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Chemistry 3 Marks Question

The d & f Block Elements · Bihar Board (BSEB) STD 12 Science (BSEB - English Medium). 62 questions.

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Question 13 Marks
Lanthanum, gadolinium and lutetium are extraordinarily stable in +3 oxidation state. Explain.
(Atomic number: La = 57, Gd = 64, Lu = 71)
Answer
$_{57}\text{La}\xrightarrow{\ \ \ \ \ \ }\ _{54}\text{La}^{3+}$
$[\text{Xe}]\ 5\text{d}1\ 6\text{s}2\ \ \ \ \ \ \ \ \ \ [\text{Xe}]\ 5\text{d}\ 0\ 6\text{s}0\\\ \ \ \ \ \ \ \ \ \ \ \ _{64}\text{Gd}\xrightarrow{\ \ \ \ \ \ \ \ \ }\ _{61}\text{Gd}^{3+}$
$[\text{Xe}]\ 4\text{f}^7\ 5\text{d}^1\ 6\text{s}^2\ \ \ \ \ \ \ \ \ [\text{Xe}]\ 4\text{f}^7\ 5\text{d}^0\ 6\text{s}^0\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ _{71}\text{Lu}\xrightarrow{\ \ \ \ \ \ \ \ \ \ }{68}\text{Lu}^{3+}$
$[\text{Xe}]\ 4\text{f}^{14}\ 5\text{d}^1\ 6\text{s}^2\ \ \ \ \ \ \ \ [\text{Xe}]\ 4\text{f}^{14}\ 5\text{d}^0\ 6\text{s}^0$
From their electronic configurations, we find that La3+, Gd3+ and Lu3+ have empty, exactly half-filled and fully filled valency shells, respectively, which make them extra stable.
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Question 23 Marks
Write down the number of 3d electrons in each of the following ions: Ti2+, V2+, Cr3+, Mn2+, Fe2+, Fe3+, Co2+, Ni2+ and Cu2+. Indicate how would you expect the five 3d orbitals to be occupied for these hydrated ions (octahedral).
Answer
Metal ion Number of d-electrons Filling of d-orbitals
Ti2+ 2 t228
V2+ 3 t328
Cr3+ 3 t328
Mn2+ 5 t328e28
Fe2+ 6 t428e28
Fe3+ 5 t328e28
CO2+ 7 t528e28
Ni2+ 8 t628e28
Cu2+ 9 t628e38
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Question 33 Marks
What are alloys? Name an important alloy which contains some of the lanthanoid metals. Mention its uses.
Answer
An alloy is a blend of metals prepared by mixing the components. Alloys may be homogeneous solid solutions in which the atoms of one metal are distributed randomly among the atoms of the other. Such alloys are formed by atoms with metallic radii that are within about 15 percent of each other. Because of similar radii and other characteristics of transition metals, alloys are readily formed by these metals.

An important alloy which contains some of the lanthanoid is mischmetall. Mischmetall consists of a lanthanoid metal (~95%) and iron (~5%) and traces of S, C, Ca and Al.

Uses:

  1. Mischmetall is used in Mg based alloy to produce bullets, shell and lighter flint.

Some individual Ln oxides are used asjphosphorus in television screens and similar fluorescing surfaces.

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Question 43 Marks
Compare the general characteristics of the first series of the transition metals with those of the second and third series metals in the respective vertical columns. Give special emphasis on the following points:

Electronic configurations:

Answer
In the 1st, 2ndand 3rdtransition series, the 3d, 4d and 5 dorbitals 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:
Cr(24) = 3d5 4s1
Cu(29) = 3d10 4s1
Similarly, there are exceptions in the second transition series. These are:
Mo(42) = 4d5 5s1
Tc(43) = 4d6 5s1
Ru(44) = 4d7 5s1
Rh(45) = 4d8 5s1
Pd(46) = 4d8 5s1
Ag(47) = 4d10 5s0
There are some exceptions in the third transition series as well. These are:
W(74) = 5d4 6s2
Pt(78) = 5d9 6s1
Au(79) = 5d10 6s1
As a result of these exceptions, it happens many times that the electronic configurations of the elements present in the same group are dissimilar.
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Question 53 Marks
What is lanthanoid contraction? What are the consequences of lanthanoid contraction?
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 4f orbital also increases. The 4f electrons 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

  1. There is similarity in the properties of second and third transition series.
  2. Separation of lanthanoids is possible due to lanthanide contraction.
  3. It is due to lanthanide contraction that there is variation in the basic strength of lanthanide hydroxides. Basic strength decreases from $\text{La(OH)}_3 \ \text{to} \ \text{Lu(OH)}_3$
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Question 63 Marks
Indicate the steps in the preparation of:
K2Cr2O7 from chromite ore.
Answer
The preparation of potassium dichromate from chromite involves the following main steps:

  1. The chromate ore is finely ground and heated strongly with molten alkali in the presence of air.

$2\text{FeCr}_2\text{O}_4+8\text{NaOH}+7/2\text{O}_2\rightarrow4\text{NaCr}_2\text{O}_4+\text{Fe}_2\text{O}_3+4\text{H}_2\text{O}\\\text{Chromite}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sodium chromate}$

  1. The solution of sodium chromate is filtered and acidified with dilute sulphuric acid so that sodium dichromate is obtained.

$2\text{Fe}_2\text{Cr}_2\text{O}_4+\text{H}_3\text{SO}_4\rightarrow\text{Na}_2\text{Cr}_2\text{O}_7+\text{Na}_2\text{SO}_4+\text{H}_2\text{O}\\\text{Sodium}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Sodium}\\ \text{Chromate} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{dichromate}$

  1. A calculated quantity of potassium chloride is added to a hot concentrated solution of sodium dichromate. Potassium dichromate is less soluble therefore it crystallizes out first.

​​​​​​​$2\text{Na}_2\text{Cr}_2\text{O}_7+2\text{KCl}\rightarrow\text{K}_2\text{Cr}_2\text{O}_7+\text{NaCl}\\\text{Sodium}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Potessium}\\\text{dichromate} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{dichromate}$

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Question 73 Marks
The chemistry of the actinoid elements is not so smooth as that of the lanthanoids. Justify this statement by giving some examples from the oxidation state of these elements.
Answer
Among the actinoids, there is a greater range of oxidation states as compared to lanthanoids. This is in part due to the fact that 5f, 6d and 7s levels are of very much comparable energies and the frequent electronic transition among these three levels is possible. This 6d-5f transition and larger number of oxidation states among actinoids make their chemistry more complicated particularly among the 3rd to 7th elements. Following examples of oxidation states of actinoids. Justify the complex nature of their chemistry.
  1. Uranium exhibits oxidation states of +3, +4, +5, +6 in its compounds. However, the dominant oxidation state in actinoides is +3.
  2. Nobelium, No is stable in +2 state because of completely filled f14 orbitals in this state.
  3. Berkelium, BK in +4 oxidation state is more stable due to f7 (exactly half filled) configuration.
  4. Higher oxidation states are exhibited in oxo ions are UO22+, PuO22+, NpO+ etc.
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Question 83 Marks
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
Chemical reactivity
Answer
Chemical reactivity: The earlier members of the lanthanoids series are quite reactive similar to calcium but, with increase in atomic number, they behave more like aluminium. The metals combine with hydrogen when. gently heated in the gas. Carbides, Ln3C, Ln2C3 and LnC2 are formed when the metals are heated with carbon. They liberate hydrogen from dilute acid and burn in halogens to form halides. They form oxides M2O3 and hydroxides M(OH)3. Actinoids are highly reactive metals, especially when finely divided. The action of boiling water on thern gives a mixture of oxide and hydride and combination with most non-metals take place at moderate temperatures. HCl attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers, alkalis have no action. Actinoids are more reactive than lanthanoids due to bigger atomic size and lower ionisation energy.
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Question 93 Marks
To what extent do the electronic configurations decide the stability of oxidation states in the first series of the transition elements? Illustrate your answer with examples.
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, Sc does not show +2 oxidation state. Its electronic configuration is 4s2 3d1. It loses all the three electrons to form Sc3+. +3 oxidation state of Sc is very stable as by losing all three electrons, it attains stable noble gas configuration, [Ar]. Ti (+4) and 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 half-filled, [Ar]3d5.
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Question 103 Marks
Indicate the steps in the preparation of:
KMnO4 from pyrolusite ore.
Answer
Pottassium Permanganate (KMnO4) is prepared from Pyrolusite ore (MnO2). The finely powdered Pyrolusite ore (MnO2) is fused with an alkali metal hydroxide like KOH in the presence of air or an oxidizing agent like KNO3 to give the dark green potassium Manganate (K2MnO4). Potassium manganate disproportionate in a neutral or acidic solution to give potassium permanganate.
$2\text{MnO}_2+4\text{KOH}+\text{O}_2\rightarrow2\text{K}_2\text{MnO}_4+2\text{H}_2\text{O}$
$3\text{MnO}^{2-}_4+4\text{H}^+\rightarrow2\text{MnO}^-_4+\text{MnO}_2+2\text{H}_2\text{O}$
Commercially potassium permanganate is prepared by the alkaline oxidative fusion of Pyrolusite ore (MnO2) followed by the electrolytic oxidation of manganate (4) ion.
$2\text{MnO}_2+4\text{KOH}+\text{O}_2\rightarrow2\text{K}_2\text{MnO}_4+2\text{H}_2\text{O}$
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Question 113 Marks
What can be inferred from the magnetic moment values of the following complex species?
Example Magnatic Moment (BM)
K4[Mn(CN)6] 2.2
[Fe(H2O)6]2+ 5.3
K2[MnCl4] 5.9
Answer
Magnetic moment (m) is given as $\mu=\sqrt{\text{n}(\text{n}+2)}$

For value n = 1, $\mu=\sqrt{1(1+2)}=\sqrt3=1.732$

For value n = 2, $\mu=\sqrt{2(2+2)}=\sqrt8=2.83$

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

For value n = 4, $\mu=\sqrt{4(4+2)}=\sqrt{24}=4.889$

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

  1. K4[Mn(CN)6]

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

$\sqrt{\text{n}(\text{n}+2)}=\sqrt8=2.2$

We can see from the above calculation that the given value is closest to n = 1. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d orbital. Hence, we can say that CN- is a strong field ligand that causes the pairing of electrons.

  1. [Fe(H2O)6]2+

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

We can see from the above calculation that the given value is closest to n = 4. Also, in this complex, Fe is in the +2 oxidation state. This means that Fe has 6 electrons in the d orbital. Hence, we can say that H2O is a weak field ligand and does not cause the pairing of electrons.

  1. K2[MnCl4]

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

We can see from the above calculation that the given value is closest to n = 5. Also, in this complex, Mn is in the +2 oxidation state. This means that Mn has 5 electrons in the d orbital. Hence, we can say that Cl- is a weak field ligand and does not cause the pairing of electrons.

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Question 123 Marks
How would you account for the following:
E° of Cu is +0.34 V while that of Zn is -0.76 V.
Answer
High ionisation enthalpy to transform Cu(s) to Cu2+(aq) is not balanced by its hydration enthalpy. However, in case of Zn after removal of electrons from 4s-orbital, stable 3d10 configuration is acquired.
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Question 133 Marks
Describe the trends in the following properties of the first series of the transition elements:
  1. Oxidation states.
  2. Atomic sizes.
  3. Magnetic behaviour of dipositive gaseous ions (M2+).
Answer
  1. As there is very little energy difference between 4s and 3d orbitals, electrons from both energy levels can be used for chemical bond formation. Therefore, all elements except Sc and Zn of the first transition series show a number of oxidation states.
  2. Atomic radii of the first transition series decrease from Sc to Cr, then remains almost constant till Ni and then increases from Cu to Zn. The reason of this variation in atomic radii has been attributed to the increase in nuclear charge in the beginning of the series. But as the electrons continue to be filled in d-orbitals, they screen the outer 4s electrons from the influence of nuclear charge. When the increased nuclear charge and the increased screening effect balance each other in the middle of transition series, the atomic radii become almost constant (Mn to Fe). Towards the end of the series, the repulsive interaction between electrons in orbitals become very dominant. As a result there is an expansion of the electron cloud; consequently, the atomic size increases.
  3. Except Zn2+, all other divalent gaseous ions of the first series of the transition elements contain unpaired electrons in their 3d-subshell and are therefore paramagnetic in nature.
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Question 143 Marks
Give reasons for the following:
  1. The transition metals generally form coloured compounds.
  2. Eo value for (Mn3+|Mn2+) is highly positive than that for (Cr3+|Cr2+) couple.
  3. The chemistry of actinoids elements is not so smooth as that of the lanthanoids.
Answer
  1. Due to presence of unpaired electrons/ d-d transition.
  2. Mn3+ is 3d4 while Cr3+ is 3d3 which in t2g half filled is extra stable.
  3. The energy difference between 5f, 6d and 7s orbitals is very less as compared to lanthanoids.
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Question 153 Marks
  1. How would you account for the following:
  1. Actinoid contraction is greater than lanthanoid contraction.
  2. Transition metals form coloured compounds.
  1. Complete the following equation:

$\text{2MnO}_{4}^{-}+\text{6H}^{+}+\text{5NO}_{2}^{-}\rightarrow$

Answer
  1. 5f orbital electrons have poor shielding effect than 4f.
  2. due to d-d transition/or the energy of excitation of an electron from lower d-orbital to' higher d-orbital lies in the visible region/presence of unpaired in the d-orbital.
  3. $\text{2 MnO}^{-}_{4}+\text{6H}^{+}+\text{5 NO}^{-}_{2}\rightarrow\text{2Mn}^{2+}+\text{3H}_{2}\text{O}+\text{5NO}^{-}_{3}$
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Question 163 Marks
Although Cr3+ and Co2+ ions have same number of unpaired electrons but the magnetic moment of Cr3+ is 3.87 B.M. and that of Co2+ is 4.87 B.M. Why?
Answer
The electronic configuration of Cr3+ and CO2+ ions are:

Due to symmetrical electronic configuration, there is no orbital contribution in Cr3+ ion. However, appreciable contribution occurs in CO2+ ion.
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Question 173 Marks
How would you account for the following:
Transition metals sometimes exhibit very low oxidation state such as +1 and 0.
Answer
+1 oxidation state is shown by element like Cu because after loss of one electron, it acquires stable configuration of 3d10. Zero oxidation state is shown in forming metal carbonyl, because p-electrons donated by ligands are accepted into the empty d-orbitals.
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Question 183 Marks
A solution of KMnO4 on reduction yields either a colourless solution or a brown precipitate or a green solution depending on pH of the solution. What different stages of the reduction do these represent and how are they carried out?
Answer
Oxidising behaviour of KMnO4 depends on pH of the solution.
 In acidic medium $\text{(pH<7)}$
$\text{MnO}_4^-+8\text{H}^++5\text{e}^-\xrightarrow{\ \ \ \ \ \ }\text{Mn}^{2+}+4\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(Colourless)}$
In alkaline medium $\text{pH<7}$
$\text{MnO}_4^-+\text{e}^-\xrightarrow{\ \ \ \ \ \ }\text{MnO}^{2-}_4\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(Green)}$
In neutral medium $\text{(pH=7)}$
$\text{MnO}_4^-+2\text{H}_2\text{O}+3\text{e}^-\xrightarrow{\ \ \ \ \ \ }\text{MnO}_{2}+4\text{OH}^-\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(Brown precipitate)}$
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Question 193 Marks
Colour of KMnO4 disappears when oxalic acid is added to its solution in acidic medium.
Answer
KMnO4 acts as oxidising agent. It oxidises oxalic acid to CO2 and itself changes to Mn2+ ion which is colourless.
$5\text{C}_2\text{O}^{2-}_4+2\text{MnO}^-_4+16\text{H}\xrightarrow{\ \ \ \ \ \ \ }2\text{Mn}^{2+}+8\text{H}_2\text{O}+10\text{CO}_2\\\ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{Coulored})\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ (\text{Colourless})$
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Question 203 Marks
How would you account for the following:
  1. The transition elements have high enthalpies of atomisation.
  2. The transition metals and their compounds are found to be good catalysts in many processes.
Answer
  1. Because of strong metallic bonds/they have stronger interatomic interaction.
  2. Because they are capable of exhibiting variables oxidation states and forming complex.
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Question 213 Marks
When a chromite ore (A) is fused with sodium carbonate in free excess of air and the product is dissolved in water, a yellow solution of compound (B) is obtained. After treatment of this yellow solution with sulphuric acid,compound (C) can be crystallised from the solution. When compound (C) is treated with KCl, orange crystals of compound (D) crystallise out. Identify A to D and also explain the reactions.
Answer
The compounds A, B, C and D are given as under:
$\text{A}=\text{FeCr}_2\text{O}_4$ $\text{B}=\text{Na}_2\text{CrO}_4$ $\text{C}=\text{Na}_2\text{Cr}_2\text{O}_7.2\text{H}_2\text{O}$ $\text{D}=\text{K}_2\text{Cr}_2\text{O}_7$
The reactions are exaplained as under:
$4\text{FeCr}_2\text{O}_7+8\text{NaCO}_3+7\text{O}_2\xrightarrow{\ \ \ \ \ \ \ \ \ }8\text{Na}_2\text{CrO}_4+2\text{Fe}_2\text{O}_3+8\text{CO}_2\\ \ \ \ \ \ \ \ ^\text{(A)}\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(B)}$
$2\text{NaCrO}_4+2\text{H}^+\xrightarrow{\ \ \ \ \ \ \ \ \ \ }\text{Na}_2\text{Cr}_2\text{O}_7+2\text{Na}^++\text{H}_2\text{O}$
$\text{Na}_2\text{Cr}_2\text{O}_7+2\text{KCl}\xrightarrow{\ \ \ \ \ \ \ \ \ }\text{K}_2\text{Cr}_2\text{O}_7+2\text{NaCl}\\ \ \ \ \ \ \ \ \ ^\text{(C)} \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ ^\text{(D)}$
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Question 223 Marks
How would you account for the following:
  1. Many of the transition elements and their compounds can act as good catalysts.
  2. The metallic radii of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second series.
  3. There is a greater range of oxidation states among the actinoids than among the lanthanoids.
Answer
  1. Due to their ability to show multiple oxidation states.
  2. Due to lanthanoid contraction.
  3. Due to comparable energies of 5f, 6d and 7s orbitals in actinoids.
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Question 233 Marks
Giving a suitable example for each, explain the following:
  1. Crystal field splitting.
  2. Linkage isomerism.
  3. Ambidentate ligand.
Answer
  1. Crystal field splitting: The splitting of d-orbitals under the influence of approaching ligand is known as crystal field splitting eg for d4, configuration is t2g3eg1 in the presence of weak field ligand.
  2. Linkage isomerism: Linkage isomers are those isomers which have same molecular formula but differ in the linkage of ligand atom to the central atom e.g. 

[Co(NH3)5 NO2]Cl2, [Co(NH3)5 ONO]Cl2.

  1. Ambident ligand: a un identate ligand which can co-ordinate to the central metal atom through more than one co-ordinating bond.e.g. NO2–, SCN–.
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Question 243 Marks
Answer the following questions:
Compare non transition and transition elements on the basis of their.
  1. Variability of oxidation states.
  2. Stability of oxidation states.
Answer
  1. Oxidation states of transition elements differ from each other by unity. In non-transition elements oxidation states normally differ by a unit of two.
  2. In transition elements higher oxidation states are favoured by heavier elements whereas in non-transition elements lower oxidation state is favoured by heavier elements.
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Question 253 Marks
How would you account for the following:
The E° value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple or Fe3+/Fe2+couple.
Answer
The E° value for the Mn3+/Mn2+ couple is much more positive than Cr3+/Cr2+ couple or Fe3+/Fe2+ couple because Mn3+ ion receiving an electron gets d-subshell half-filled which is highly stable, while in case of Fe3+, d-subshell is already half-filled, so it does not receive electron easily.
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Question 263 Marks
Compare the chemistry of actinoids with that of the lanthanoids with special reference to:
Chemical reactivity
Answer
Chemical reactivity: The earlier members of the lanthanoids series are quite reactive similar to calcium but, with increase in atomic number, they behave more like aluminium. The metals combine with hydrogen when. gently heated in the gas. Carbides, Ln3C, Ln2C3 and LnC2 are formed when the metals are heated with carbon. They liberate hydrogen from dilute acid and burn in halogens to form halides. They form oxides M2O3 and hydroxides M(OH)3. Actinoids are highly reactive metals, especially when finely divided. The action of boiling water on thern gives a mixture of oxide and hydride and combination with most non-metals take place at moderate temperatures. HCl attacks all metals but most are slightly affected by nitric acid owing to the formation of protective oxide layers, alkalis have no action. Actinoids are more reactive than lanthanoids due to bigger atomic size and lower ionisation energy.
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Question 273 Marks
Ionisation enthalpies of Ce, Pr and Nd are higher than Th, Pa and U.
Answer
It is because in the beginning, when 5f-orbitals begin to be occupied, they will penetrate less into the inner core of electrons. The 5f-electrons will therefore, be more effectively shielded from the nuclear charge than 4f-electrons of the corresponding lanthanoids. Therefore, outer electrons are less firmly held and they are available for bonding in the actinoids.
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Question 283 Marks
[Ti(H2O)]3+ is coloured while [Sc(H2O)6]3+ is colourless.
Answer
This is due to d-d transition of electron in [Ti(H2O)]3+ complex Ti3+ has one electron in d-orbital (3d1) which absorb energy corresponding to blue-green region and jumps from t2g to eg set of d-orbitals $\big(\text{t}^1_{2\text{g}}\text{ e}^0_{\text{g}}\xrightarrow{\ \ \ \ \ }\text{t}^0_{2\text{g}}\text{ e}^1_{\text{g}}\big).$ But Sc3+ has no electron in the d-orbital.
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Question 293 Marks
How would you account for the following:
There is a gradual decrease in the atomic sizes of transition elements in a series with increasing atomic numbers.
Answer
There is a gradual decrease in the atomic sizes of transition elements in a series with increasing atomic numbers due to poor shielding effect of d-electrons, the net electrostatic attraction between the nucleus and the outermost electrons increases.
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Question 303 Marks
How would you account for the following?
  1. Transition metals exhibit variable oxidation states.
  2. Zr(z=40) and Hf (Z=72) have almost identical redii.
  3. Transition metals and their compounds act as catalyst.
Answer
  1. Due to incomplete filling of d-orbitals, transition metals show variable oxidation states.
  2. Because of Lanthanoid Contraction.
  3. Because of their ability to show multiple/variable oxidation states.
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Question 313 Marks
E° values of Mn, Ni and Zn are more negative than expected.
Answer
Negative E° values for Mn2+ and Zn2+ are related to stabilities of half-filled and fully filled configurations, respectively. But for Ni2+, E° value is related to the highest negative enthalpy of hydration.
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Question 323 Marks
Give reasons:
  1. E° value for Mn3+/ Mn2+ couple is much more positive than that for Fe3+/ Fe2+.
  2. Iron has higher enthalpy of atomization than that of copper.
  3. Sc3+ is colourless in aqueous solution whereas Ti3+ is coloured.
Answer
  1. E°Mn3+/ Mn2+ = more positive

E°Fe3+/ Fe2+ = less positive

Mn2+ = 3d5 4s°

Mn3+ = 3d4

Mn3+ will have a tendency to go in the +2 state because Mn2+ is more stable than Mn3+, whereas in the case of Fe, Fe3+ is more stable than Fe2+.

  1. Iron has more number of unpaired electrons. Hence, it will have stronger interatomic interaction than copper.
  2. Sc3 is colourless as it does not have unpaired electrons

Ti3+ has one unpaired electron. Hence, it is coloured.

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Question 333 Marks
Give reasons:
  1. Mn shows the highest oxidation state of +7 with oxygen but with fluorine it shows the highest oxidation state of +4.
  2. Transition metals show variable oxidation states.
  3. Actinoids show irregularities in their electronic configurations.
Answer
  1.  Ability of oxygen to form multiple bond/ $p\pi-d\pi$ bond.
  2. Partially filled d orbitals/due to comparable energies of ns and (n-1) d orbitals.
  3. Due to relative stabilities of the fo,f7 and f14 occupancies of the 5f orbitals/Comparable energies of 7s,6d,5f orbitals. 
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Question 343 Marks
Give reasons for the following:
  1. Transition elements and their compounds act as catalysts.
  2. E0 value for (Mn2+|Mn) is negative whereas for (Cu2+|Cu) is positive.
  3. Actinoids show irregularities in their electronic configuration.
Answer
  1. Transition elements show catalytic property due to the presence of vacant orbitals that is why they have the tendency to form variable oxidation states. Hence, they can easily form intermediates with suitable reactants.
  2. E0 value of Cu2+/ Cu is positive that is why copper is least reactive metal among first transition metal this is because copper has the high enthalpy of atomization and enthalpy of ionization in comparison to Mn2+|Mn.
  3. This happens because of less energy difference between 5f, 6d and 7f subshell of actinides and hence electrons can easily be transferred from one subshell to another.
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Question 353 Marks
Match the properties given in Column I with the metals given in Column II.

 
Column I (Property)
 
Column II (Metal)
(i)
An element which can show +8 oxidation state
(a)
Mn
(ii)
3d block element that can show upto +7 oxidation state
(b)
Cr
(iii)
3d block element with highest melting point
(c)
Os
   
(d)
Fe

 

Answer
 
Column I (Property)
 
Column II (Metal)
(i)
An element which can show +8 oxidation state
(c)
Os
(ii)
3d block element that can show upto +7 oxidation state
(a)
Mn
(iii)
3d block element with highest melting point
(b)
Cr

Explanation:

  1. Osmium is an element which show +8 oxidation state.
  2. 3d block element that can show up to +7 oxidation state is manganese.
  3. 3d block element with highest melting point is chromium.
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Question 363 Marks
Answer the following questions:
Give reasons:
  1. Among transition metals, the highest oxidation state is exhibited in oxoanions of a metal.
  2. Ce4+ is used as an oxidising agent in volumetric analysis.
  3. Zn2+ salts are white while Cu2+ salts are blue.
Answer
  1. In these oxoanions the oxygen atoms are directly bonded to the transition metal. Since oxygen is highly electronegative, the oxoanions bring out the highest oxidation state of the metal.
  2. Ce4+ has the tendency to attain +3 oxidation state which is more stable and so it is used as an oxidising agent in volumetric analysis.
  3. Zn2+ ion has all its orbitals completely filled whereas in Cu2+ ion there is one half-filled 3d-orbital. Therefore, due to d-d transition Cu2+ has a tendency to form coloured salts whereas Zn2+ has no such tendency.
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Question 373 Marks
How would you account for the following? 
  1. Many of the transition elements are known to form interstitial compounds.
  2. The metallic radii of the third (5d) series of transition metals are virtually the same as those of the corresponding group members of the second (4d) series.
  3. Lanthanoids form primarily +3 ions, while the actinoids usually have higher oxidation states in their compounds, +4 or even +6 being typical.
Answer
  1. Because small atoms like H, C or N are trapped inside the crystal lattices of transition metals,
  2. Because of lanthanoid contraction. 
  3. This is due to comparable energies of 5f, 6d and 7s orbitals in actinoids.
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Question 383 Marks
Account for the following:
  1. CuCl2 is more stable than Cu2Cl2.
  2. Atomic radii of 4d and 5d series elements are nearly same.
  3. Hydrochloric acid is not used in permanganate titrations.
Answer
  1. In CuCl2, Cu is in +2 oxidation state which is more stable due to high hydration enthalpy as compared to Cu2Cl2 in which Cu is in +1 oxidation state.
  2. Due to lanthanoid contraction.
  3. Because HCl is oxidised to chlorine.
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Question 393 Marks
How would you account for the following:
  1. Among lanthanoids, Ln (III) compounds are predominant. However, occasionally in solutions or in solid compounds, +2 and +4 ions are also obtained.
  2. The EoM2+/M for copper is positive (0.34V). Copper is the only metal in the first series of transition elements showing this behaviour.
  3. The metallic radii of the third (5d) series of transition metals are nearly the same as those of the corresponding members of the second series.
Answer
  1. Lanthanoid Metals show +2 and +4 oxidation states to attain stable f0 and f7 configurations.
  2. Because of high enthalpy of atomization and ionisation which is not compensated by the enthalpy of hydration of Cu2+.
  3. Due to lanthanoid contraction.
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Question 403 Marks
How would you account for the following?
  1. The atomic radii of the metals of the third (5d) series of transition elements are virtually the same as those of the corresponding members of the second (4d) series.
  2. The Eo value for the Mn3+/Mn2+ couple is much more positive than that for Cr3+/Cr2+ couple or Fe3+/Fe2+ couple.
  3. The highest oxidation state of a metal is exhibited in its oxide or fluoride.
Answer
  1. Due to Lanthanoid Contraction/or its meaning.
  2. Due to stable half – filled 3d5 configuration of Mn2+/high 3rd ionisation enthalpy of Mn.
  3. Becuase Oxygen or Fluorine is highly electronegative and small size element.
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Question 413 Marks
Complete the following chemical equations:
  1. $\text{MnO}_{4}^{-}+\text{C}_{2}\text{O}_{4}^{2-}+\text{H}+\rightarrow$
  2. $\text{KMnO}_{4}\xrightarrow{\text{heated}}$
  3. $\text{Cr}_{2}\text{O}_{4}^{2-}+\text{H}_{2}\text{S}+\text{H}^{+}+\rightarrow$
Answer

  1. $\text{5C}_{2}\text{O}_{4}^{2-}+\text{2MnO}_{4}^{-}+\text{16H}^{4+}\xrightarrow{\ \ \ \ \ }\ \text{2Mn}^{2+}+\text{8H}_{2}\text{O}+\text{10CO}_{2}.$

  2. $\text{KMnO}_{4}\xrightarrow{\text{heat}}\ \text{K}_{2}\text{MnO}_{2}+\text{MnO}_{2}+\text{O}_{2}$

  3. $\text{Cr}_{2}\text{O}_{7}^{2-}+\text{3H}_{2}\text{S}+\text{8H}^{+}\xrightarrow{\ \ \ \ \ \ }\ \text{2Cr}^{3+}+\text{3S}+\text{7H}_{2}\text{O}$.

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Question 423 Marks
How would you account for the following situations?
  1. The transition metals generally form coloured compounds.
  2. With 3d4 configuration, Cr2+ acts as a reducing agent but Mn3+ acts as an oxidising agent. (Atomic Numbers, Cr = 24, Mn = 25).
  3. The actinoids exhibit a larger number of oxidation states than the corresponding lanthanoids.
Answer
  1. Transition metal contain unpaired electrons and are excited to higher energy levels/d-d transition/absorption in visible region.
  2. Cr2+ is reducing as its configuration changes from d4 to d3, the latter having half filled t2g level whereas Mn3+ to Mn2+ results in half filled configuration.
  3. Because 5f, 6d, and 7s are of comparable energy.
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Question 433 Marks
Complete the following chemical reaction equations:
$\text{Cr}_2\text{O}^{2-}_7(\text{aq})+\text{Fe}^{2+}(\text{aq})+\text{H}^+(\text{aq})\xrightarrow{\ \ \ \ \ \ \ \ \ }$
Answer
$\ \ \ \ \ \ \ \ \ \ \ \ \ \text{Cr}_2\text{O}^2_7+14\text{H}^+6\text{e}^-\xrightarrow{\ \ \ \ \ \ }2\text{Cr}^{3+}+7\text{H}_2\text{O}\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{Fe}^{2+}\xrightarrow{\ \ \ \ \ \ \ }\text{Fe}^{3+}+\text{e}^-\times6\\\underline{\overline{\text{Cr}_2\text{O}^{2-}_7+6\text{Fe}^{2+}+14\text{H}^+\xrightarrow{\ \ \ \ \ \ }2\text{Cr}^{3+}+6\text{Fe}^{3+}+7\text{H}_2\text{O}}}$
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Question 443 Marks
Although fluorine is more electronegative than oxygen, but the ability of oxygen to stabilise higher oxidation states exceeds that of fluorine. Why?
Answer
The electronic configuration of fluorine is 1s2 2s2 2p5. Thus it can form only one bond as it has only one unpaired electron. Electronic configuration of oxygen is 1s2 2s2 2p6 3s2 3p6.
It may be mentioned that oxygen also has vacant d-orbitals along with two 3p orbitals containing single electron. Thus, oxygen has greater bond formation capacity. In other words, it has greater ability to stabilize higher oxidation states.
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Question 453 Marks
Identify the metal and justify your answer.
  1. Carbonyl M (CO)5
  2. MO3F
Answer
It is = 0

x = +7M is in +7 oxidation state so that the given compound is MnO3F

  1. Fe(CO)5 by EAN rule

EAN = x + 2 × 5 = 36 (Kr is the nearest inert gas)

x = 26 (atomic no. of the metal) so the metal is iron.

  1.  MO3F is MnO3F

In MO3F let us assume M = x

x + 3 × (-2) + (-1)

x = +7

M is in + 7 osidation state so that the given compound is MnO3F

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Question 463 Marks
Explain the following observations:
There is a close similarity in physical and chemical properties of the 4d and 5d series of the transition elements, much more than expected on the basis of usual family relationship.
Answer
This is because 5d and 4d-series elements have virtually the same atomic and ionic radii due to lanthanide contraction. Due to equality in size of Zr and Hf, Nb and Ta, Mo and W, etc., the two elements of each pair have the same properties.
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Question 473 Marks
Complete the following chemical reaction equations:
$\text{MnO}^-_4(\text{aq})+\text{C}_2\text{O}^{2-}_4(\text{aq})+\text{H}^+(\text{aq})\xrightarrow{\ \ \ \ \ \ \ \ \ }$
Answer
$\ \ \ \ \ \ \ [\text{MnO}^-_4+8\text{H}^++5\text{e}\xrightarrow{\ \ \ \ \ \ }\text{Mn}^{2+}+4\text{H}_2\text{O}\times2]\\ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \ \text{[C}_2\text{O}^2_4\xrightarrow{\ \ \ \ \ \ \ \ }2\text{CO}^2_4+2\text{e}^-]\times5\\\underline{\overline{\text{MnO}_4+5\text{C}_2\text{O}^{2-}_{4}+16\text{H}^+\xrightarrow{\ \ \ \ \ \ \ }2\text{Mn}^{2+}+10\text{CO}_2+8\text{H}_2\text{O}}}$
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Question 483 Marks
While filling up of electrons in the atomic orbitals, the 4s orbital is filled before the 3d orbital but reverse happens during the ionisation of the atom. Explain why?
Answer
According to n + l rule:
For
3d = n + l = 5
4s = n + l = 4
Therefore, the electron will enter in 4s orbital first and then in 3d orbitals. Ionisation enthalpy is responsible for the ionisation of atom. 4s electrons are loosely held by the nucleus. So electrons are removed from 4s orbital prior to 3d.
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Question 493 Marks
Answer the following questions:
In the titration of FeSO4 with KMnO4 in the acidic medium, why is dil. H2SO4 used instead of dil. HCl?
Answer
Dil. H2SO4 is an oxidising agent and oxidises FeSO4 to Fe2(SO4)3. Dil. HCl is a reducing agent and liberates chlorine on reacting with KMnO4 solution.
Hence, the part of the oxygen produced from KMnO4 is used up by HCl.
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Question 503 Marks
A violet compound of manganese (A) decomposes on heating to liberate oxygen and compounds (B) and (C) of manganese are formed. Compound (C) reacts with KOH in the presence of potassium nitrate to give compound (B). On heating compound (C) with conc. H2SO4 and NaCl, chlorine gas is liberated and a compound (D) of manganese along with other products is formed. Identify compounds A to D and also explain the reactions involved.
Answer
A = KMnO4, B = K2MnO4, C = MnO2, D = MnCl2

Explanation:

A violet compound of manganese which is potassium permanganate (KMnO4), decomposes to liberate potassium manganate(K2MnO4) and manganese dioxide(MnO2) along with oxygen.

KMnO4(A) → K2MnO4(B) +MnO2(C) + O2

Manganese dioxide (MnO2)reacts with KOH to give potassium manganate (K2MnO4)

MnO2(C) + KOH + O2→2K2MnO4(B) +2H2O

On heating Manganese dioxide (MnO2) with NaCl and H2SO4, we get Manganese(II) chloride(MnCl2), chlorine gas and other products,

MnO2(C) + 4NaCl + 4H2SO4→ MnCl2(D) +4NaHSO4+2H2O +Cl2

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