NCERT Solutions for Class 12 Science Chemistry Chapter 8 - The D And F Block Elements

Answer:

X is in +3 oxidation state that means the X³⁺ has been formed by the less of 3 electrons. As a result, the ground state electronic configuration is [Ar]3d⁵

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

Answer:

Cu(II) is more stable than Cu(I). As it is known that, Cu(I) has 3d¹⁰ stable configuration whereas Cu(II) has 3d⁹ configuration. Cu(II) is more stable than Cu(II) due to greater effective nuclear charge. It releases more energy and therefore is more stable.

Hence, the correct answer is option (i).

Answer:

On moving left to right, mass increases while metallic radius decreases. Decrease in metallic radius coupled with an increase in atomic mass results in an increase in the density of the metal.
Among the given choices, Cu belongs to the right side of periodic table in transition metal, and it must have the highest density.

Hence, the correct answer is option (iv).

Answer:

The transition elements form coloured salt due to the presence of unpaired electrons. CuF₂ Cu(II) contain one unpaired electron. CuF₂ is coloured in sold state.

Hence, the correct answer is option (ii).

Answer:

We get Mn₂O₇ which arises as a green oil by the addition of concentrated H₂SO₄ to KMnO₄.
The reaction initially produces permanganic acid, HMnO₄, which is dehydrated by cold sulfuric acid to form its anhydride, Mn₂O₇.
$2{\mathrm{KMnO}}_4+2{\mathrm{H}}_2{\mathrm{SO}}_4\left(\mathrm{Conc}.\right)\stackrel{ }{\to }{\mathrm{Mn}}_2{\mathrm{O}}_7+2{\mathrm{KHSO}}_4+{\mathrm{H}}_2\mathrm{O}$

Hence, the correct answer is option (i).

Answer:

As the no. of unpaired electron increases, its value of magnetic moment also increases. Since, 3d⁵ has 5 unpaired electrons so its shows highest magnetic moment.

$\begin{array}{rcl}\mu & =& \sqrt{5\left(5+2\right)}\\ & =& \sqrt{35}\\ & =& 5.95 \mathrm{BM}\end{array}$

Hence, the correct answer is option (ii).

Answer:

The elements of lanthanide series usually show the oxidation state of +3.
Metals of lanthanides show the oxidation state of +2 in the solutions of their their complex compounds. Sometime lanthanides also exhibit oxidation State of +4 occasionally.
These metals have have this unequal distribution of oxidation state due to the high stability of empty, half-filled or fully filled f-subshells.
Hence, the correct answer is option (ii).

Answer:

Disproportionation reaction are those reactions in which the same element/compound get oxidized and reduced simultaneously.

3Mn${\mathrm{O}}_4^{2-}$ + 4H⁺ → 2Mn${\mathrm{O}}_4^{-}$ + MnO₂ + 2H₂O
In the above reaction, Mn is in +6 oxidation state in MnO₄^2- and changes to +7 in MnO₄^- and +4 in MnO₂.

Hence, the correct answer is option (i).

Answer:

It slow down initially because Mn${\mathrm{O}}_4^{-}$ is getting converted to Mn²⁺ but once it forms Mn²⁺, the reaction becomes faster due to the catalytic action.

Hence, the correct answer is option (iv).

Answer:

Thulium (Tm) has atomic no. 69, and actinoid series has elements from atomic no. 90–103, so it does not belong to this series.
Thulium belong to lanthanoid series.

Hence, the correct answer is option (iii).

Answer:

KMnO₄ acts as an oxidising agent in acidic medium in the following reaction–

$\frac{{\mathrm{H}}_2\mathrm{S}+\left[\mathrm{O}\right]\to {\mathrm{H}}_2\mathrm{O}+\mathrm{S}]\times 5}{2{\mathrm{KMnO}}_4+3{\mathrm{H}}_2{\mathrm{SO}}_4+5{\mathrm{H}}_2\mathrm{S}\to {\mathrm{K}}_2{\mathrm{SO}}_4+2{\mathrm{MnSO}}_4+8{\mathrm{H}}_2\mathrm{O}+5\mathrm{S}}$

with 2 moles of KMnO₄, 5 moles of S^2– ions react. Likewise, 1 mole of S^2– ion will react with $\frac{2}{5}$ moles of KMnO₄.
Hence, the correct answer is option (i).

Answer:

From the given options, V₂O₅ and Cr₂O₃ are amphoteric oxide as they both react with alkalies as well as acids.

Hence, the correct answer is option (i).

Answer:

The correct electronic configuration of gadolinium with atomic no 64 is
[Xe]4f⁷ 5d¹ 6s²
It has more stability because of half-filled 4f subshell.

Hence, the correct answer is option (i).

Answer:

Interstitial compounds are very hard and can retain metallic conductivity. As they are chemically inert which means they not reactive. They also have a high meeting point which is higher than those of pure metals.

Hence, the correct answer is option (iv).

Answer:

Cr⁺³ has an electronic configuration of [Ar] 3d³ 4s⁰, having unpaired electrons as 3.

Therefore, magnetic moment of Cr⁺³ is $\sqrt{3\left(3+2\right)}$

Therefore, magnetic moment is = 3.87 BM

Hence, the correct answer is option (ii).

Answer:

When alkaline KMnO₄ is treated with KI, iodine ion is oxidised to ${\mathrm{IO}}_3^{-}$

Reaction:- $2{\mathrm{KMnO}}_4+{\mathrm{H}}_2\mathrm{O}+\mathrm{KI}\stackrel{ }{\to }2{\mathrm{MnO}}_4+2\mathrm{KOH}+{\mathrm{KIO}}_3$

Hence, the correct answer is option (iii).

Answer:

In the electrochemical series, Cu lies below hydrogen, therefore it does not liberate H₂ from acids.

Hence, the correct answer is option (i).

Answer:

When acidified K₂Cr₂O₇ solution is added to Sn²⁺ salt, then Sn²⁺ will change to Sn⁴⁺. To know that, the reaction is given below.


Hence, the correct answer is option (iii).

Answer:

Higher oxidation is related to the no. of bond formation in the compound.
Here, Highest oxidation state of manganese in fluoride is +4 (MnF₄ ) whereas in oxides, the highest oxidation state is +7 (Mn₂O₇). It is because in covalent compounds fluorine can form single bond while oxygen forms double bond.

Hence, the correct answer is option (iv).

Answer:

Even though Zirconium and hafnium both belongs to different transition series but they show similar physical and chemical properties because both have similar atomic radius. Due to lanthanide contraction, Zr and Hf possess nearly the same atomic and ionic radii.

Hence, the correct answer is option (iii).

Answer:

HCl is not used to make the medium acidic in oxidation reactions of KMnO₄ in acidic medium because if HCl is used, the oxygen produced from KMnO₄ + HCl will be partly utilised in oxidising HCl to Cl₂, which itself acts as an oxidising agent and partly oxidises the reducing agent.

Hence, the correct answer is option (ii).

Answer:

From the following compounds, KMnO₄ and Ce(SO₄)₂ are coloured compounds.
KMnO₄ – It is purple in colour due to charge spectra not because of unpaired electrons.
Ce(SO₄)₂ is yellow coloured due to charge transfer and not because of f – f transition.

Hence, the correct answers are option (i) and (ii).

Answer:

Co²⁺ and Cr³⁺ have almost same spin magnetic moment.

• Electronic configuration of Co²⁺ = [Ar] 3d⁷. It has 3 no. of unpaired no. of electrons.

• Electronic configuration of Cr²⁺ = [Ar] 3d⁴. It has 4 no. of unpaired no. of electrons.

• Electronic configuration of Mn²⁺ = [Ar] 3d⁵. It has 5 no. of unpaired no. of electrons.

• Electronic configuration of Cr³⁺ = [Ar] 3d³. It has 3 no. of unpaired no. of electrons.

Hence, the correct answers are option (i) and (iv).

Answer:

In d-block elements, the higher oxidation states are more stable for heavier elements. Therefore, Mo(VI) and W(VI) are more stable than Cr(VI). This is why, in acidic medium, Cr(VI) in the form of dichromate is a strong oxidizing agent when MoO₃ and WO₃ are not.

Hence, the correct answers are option (ii) and (iii).

Answer:

Pu and Np show oxidation state upto +7. The oxidation state of following actinoids are:-

• Americium (Z = 95), its electronic configuration

Answer:

From the given actinoides, U and Np each have one electron in 6d orbital.
General electronic configuration of actinoids is
(n – 1)f^{1 – 14} (n – 1)d^{0 – 2} ns².

Hence, the correct answers are option (i) and (ii).

Answer:

From the given lanthanoids Eu and Yb show +2 oxidation state besides the characteristic oxidation state +3 of lanthanoids.

• Cerium (Z = 57) ⇒ It's electronic configuration

Answer:

From the following ions Mn²⁺ and Fe²⁺ will show higher spin only magnetic values as:–

$$\begin{gathered} {\mathrm{Ti}}^{3+}=\left[\mathrm{Ar}\right] 3d^1 \\ {\mathrm{Mn}}^{2+}=\left[\mathrm{Ar}\right] 3d^5 \left(t_{2g}^3{e}_g^2\right) \\ {\mathrm{Fe}}^{2+}=\left[\mathrm{Ar}\right] 3d^6 \left(t_{2g}^4{e}_{2g}^2\right) \\ {\mathrm{Co}}^{3+}=\left[\mathrm{Ar}\right] 3d^6 \left(t_{2g}^6{e}_g^0\right) \end{gathered}$$

Hence, the correct answers are options (ii) and (iii).

Answer:

From the given transition elements, Cr and Co will form MF₃ type compounds with halogens i.e., CrF₃ and CoF₃ on the other hand, Cu and Ni do not form CuF₃ and NiF₃.

Hence, the correct answers are option (i) and (ii).

Answer:

Only species can act as oxidising agent when metal is present in high oxidation state but lower oxidation state show stability.
Since, W and Mo have more stable higher oxidations states so they will not as oxidising agents.

Hence, the correct answers are option (ii) and (iii).

Answer:

The electronic configuration of ₅₈Ce = ₅₄[Xe]4f2 5d⁰ 6s².
Therefore, the electronic configuration of Ce⁴⁺ = ₅₄[Xe]4f⁰.
Thus, it has a tendency to attain noble gas configuration as well as it can attain if configuration.

Hence, the correct answers are options (ii) and (iii).

Answer:

Cu is less reactive than hydrogen that is why it cannot replace hydrogen from acids. Cu has a positive E° value which is less reactive than hydrogen which has an electrode potential of 0 V.

Answer:

A negative E^⊖ value starts that the oxidizes species is more stable than the reduced species, metals will easily lose electrons and can get oxidized. Negative value of Mn²⁺and Zn²⁺ are related to stabilities of half filled and completely filled configuration. In case of Ni2, E^⊖ value is related to the highest negative enthalpy of hydration. Therefore, E^⊖ value for Mn, Ni and Zn are more negative than expected.

Answer:

Cr has 3d⁵ 4s¹ configuration that is why ionization enthalpy of Cr is less than that of Zn.
In case of zinc, the electrons comes out from completely filled s orbital. This is the reason why the removed electron from zinc consumes more energy as compared to the chromium.

Answer:

Transition elements have electrons in the d orbital that participate in metallic bonding. Electrostatic force between conduction electrons and positively charged metal ions arises metallic bonding.
Due to the involvement of greater number of electrons in the interatomic bonding from (n – 1) d-orbitats in addition to ns electrons in forming metallic bond, the transition elements show high melting point.

Answer:

When Cu²⁺ ion is treated with KI, it will form Cu₂I₂ which is a white precipitate is the final product.
The reaction will be as follows:
$2{\mathrm{Cu}}^{2+}+4{\mathrm{I}}^{-}\stackrel{}{\to }\underset{\left(\mathrm{white} \mathrm{ppt}\right)}{{\mathrm{Cu}}_2{\mathrm{I}}_2+{\mathrm{I}}_2}$
We get CuI₂ in this reaction which being unstable, dissociates into Cu₂I₂ and I₂.

Answer:

CuCl₂ is more stable than Cu₂Cl₂ as it has a higher electron density than Cu⁺.
Stability of Cu²⁺(aq) is more than than Cu⁺(aq) is because of much more negative value of hydration enthalpy of Cu⁺² than Cu⁺ which more than compensates for the second ionization enthalpy of Cu.

Answer:

MnO₂ is the brown compound of Mn which reacts with HCl to give Cl₂ gas.
This gas when treated with NH₃ can form an explosive compound i.e NCl₃.
Thus, A = MnO₂, B = Cl₂ and C = NCl₃.
We can get the following reactions - 
(i) $\underset{\left(\mathrm{A}\right)}{{\mathrm{MnO}}_2+4\mathrm{HCl}\stackrel{}{\to }}\underset{\left(\mathrm{B}\right)}{{\mathrm{MnCl}}_2+{\mathrm{Cl}}_2+2{\mathrm{H}}_2\mathrm{O}}$
(ii) $\underset{\left(\mathrm{Excess}\right)}{{\mathrm{NH}}_3+3{\mathrm{Cl}}_2}\stackrel{}{\to }\underset{\left[\mathrm{C}\right)}{{\mathrm{NCl}}_3+3\mathrm{HCl}}$

Answer:

Oxygen has the ability to form multiple bonds with metals whereas fluorine can't form multiple bonds with metals. Therefore, oxygen has more ability to stabilize a higher oxidation state than fluorine.

Answer:

We can determine the magnetic moment of any metal ion on the basis of spin as well as the orbital contribution of electrons. There is no orbital contribution in Cr³⁺ ion due to symmetrical electronic configuration. Therefore, Cr⁺³ has less magnetic moment than Co⁺².

Answer:

Ce, Pr and Nd are lanthanides and they consist of an incomplete 4f subshell. Th, Pa and U are actinoids and they consist of incomplete 5f shell. Initially, when 5f-orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The 5f-electrons will hence, be more effectively shielded from the nuclear charge than 4f electrons of corresponding lanthanoids. Therefore, outer electrons are less firmly held and they are available for bonding in the actinoids.

Answer:

Even though Zr belongs to 4d and Hf belongs to the 5d transition series, it is quite difficult to separate them due to lanthanoid contraction.
Because of lanthanoid contraction, they have almost similar sizes (Zr = 160 pm and Hf = 159 pm) and therefore, the same chemical properties. This is why it is hard to separate them by chemical method.

Answer:

It is because of the fact that after losing one more electron Ce acquires stable 4f⁰ electronic configuration. As a result, Ce shows +4 oxidation state also along with +3 oxidation state.

Answer:

The colour of acidic solution of KMnO₄ disappear when oxalic acid is added to it, this happens due to the reduction of MnO₄ ion to Mn²⁺.
Following is the chemical reaction occurring during this neutralization reaction - 
$\underset{\left(\mathrm{Coloured}\right)}{5{\mathrm{C}}_2{\mathrm{O}}_4^{2-}+2{\mathrm{mnO}}_4^{-}+16{\mathrm{H}}^{+}}\stackrel{}{\to }\underset{\left(\mathrm{Colourless}\right)}{2{\mathrm{Mn}}^{2+}+8{\mathrm{H}}_2\mathrm{O}+10{\mathrm{CO}}_2}$

Answer:

When an orange solution containing ${\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}$ ion is treated with alkali, a yellow solution is formed. It is called ${\mathrm{CrO}}_4^{2-}$. In the same way, when H⁺ ions are added to yellow solutions, orange solution is formed. Due to the interconversion of ions, an orange solution is obtained.
$\underset{\underset{\left(\mathrm{orange}\right)}{\mathrm{Dischromate}}}{{\mathrm{Cr}}_2{\mathrm{O}}_7^{2-}}\underset{{\mathrm{H}}^{+}}{\overset{{\mathrm{OH}}^{-}}{\to }}\underset{\underset{\left(\mathrm{yellow}\right)}{\mathrm{Chromate}}}{{\mathrm{CrO}}_4^{2-}}$

Answer:

The oxidising behaviour of KMnO₄ depends on p^H of the solution.
In acidic medium (pH < 7)
${\mathrm{MnO}}_4^{-}+8{\mathrm{H}}^{+}+5{\mathrm{e}}^{-}\stackrel{}{\to }\underset{\left(\mathrm{Colorless}\right)}{{\mathrm{Mn}}^{2+}+4{\mathrm{H}}_2\mathrm{O}}$
In Alkaline medium (pH > 7)
${\mathrm{MnO}}_4^{-}+{\mathrm{e}}^{-}\stackrel{}{\to }\underset{\left(\mathrm{green}\right)}{{\mathrm{MnO}}_4^{2-}}$
In neutral medium (pH = 7) 
${\mathrm{MnO}}_4^{-}+2{\mathrm{H}}_2\mathrm{O}+3{\mathrm{e}}^{-}\stackrel{}{\to }\underset{\left(\mathrm{Brown} \mathrm{ppt}\right)}{{\mathrm{MnO}}_2+4{\mathrm{OH}}^{-}}$

Answer:

The atomic radii of the second and third transition elements are almost similar due to lanthanoid contraction. This is why they resemble each other more and show similar characters, unlike first row elements.

Answer:

${\mathrm{E}}^{\ominus }$ value of Cu is positive as the sum of sublimation enthalpy and ionisation enthalpy to convert Cu(s) to Cu²⁺(aq) is so high that it is not balanced by its hydration enthalpy.
Negative ${\mathrm{E}}^{⊝}$ for Zn is because of the fact that after removal of electrons from 4s orbital stable 3d¹⁰ configuration is obtained.

Answer:

The size of the ion of transition elements decreases with the increase in the oxidation state. According to Fajan's rule, as the size of the metal ion decreases, the covalent character of the bond formed increases.
Therefore, the halide of transition elements becomes more covalent with the increasing oxidation state of the metal.

Answer:

During filling up of electrons in atomic orbitals follows (n + 1) rule. As compared to 3d orbitals 4s orbita has lower energy. After the orbitals are filled, 4s goes beyond 3d i.e it gets farther from the nucleus than 3d. This is why the electron from 4s is removed earlier. Atomic orbitals are filled in order of increasing energies.

Answer:

Reactivity of transition elements depends mostly on ionisation enthalpies. Ionisation enthalpies increase almost regularly as we move from left to right in the periodic table (Sc to Cu).
This is why, the reactivity of transition element decreases almost regularly from Sc to Cu.

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Answer:

Cu²⁺ oxidises I to iodine
Cu²⁺ iodide is not known due to the fact that Cu²⁺ oxidises I to iodine.
Hence, the correct answer is option (i).

Answer:

It is true that the separation of Zr and Hf is difficult. This is because of lanthanoid contraction which causes almost similar radii of both of them. It is not due to the fact that they lie in the same group of periodic table.

Hence, the correct answer is option (ii).

Answer:

Actinoids form more stable complexes compared to lanthanoids. It is because to, actinoids can utilise their 5f orbitals along with 6d orbitals in bonding whereas lanthanoids do not use their 4f orbitals for bonding.

Hence, correct answer is option (iii).

Answer:

It is true not Cu cannot liberate hydrogen from acids because it has positive electrode potential.
Metals that have negative value of electrode potential, they liberate H₂ gas.
Hence, the correct answer is option (i).

Answer:

The highest oxidation state of osmium is +8 because of its ability to expand its octet by using all its 8 electrons (2 from 6s and 6 from 5d).
The electronic configuration of Osmium is 5d⁶ 6s². As 6d and 6s are close in energy, all the 8 electrons can participate in bonding.

Answer:

The substances from A to E are as following:
A = Cu ; B = Cu(NO₃)₂
C = [Cu(NH₃)₄]²⁺ ; D = CO₂ ; E = CaCO₃
Their reactions are:
(i) ${\mathrm{CuCO}}_3 \stackrel{∆}{\to } \mathrm{CuO}+\mathrm{CO}2]\times 2$
(ii) $2\mathrm{CuO} + \mathrm{CuS} \stackrel{}{\to } 3\mathrm{Cu}+{\mathrm{SO}}_2\left[\mathrm{A}\right]$
$$\begin{gathered} \left(\mathrm{iii}\right) \mathrm{Cu}+4{\mathrm{HNO}}_3(\mathrm{Conc}.) \stackrel{}{\to } \mathrm{Cu}{\left({\mathrm{NO}}_3\right)}_2+2\mathrm{No}+2{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{iv}\right) {\mathrm{Cu}}^{2+}+2{\mathrm{NH}}_3 \stackrel{}{\to } \left[\mathrm{Cu}{\left({\mathrm{NH}}_3\right)}_4\right] \\ \left[\mathrm{B}\right] (\mathrm{Blue} \mathrm{solution}) \\ \left(\mathrm{v}\right) \mathrm{Ca}{\left(\mathrm{OH}\right)}_2 + {\mathrm{CO}}_2 \stackrel{}{\to } {\mathrm{CaCO}}_3 + {\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{milky}\right) \\ \left(\mathrm{vi}\right) {\mathrm{CaCO}}_3 + {\mathrm{H}}_2\mathrm{O} + {\mathrm{CO}}_2 \stackrel{}{\to } \mathrm{Ca}{\left({\mathrm{HCO}}_3\right)}_2 \end{gathered}$$
 

Answer:

The orange compound is k₂ Cr₂ O7. When Na₂ Cr₂ O₇ reacts with KCl, we get the orange compound i.e. k₂ Cr₂ O₇.
In acidic medium, yellow coloured CrO₄^2–(Chromate ion) change into dichromate.
We can prepare potassium dichromate from chromate by the given process.
A = FeCr₂O₄; B = Na₂ CrO₄ ; C = Na₂Cr₂O₇; D = k₂Cr₂O₇.
$$\begin{gathered} \left(\mathrm{i}\right) 4{\mathrm{FeCr}}_2{\mathrm{O}}_4+8{\mathrm{Na}}_2{\mathrm{CO}}_3+7{\mathrm{O}}_2 \stackrel{}{\to } 8{\mathrm{Na}}_2 {\mathrm{CrO}}_4+2{\mathrm{Fe}}_2{\mathrm{O}}_3+8{\mathrm{CO}}_2 \\ \left[\mathrm{A}\right] \left[\mathrm{B}\right] \\ \left(\mathrm{ii}\right) 2{\mathrm{Na}}_2{\mathrm{CrO}}_4+2{\mathrm{H}}^{+}\stackrel{}{\to } {\mathrm{Na}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+2{\mathrm{N}}^{\mathrm{a}+}+{\mathrm{H}}_2\mathrm{O} \\ \left(\mathrm{iii}\right) {\mathrm{Na}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+2\mathrm{kcl} \stackrel{}{\to }{\mathrm{k}}_2{\mathrm{Cr}}_2{\mathrm{O}}_7+2\mathrm{NaCl} \\ \left[\mathrm{C}\right] \left[\mathrm{D}\right] \end{gathered}$$

Answer:

It is the method of preparation of potassium permanganate (purple).
Here, A = MnO₂         C = KMnO₄
          B = k₂MnO₄     D = KlO₃
Reactions are as follows–
2MnO₂ + 4kOH + O₂ → 2k₂MnSO₄ + 2H₂O
        [A]                                      [B]
When [A] is fused with kOH, it gives a [B].
Potassium manganate dill permanganate to give purple potassium permanganate i.e. [C].
$\to 3{\mathrm{MnO}}_4^{2-}+4^{\mathrm{H}+}\stackrel{}{\to }\underset{\left[\mathrm{C}\right]}{2{\mathrm{MnO}}_4^{-}}+{\mathrm{MnO}}_2+2{\mathrm{H}}_2\mathrm{O}$
Potassium Permanganate on reacting with KI to give [D] & manganese oxide.
$2{\mathrm{MnO}}_4^{-}+{\mathrm{H}}_2\mathrm{O}+\mathrm{KI}\stackrel{}{\to }\underset{\left[\mathrm{A}\right]}{2{\mathrm{MnO}}_2}+2{\mathrm{OH}}^{–}+ \underset{\left[\mathrm{D}\right]}{{\mathrm{KlO}}_3}$

Answer:

(i) We know that as the size decrease covalent character increases. Hence, La₂O₃ is more ionic and Lu2O3 is more covalent.
(ii) Stability of oxo salts decreases as the size decreases from La to Lu.
(iii) As the size of lanthanoids decreases, the stability of complexes increase.
(iv) The Radii of 4d and 5d-Block elements will be almost similar.
(v) The Acidic character of lanthanoid oxides increases from La to Lu.

Answer:

(a) (i) Cu has highest second ionisation enthalpy because the electronic configuration of Cu is 3d¹⁰4s¹. So, second electron needs to be removed from completely filled d-orbital that is very difficult.
(ii) Zinc, as its electronic configuration is 3d¹⁰ 4s² and Zn²⁺ = 3d¹⁰ which is fully filled. Therefore, it is very stable. To remove the third electron, would require very high energy.
(iii) Zinc, as it has fully filled 3d subshell and no unpaired electron is available for metalic bonding.

(b) (i) The given metal is:
Carbonyl M(CO)₅ is Fe (CO)₅
Acc. to EAN = no. of electrons in metal + 2 × (Co)
= atomic no. of nearest inert you.
In M(CO)5 = x + 2 × (5) = 36 (kr is the nearest insert gas)
x = 26 (atomic no. of metal)
The metal is Fe (iron).
(ii) The given metal is
MO₃F is MnO₃F.
Suppose the oxidation state of M is x.
= x + 3 × (–2) + (–1) = 0
Or, x = +–7 i.e., M is in +7 oxidation state of +7.
Therefore, MnO₃F is the given compound.

Answer:

Interstitial compounds are formed when small atoms like H1 C and N gets trapped inside the crystal lattice of transition metals.
They posses following properties:
→ They tend to have High meeting points even more higher than pure metals.
→ They are hard in nature.
→ Have the ability to retain metallic conductivity.
→ They are chemically inert.

Answer:

(a) The reaction between iodine and persulphate ions is as follows:
$2{\mathrm{I}}^{–}+5_2{\mathrm{O}}_8^{2-}\stackrel{ \mathrm{Fe}\left(\mathrm{iii}\right) }{\to }{\mathrm{I}}_2+2\mathrm{SO}{4}_4^{2-}$
Role of Fe(III) pone to catalyse the resction + between iodine and persulphate ions.
$$\begin{gathered} 2{\mathrm{Fe}}^3 + 2{\mathrm{I}}^{–} \stackrel{ }{\to } 2{\mathrm{Fe}}^2+{\mathrm{I}}_2 \\ 2{\mathrm{Fe}}^{2+} + {\mathrm{S}}_2{\mathrm{O}}_8^{2-} \stackrel{ }{\to } 2\mathrm{Fe}{2}^{3+}+2{\mathrm{SO}}_4^{2-} \end{gathered}$$
(b) Three processes where transition metals acts as catalysts are:
* Use of Vanadium (V) oxide in contact process for oxidation of SO₂ to SO₃.
* Finely divided iron in Haber's process in conversion of N₂ and H₂ to NH₃.
* MnO₂ in preparation of oxygen from kC10₃.

Answer:

The composed from A to D are as follows
A = KMnO₄     B = K₂MnO₄
C = MnO₂       D = MnCl₂
The compound (C) on treating with cone. H₂SO₄ and NaCl gives Cl₂ gas, Thus it is manganese dioxide (MnO₂). It is obtained alongwith MnO₄^2– when KMnO₄ (violet) is heated.
The following reaction takes place.
$$\begin{gathered} \underset{\left[\mathrm{A}\right]}{2{\mathrm{KMnO}}_4}\stackrel{ ∆ }{\to }\underset{\left[\mathrm{B}\right]}{{\mathrm{K}}_2{\mathrm{MnO}}_4} + \underset{\left[\mathrm{C}\right]}{{\mathrm{MnO}}_2+{\mathrm{O}}_2} \\ 2{\mathrm{MnO}}_2 + 4\mathrm{KOH} + {\mathrm{O}}_2 \stackrel{ ∆ }{\to } 2{\mathrm{K}}_2{\mathrm{MnO}}_4 + 2{\mathrm{H}}_2\mathrm{O} \end{gathered}$$
MnO₂ reacts with KOH to give K₂MnO₄
${\mathrm{MnO}}_2+4\mathrm{NaCl}+4{\mathrm{H}}_2{\mathrm{SO}}_4\stackrel{ }{\to }\underset{\left[\mathrm{D}\right]}{{\mathrm{MnCl}}_2}+4{\mathrm{NaSO}}_4+2{\mathrm{H}}_2\mathrm{O}+{\mathrm{Cl}}_2$
On heating manganese dioxide (MnO₂) with NaCl and H₂SO₄, we get (MnCl₂), chlorine gas and other products.