm{I}^{-}+6 \mathrm{OH}^{-} \longrightarrow \mathrm{IO}_3^{-}+3 \mathrm{H}_2 \mathrm{O}+6 \mathrm{e}^{-}$

Answer

(a) Copper lies below hydrogen in the electrochemical series and hence does not liberate $H_2$ from acids. Therefore, option (a) is not correct.

Other three options ( $b, c, d)$ are correct.

Answer

(c) When acidified $K_2 Cr_2 O_7$ solution is added to $Sn^{2+}$ salts then $Sn^{2+}$ changes to $Sn^{4+}$. The reaction is given below

Answer

(d) Highest oxidation state of manganese in fluoride is $+4(MnF_4)$ but highest oxidation state in oxides is $+7(Mn_2 O_7)$. The reason is that in covalent compounds fluorine can form single bond while oxygen forms double bond.

Answer

(c) Due to lanthanoide contraction, $Zr$ and $Hf$ possess nearly same atomic and ionic radii i.e., $Zr=160 pm$ and $Hf=159 pm, Zr^{4+}=79 pm$ and $Hf^{4+}=78 pm$. Therefore, these two elements show similar properties (physical and chemical properties).

Answer

(b) $HCl$ is not used to make the medium acidic in oxidation reactions of $KMnO_4$ in acidic medium. The reason is that if $HCl$ is used, the oxygen produced from $KMnO_4+HCl$ is partly utilised in oxidising $HCl$ to $Cl$, which itself acts as an oxidising agent and partly oxidises the reducing agent.

Answer

$(a, b)$

$KMnO_4$ is coloured due to the charge transfer and not because of the presence of unpaired electrons. Similarly, oxidation state of $Ce$ in $Ce(SO_4)_2$ is +4 with $4 f^{\circ}$ electronic configuration. It is also coloured (yellow) due to charge transfer and not due to $f-f$ transition.

Answer

$(a, d)$

Electronic configuration of $Co^{2+}=[Ar] 3 d^{7}$; Number of unpaired electrons $=3$

Electronic configuration of $Cr^{2+}=[Ar] 3 d^{4}$; Number of unpaired electrons $=4$

Electronic configuration of $Mn^{2+}=[Ar] 3 d^{5}$; Number of unpaired electrons $=5$

Electronic configuration of $Cr^{3+}=[Ar] 3 d^{3}$; Number of unpaired electrons $=3$

Hence, it is clearly seen that both $Co^{2+}$ and $Cr^{3+}$ have same number of unpaired electrons. i.e., 3 .

Answer

$(b, c)$

In d-block elements, for heavier elements, the higher oxidation states are more stable. Hence, $Mo(VI)$ and $W(VI)$ are more stable than $Cr(VI)$. Thats why, $Cr(VI)$ in the form of dichromate is a stronger oxidising agent in acidic medium whereas $MoO_3$ and $WO_3$ are not.

Answer

$(b, d)$

The oxidation states of the following actinoids are

(a) Americium $(Z=95)$; Electronic configuration $=[R n] 5 f^{7} 6 d^{0} 7 s^{2}$ Oxidation states shown by $A m=+3,+4,+5,+6$.

(b) Plutonium $(Z=94)$; Electronic configuration $=[Rn] 5 f^{6} 6 d^{0} 7 s^{2}$ Oxidation states shown by $Pu=+3,+4,+5,+6,+7$.

(c) Uranium ( $Z=92$ ); Electronic configuration $=[Rn] 5 f^{3} 6 d^{1} 7 s^{2}$ Oxidation states shown by $U=+3,+4,+5,+6$.

(d) Neptunium ( $Z=93$ ); Electronic configuration $=[Rn] 5 f^{4} 6 d^{1} 7 s^{2}$ Oxidation states shown by $Np=+3,+4,+5,+6,+7$.

Answer

( $a, b$ )

General electronic configuration of actinoids is $(n-1) f^{1-14}(n-1) d^{0-2} n s^{2} . U$ and $N p$ each have one electron in $6 d$ orbital. (Also, refer to Q. 25)

Answer

$(b, c)$

(a) Cerium $(Z=57) \Rightarrow$ Electronic configuration $=[Xe] 4 f^{5} 5 d^{0} 6 s^{2}$ Oxidation state of $Ce=+3,+4$

(b) Europium $(Z=63) \Rightarrow$ Electronic configuration $=[Xe] 4 f^{7} 5 d^{0} 6 s^{2}$ Oxidation state of $Eu=+2,+3$

(c) Ytterbium $(Z=70) \Rightarrow$ Electronic configuration $=[Xe] 4 f^{14} 5 d^{0} 6 s^{2}$ Oxidation state of $Yb=+2,+3$

(d) Holmium $(Z=67) \Rightarrow$ Electronic configuration $=[Xe] 4 f^{11} 5 d^{0} 6 s^{2}$ Oxidation state of $Ho=+3$

Answer

$(b, c)$

As,

$$ \begin{aligned} Ti^{3+} & =[Ar] 3 d^{1} \\ Mn^{2+} & =[Ar] 3 d^{5},(t_{2 g}^{3} e_g^{2}) \\ Fe^{2+} & =[Ar] 3 d^{6}(t_{2 g}^{4} e_{2 g}^{2}) \\ Co^{3+} & =[Ar] 3 d^{6}(t_{2 g}^{6} e_g^{0}) \end{aligned} $$

Crystal field splitting energy (CFSE) is high in $Co^{3+}$, thus electrons pair up in $t_{2 g}$. Hence, only $Fe^{2+}$ and $Mn^{2+}$ show higher spin magnetic moment value.

Answer

(a, b)

Transition elements such as $Cr$ and $Co$ form binary compounds with halogens, i.e., $CrF_3$ and $CoF_3$ whereas $Cu$ and $Ni$ do not form $CuF_3$ and $NiF_3$.

Answer

$(b, c)$

A species can act as oxidising agent only when metal is present in high oxidation state but lower oxidation state show stability. As higher oxidations states of $W$ and $Mo$ are more stable, therefore they will not act as oxidising agents.

Answer

$(b, c)$

Electronic configuration of $ _{58} Ce= _{54}[Xe] 4 f^{2} 5 d^{0} 6 s^{2}$.

Therefore, electronic configuration of $Ce^{4+}= _{54}[Xe] 4 f^{0}$.

Thus, it has a tendency to attain noble gas configuration and attain $f^{0}$ configuration.

Answer

Copper not replace hydrogen from acids because $Cu$ has positive $E^{\circ}$ value, i.e., less reactive than hydrogen which has electrode potential $0.00 V$.

Answer

Negative values of $Mn^{2+}$ and $Zn^{2+}$ are related to stabilities of half-filled and completely filled configuration respectively. But for $Ni^{2+}, E^{\circ}$ value is related to the highest negative enthalpy of hydration.

Hence, $E^{s}$ values for $Mn, Ni$ and $Zn$ are more negative than expected.

Answer

Ionisation enthalpy of $Cr$ is less than that of $Zn$ because $Cr$ has stable $d^{5}$ configuration. In case of zinc, electron comes out from completely filled $4 s$-orbital. So, removal of electron from zinc requires more energy as compared to the chromium.

Answer

Transition elements show high melting point, due to involvement of greater number of electrons in the interatomic bonding from $(n-1) d$-orbitals in addition to $n s$ electrons in forming metallic bond. Thus, large number of electrons participate forming large number of metallic bond.

Answer

When $Cu^{2+}$ ion is treated with $KI$, it produces $Cu_2 I_2$ white precipitate in the final product.

$$ 2 Cu^{2+}+4 I^{-} \longrightarrow \underset{\text { (White ppt.) }}{Cu_2 I_2}+I_2 $$

(In this reaction, $CuI_2$ is formed which being unstable, dissociates into $Cu_2 I_2$ and $I_2$ ).

Answer

Among $Cu_2 Cl_2$ and $CuCl_2, CuCl_2$ is more stable. Stability of complex can be explained on the basis of stability of oxidation state of copper. Stability of $Cu^{2+}(a q)$ rather than $Cu^{+}(a q)$ is due to much more negative value of $[\Delta_{hyd} H^{s}.$ of $.Cu^{2+}(a q)]$ than $Cu^{+}$which more than compensates for the second ionisation enthalpy of $Cu$.

Thinking Process

This problem is based on the properties of $MnO_2$ and preparation of $NCl_3$.

Answer

$MnO_2$ is the brown compound of $Mn$ which reacts with $HCl$ to give $Cl_2$ gas. This gas forms an explosive compound $NCl_3$ when treated with $NH_3$. Thus, $A=MnO_2 ; B=Cl_2 ; C=NCl_3$ and reactions are as follows

(i) $\underset{[A]}{MnO_2+4 HCl} \longrightarrow MnCl_2+\underset{[B]}{Cl_2}+2 H_2 O$

(ii) $NH_3+\underset{\text { (Excess) }}{3 Cl_2} \longrightarrow \underset{\text { [C] }}{NCl_3}+3 HCl$

Answer

Oxygen can form multiple bonds with metals, while fluorine can’t form multiple bond with metals. Hence, oxygen has more ability to stabilize higher oxidation state rather than fluorine

Answer

Magnetic moment of any metal ion can be decided on the basis of spin as well as orbital contribution of electron. Due to symmetrical electronic configuration, there is no orbital contribution in $Cr^{3+}$ ion. However, appreciable orbital contribution takes place in $Co^{2+}$ ion.

Answer

$Ce, Pr$ and $Nd$ are lanthanoids and have incomplete $4 f$ shell while $Th, Pa$ and $U$ are actinoids and have $5 f$ shell incomplete.

In the beginning, when 5 -orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The $5 f$-electrons will therefore, be more effectively shielded from the nuclear charge than $4 f$ electrons of the corresponding lanthanoids.

Therefore, outer electrons are less firmly held and they are available for bonding in the actinoids

Answer

Separation of $Zr$ and $Hf$ are quite difficult because of lanthanoid contraction. Due to lanthanoid contraction, they have almost same size $(Zr=160 pm$ and $Hf=159 pm)$ and thus, similar chemical properties. That’s why it is very difficult to separate them by chemical methods.

Answer

It is due to the fact that after losing one more electron Ce acquires stable $4 f^{\circ}$ electronic configuration. So, Ce shows +4 oxidation state also alongwith +3 oxidation state.

Answer

When oxalic acid is added to acidic solution of $KMnO_4$, its colour disappear due to reduction of $MnO_4^{-}$ion to $Mn^{2+}$. Chemical reaction occurring during this neutralisation reaction is as follows

$$ 5 C_2 O_4^{2-}+\underset{\text { (Coloured) }}{2 MnO_4^{-}}+16 H^{+} \longrightarrow \underset{\text { (Colourless) }}{2 Mn^{2+}}+8 H_2 O+10 CO_2 $$

Answer

When orange solution containing $Cr_2 O_7^{2-}$ ion is treated with an alkali, a yellow solution of $\mathrm{CrO}_4^{2-}$ is obtained. On the same way, $\underset{\substack{\text { Dischromate } \\ \text { (orange) }}}{\mathrm{Cr}_2 \mathrm{O}_7^{2-}} \xrightarrow[\mathrm{H}^{+}]{\mathrm{OH}^{-}} \underset{\begin{array}{l}\text { Chromate } \\ \text { (yellow) }\end{array}}{\mathrm{CrO}_4{ }^{2-}}$

when $H^{+}$ions are added to yellow solution, an orange solution is obtained due to interconversion.

Answer

Oxidising behaviour of $KMnO_4$ depends on $pH$ of the solution.

In acidic medium ( $pH<7$ )

$$ MnO_4^{-}+8 H^{+}+5 e^{-} \longrightarrow \underset{\text { Colourless }}{Mn^{2+}}+4 H_2 O $$

In alkaline medium $(pH>7)$

$$ MnO_4^{-}+e^{-} \longrightarrow \underset{(\text { Green })}{MnO_4^{2-}} $$

In neutral medium $(pH=7)$

$$ MnO_4^{-}+2 H_2 O+3 e^{-} \longrightarrow \underset{\text { (Brown ppt) }}{MnO_2}+4 OH^{-} $$

Answer

Due to lanthanoid contraction, the atomic radii of the second and third row transition elements is almost same. So, they resemble each other much more as compared to first row elements and show similar character.

Answer

$E^{S}$ value of $Cu$ is positive because of the fact that sum of sublimation enthalpy and ionisation enthalpy to convert $Cu(s)$ to $Cu^{2+}(a q)$ is so high that it is not compensate by its hydration enthalpy. $E^{\circ}$ value for $Zn$ is negative because of the fact that after removal of electrons from $4 s$ orbital, stable $3 d^{10}$ configuration is obtained.

Thinking Process

This problem is based on concept of Fajan’s rule and its application.

Answer

As the oxidation state increases, size of the ion of transition element decreases. As per Fajan’s rule, as the size of metal ion decreases, covalent character of the bond formed increases.

Therefore, the halide of transition elements become more covalent with increasing oxidation state of the metal.

Answer

During filling up of electrons follow $(n+l)$ rule. Here $4 s$ has lower energy than $3 d$ orbital. After the orbitals are filled $4 s$ goes beyond $3 d$, i.e., $4 s$ is farther from nucleus than $3 d$. So, electron from $4 s$ is removed earlier than from $3 d$.

Answer

Reactivity of transition elements depends mostly upon their ionisation enthalpies. As we move from left to right in the periodic table ( $Se$ to $Cu$ ), ionisation enthalpies increase almost regularly.

Hence, their reactivity decreases almost regularly from $Se$ to $Cu$.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(4)$

C. $\rightarrow$ (2)

E. $\rightarrow(1)$

Answer

A. $\rightarrow(2)$

B. $\rightarrow(1)$

C. $\rightarrow(4)$

D. $\rightarrow(5)$

E. $\rightarrow(3)$

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(2)$

A. Osmium is an element which can show +8 oxidation state.

B. $3 d$ block element that can show upto +7 oxidation state is manganese.

C. $3 d$ block element with highest melting point is chromium.

Answer

A. $\rightarrow(3)$

B. $\rightarrow(1)$

C. $\rightarrow(5)$

D. $\rightarrow(2)$

A. Oxidation state of $Mn$ in $MnO_2$ is +4 .

B. Most stable oxidation state of $Mn$ is +2 .

C. Most stable oxidation state of $Mn$ in oxides is +7 .

D. Characteristic oxidation state of lanthanoids is +3 .

Answer

A. $\rightarrow(4)$

B. $\rightarrow(1) \quad$

C. $\rightarrow(2) \quad$

D. $\rightarrow(5)$

E. $\rightarrow(6)$

Answer

A. $\rightarrow(2)$

B. $\rightarrow(4)$

C. $\rightarrow(1)$

D. $\rightarrow(5)$

E. $\rightarrow(3)$

A. Lanthanoid which shows +4 oxidation state is cerium.

$ _{58} Ce=[Xe] 4 f^{2} 5 d^{0} 6 s^{2} ;$ Oxidation state $=+3,+4$

B. Lanthanoid which can show +2 oxidation state is europium.

$ _{63} Eu=[Xe] 4 f^{7} 5 d^{0} 6 s^{2} ;$ Oxidation state $=+2,+3$

C. Radioactive lanthanoid is promethium. It is the only synthetic (man-made) radioactive lanthanoid.

D. Lanthnoid which has $4 f^{7}$ electronic configuration in +3 oxidation state is gadolinium.

$ _{64} Gd=[Xe] 4 f^{7} 5 d^{1} 6 s^{2} ;$ Oxidation state $=+3$

E. Lanthanoid which has $4 f^{14}$ electronic configuration in +3 oxidation state is lutetium

$ _{71} Lu=[Xe] 4 f^{14} 5 d^{1} 6 s^{2} ;$ Oxidation state $=+3$

Answer

A. $\rightarrow(3)$

B. $\rightarrow(4)$

C. $\rightarrow(2)$

D. $\rightarrow(1)$

A. $Cu^{+}=3 d^{10}$ which is very stable configuration due to full-filled orbitals. Hence, removal of second electron requires very high energy.

B. $Zn^{2+}=3 d^{10}$ which is very stable configuration. Hence, removal of third electron requires very high energy.

C. Metal carbonyl with formula $M(CO)_6$ is $Cr(CO)_6$.

D. Nickel is the element with highest heat of atomisation.

Answer

(a) Assertion and reason both are correct and reason is the correct explanation of assertion.

Copper (II) iodide $(CuI_2)$ is not known because $Cu^{2+}$ oxidises $I^{-}$to lodine.

Answer

(b) Assertion and reason are true but reason is not correct explanation of assertion.

Separation of $Zr$ and $Hf$ is difficult; it is not because of they lie in the same group of Periodic Table. This is due to lanthanoid contraction which causes almost similar radii of both of them.

Answer

(c) Assertion is not true but reason is true.

Actinoids form relatively more stable complexes as compared to lanthanoids because of actinoids can utilise their $5 f$ orbitals along with $6 d$ orbitals in bonding but lanthanoids do not use their $4 f$ orbitals for bonding.

Answer

(a) Assertion and reason both are correct and reason is correct explanation of assertion. Cu can not liberate hydrogen from acids because it has positive electrode potential. Metals having negative value of electrode potential liberate $H_2$ gas.

Answer

(b) Assertion and reason both are correct but reason is not the correct explanation of assertion.

The highest oxidation state of osmium is +8 due to its ability to expand their octet by using its all 8 electrons ( 2 from 6 s and 6 from $5 d$ ).

Answer

The substances from $A$ to $E$ are

$$ A=Cu ; B=Cu(NO_3)_2 ; C=[Cu(NH_3)_4]^{2+} ; D=CO_2 ; E=CaCO_3 $$

Reactions:

(i) $\mathrm{{CuCO_3 \xrightarrow{\Delta} CuO+CO_2] \times 2}}$

(ii) $\mathrm{2 CuO+CuS \longrightarrow \underset{\text{[A]}}{3 Cu}+SO_2}$

(iii) $\mathrm{\underset{\text{[A]}}{Cu}+4 HNO_3 (conc.) \longrightarrow \underset{\text{[B]}}{Cu(NO_3)_2}+2 NO+2 H_2 O}$

(iv) $\mathrm{\underset{\text{[B]}}{Cu^{2+}}+NH_3 \longrightarrow \underset{\substack{ \text{[C]} \\ \text { (Blue Solution) }}}{[Cu(NH_3)_4]}}$

(v) $\mathrm{Ca}(\mathrm{OH})_2+ \underset{\text{[D]}}{\mathrm{CO}_2} \longrightarrow \underset{\substack{\text{[E]} \\ \text{(Milky)}}}{\mathrm{CaCO}_3}+\mathrm{H}_2 \mathrm{O}$

(vi) $\mathrm{CaCO_3+H_2 O+CO_2 \longrightarrow Ca(HCO_3)_2}$

Answer

$K_2 Cr_2 O_7$ is an orange compound. It is formed when $Na_2 Cr_2 O_7$ reacts with $KCl$. In acidic medium, yellow coloured $CrO_4^{2-}$ (chromate ion) changes into dichromate.

The given process is the preparation method of potassium dichromate from chromite ore.

$$ \mathrm{A=FeCr_2 O_4 ; B=Na_2 CrO_4 ; C=Na_2 Cr_2 O_7 ; D=K_2 Cr_2 O_7 .} $$

(i) $\mathrm{\underset{\text{[A]}}{4 FeCr_2 O_4} + 8 Na_2 CO_3+7 O_2 \longrightarrow \underset{\text{[B]}}{8 Na_2 CrO_4}+2 Fe_2 O_3+8 CO_2}$

(ii) $\mathrm{2 Na_2 CrO_4+2 H^{+} \longrightarrow Na_2 Cr_2 O_7+2 Na^{+}+H_2 O}$

(iii) $\mathrm{\underset{\text{[C]}}{Na_2 Cr_2 O_7}+2 KCl \longrightarrow \underset{\text{[D]}}{K_2 Cr_2 O_7}+2 NaCl}$

Thinking Process

This problem based on the concept of preparation and properties of potassium permanganate.

Answer

It is the method of preparation of potassium permanganate (purple). Thus, $(A)=MnO_2$

$(C)=KMnO_4$

$(B)=K_2 MnO_4$

$(D)=KIO_3$

$$ \begin{aligned} & \underset{[A]}{2 MnO_2}+4 KOH+O_2 \longrightarrow \underset{[B]}{2 K_2 MnO_4}+2 H_2 O \\ & 3 MnO_4^{2-}+4 H^{+} \longrightarrow \underset{[C]}{2 MnO_4^{-}}+MnO_2+2 H_2 O \\ & 2 MnO_4^{-}+H_2 O+KI \longrightarrow \underset{[A]}{2 MnO_2}+\underset{[D]}{2 OH^{-}}+KIO_3 \end{aligned} $$

Answer

(i) As the size decreases covalent character increases. Therefore, $La_2 O_3$ is more ionic and $Lu_2 O_3$ is more covalent.

(ii) As the size decreases from La to Lu, stability of oxo salts also decreases.

(iii) Stability of complexes increases as the size of lanthanoids decreases.

(iv) Radii of $4 d$ and $5 d$-block elements will be almost same.

(v) Acidic character of oxides increases from La to Lu.

Answer

(a) (i) $Cu$, because the electronic configuration of $Cu$ is $3 d^{10} 4 s^{1}$. So, second electron needs to be removed from completely filled $d$-orbital which is very difficult.

(ii) Zinc, because of electronic configuration of $Zn=3 d^{10} 4 s^{2}$ and $Zn^{2+}=3 d^{10}$ which is fully filled and hence is very stable. Removal of third electron requires very high energy.

(iii) Zinc, because of it has completely filled $3 d$ subshell and no unpaired electron is available for metallic bonding.

(b) (i) Carbonyl $M(CO)_5$ is $Fe(CO)_5$

According to EAN rule, the effective number of a metal in a metal carbonyl is equal to the atomic number of nearest inert gas EAN is calculated as

EAN $=$ number of electrons in metal $+2 \times(CO)$ $=$ atomic number of nearest inert gas

$\ln M(CO)_5=x+2 \times(5)=36$ ( $Kr$ is the nearest inert gas)

$x=26$ (atomic number of metal)

So, the metal is $Fe$ (iron).

(ii) $MO_3 F$ is $MnO_3 F$.

In $MO_3 F$

Let us assume that oxidation state of $M$ is $x$

$$ x+3 \times(-2)+(-1)=0 $$

or, $x=+7$ i.e., $M$ is in +7 oxidation state of +7 . Hence, the given compound is $MnO_3 F$.

Answer

When small atoms like $H, C$ and $N$ get trapped inside the crystal lattice of transition metals.

(a) Such compounds are called interstitial compounds.

(b) Their characteristic properties are;

(i) They have high melting points, higher than those of pure metals.

(ii) They are very hard.

(iii) They retain metallic conductivity.

(iv) They are chemically inert.

Answer

(a) Reaction between iodide and persulphate ions is

$ \begin{array}{cccc} & 2 \mathrm{I}^{-}+\mathrm{S}_2 \mathrm{O}_8^{2-} & \xrightarrow{\mathrm{Fe}(\mathrm{III})} & \mathrm{ I_2 + 2SO_4^{2-}} \\ \text { Role of } \mathrm{Fe}(\mathrm{III}) \text { ions } & \mathrm{Fe}^{3+}+2 \mathrm{I}^{-} & \longrightarrow & 2 \mathrm{Fe}^{2+}+\mathrm{I}_2 \\ &2 \mathrm{Fe}^{2+}+\mathrm{S}_2 \mathrm{O}_8^{2-} & \longrightarrow & 2 \mathrm{Fe}^{3+}+2 \mathrm{SO}_4^{2-} \end{array} $

(b) (i) Vanadium ( $V$ ) oxide used in contact process for oxidation of $SO_2$ to $SO_3$.

(ii) Finely divided iron in Haber’s process in conversion of $N_2$ and $H_2$ to $NH_3$.

(iii) $MnO_2$ in preparation of oxygen from $KClO_3$.

Thinking Process

This problem is based on preparation and properties of $KMnO_4 K_2 MnO_4$ and $MnO_2$.

Answer

Since, compound $(C)$ on treating with conc. $H_2 SO_4$ and $NaCl$ gives $Cl_2$ gas, so it is manganese dioxide $(MnO_2)$. It is obtained alongwith $MnO_4^{2-}$ when $KMnO_4$ (violet) is heated.

Thus,

$(A)=KMnO_4$

$(B)=K_2 MnO_4 $

$(C)=MnO_2$

$(D)=MnCl_2$

$$ \begin{matrix} \underset{[A]}{KMnO_4} \stackrel{\Delta}{\longrightarrow} \underset{[B]}{K_2 MnO_{4}}+\underset{[C]}{MnO_2}+O_2 \\ \\ 2 MnO_2+4 KOH+O_2 \longrightarrow 2 K_2 MnO_4+2 H_2 O \\ \\ MnO_2+4 NaCl+4 H_2 SO_4 \longrightarrow MnCl_2+4 NaHSO_4+2 H_2 O+Cl_2 \end{matrix} $$