CBSE Class 12 Chemistry Transition Elements Worksheet Set 01

Question. The transitional metal which forms green compound in +3 oxidation state and yellow orange compound in +6 oxidation state is?
a. Fe
b. Ni
c. Cr
d. Co
Answer : C

Question. In a reaction the ferrous (Fe⁺⁺) iron is oxidised to ferric (Fe++) ion. The equivalent weight of the ion in the above reaction is equal to:
a. Half of the atomic weight
b. 1/5 of the atomic weight
c. The atomic weight
d. Twice the atomic weight
Answer : C

Question. Which is most reactive metal?
a. Fe
b. Pt
c. Ni
d. Co
Answer : A

Question. The decrease in atomic volume from Cr to Cu is very negligible because:
a. Increase in nuclear change
b. Screening effect
c. Unpaired electrons of Cr
d. None
Answer : B

Question. The number of oxidation states of manganese is:
a. 4
b. 6
c. 3
d. 8
Answer : B

Question. The tendency towards complex formation is maximum in:
a. s - block elements
b. p - block elements
c. d - block elements
d. f - block elements
Answer : C

Question. Which is not true for transition elements?
a. They are all metals
b. They show variable valency
c. They form coloured ions
d. They do not form co-ordinate compounds
Answer : D

Question. Which element gives maximum balanced oxide?
a. V
b. Cr
c. Mn
d. Fe
Answer : C

Question. In which of the following metallic bond is strongest?
a. Fe
b. Sc
c. V
d. Cr
Answer : D

Question. Which of the following elements is alloyed with copper to form brass?
a. Lead
b. Silver
c. Zinc
d. Antimony
Answer : C

Question. How many unpaired electrons are there in 2 Ni + ?
a. 2
b. 4
c. 5
d. 0
Answer : A

Question. Which of the following have maximum number of unpaired electrons?
a. Fe³⁺
b. Fe²⁺
c. Co²⁺
d. Co³⁺
Answer : A

Question. Which of the following element has maximum first ionisation potential?
a. V
b. Ti
c. Cr
d. Mn
Answer : D

Question. Electronic configuration of Cu(Z=29) is:
a. [Ar]3d⁹ 4s²
b. [Ar]3d¹⁰ 4s²
c. [Ar]3d² 4s²
d. [Ar]3d⁶ 4s²
Answer : B

Question. The magnetic moment of a metal ion of first transition series is 2.83BM. Therefore it will have unpaired electrons:
a. 6
b. 4
c. 3
d. 2
Answer : D

Question. The valence shell electronic configuration of Cr²⁺ ion is:
a. o 4 4s 3d
b. 2 2 4s 3d
c. 2 0 4s 3d
d. 6 2 3p 4s
Answer : A

Question. Transition metals are related to which block?
a. s-block
b. p-block
c. d-block
d. None of these
Answer : B

Question. The oxidation number of iron in potassium ferrocyanide i
a. +2
b. +3
c. +4
d. Zero
Answer : A

Question. The atomic radii of the elements are almost same of which series?
a. Fe − Co − Ni
b. Na − K − Rb
c. F − Cl − Br
d. Li − Be − B
Answer : A

Question. Which of the following structure is that of a coinage metal?
a. 2, 8, 1
b. 2, 8, 18, 1
c. 2, 8, 8
d. 2, 18, 8, 3
Answer : B

Question. Which of the following transition metal cation has maximum unpaired electrons?
a. Mn⁺²
b. Fe⁺²
c. Co²⁺
d. Ni²⁺
Answer : A

Question. Which of the following statements is not true in regard to transition elements?
a. They readily form complex compounds
b. They show variable valency
c. All their ions are colourless
d. Their ions contain partially filled d -electron levels
Answer : C

Question. Which belongs to the actinides series?
a. Ce
b. Cf
c. Ca
d. Cs
Answer : B

Question. Effective magnetic droment of Sc+3 ion is:
a. 1.73
b. 0
c. 5.92
d. 2.83
Answer : B

Question. On the basis of position in the electrochemical series, the metal which does not displace hydrogen from water and acid is?
a. Hg
b. Al
c. Pb
d. Ba
Answer : A

Question. Which of the following represents the electronic configuration of a transition element?
a. 1s² ,2s² p⁶ .......ns² p³
b. 1s² ,2s² p⁶ .......ns² p⁶ d³ , (n +1)s²
c. 1s² ,2s² p⁶ .......ns² p⁶ d¹⁰ , (n +1)s² p¹
d. 1s² , 2s² p⁶ .......ns² p⁶
Answer : B

Question. Which of the following is not an actinide?
a. Curium
b. Californium
c. Uranium
d. Terbium
Answer : D

Question. Which of the following is not true for transition metals?
a. They are malleable and ductile
b. They have high boiling and melting points
c. They crystallize with body centred cubic and hexagonal close-packed structures only
d. They show variable oxidation states although not always
Answer : C

Question. Which of the following ions is paramagnetic?
a. Cu⁺
b. Zn⁺²
c. Ti⁺³
d. Ti⁺⁴
Answer : C

Question. Iron is rendered passive by the action of:
a. Conc. H₂SO₄
b. Conc. H₃PO₄
c. Conc. HCl
d. Conc. HNO₃
Answer : D

Question. From +6 to +1 oxidation state is shown by the element of....... group:
a. V-B
b. VI-B
c. VII-B
d. VIII
Answer : B

Question. Fe²⁺ shows:
a. Ferromagnetism
b. Paramagnetism
c. Diamagnetism
d. None of these
Answer : B

Question. Rare-earth elements are exhibited by:
a. At. No. 58 to 71
b. At. No. 21 to 30
c. At. No. 39 to 71
d. At. No. 81 to 91
Answer : A

Question. The tendency of 3d-metal ions to form stable complexes is due to their:
a. Variable oxidation state
b. Strong electronegative nature
c. High charge/size ratio and vacant d-orbitals
d. Very low ionization energies
Answer : C

Question. Fe, Co and Ni have valuable catalytic properties in process involving:
a. Organic compound
b. Oxidation
c. Hydrogenation
d. Compounds of hydrogen f-Block Elements
Answer : C

Question. Which of the following oxides of manganese is amphoteric?
(a) MnO₂
(b) Mn₂O₃
(c) Mn₂O7
(d) MnO
Answer : A

Question. In which of the following compounds manganese has oxidation number equal to that of iodine in KIO₄ ?
(a) Potassium manganate
(b) Potassium permanganate
(c) Manganous chloride
(d) Manganese chloride
Answer : B

Question. Which of the following statement is not true about Mohr's salt?
a. It decolourises KMnO₄
b. It is a primary standard
c. It is a double salt
d. Oxidation state of iron is +3 in this salt
Answer : D

Question. Which of the following transition metal is present in misch metal?
a. La
b. Sc
c. Ni
d. Cr
Answer : A

Question. The hardness of Cr is due to ...... and metallic lusture is due to .......
a. Covalent bond, metallic bond
b. Covalent bond, hydrogen bond
c. Metallic bond, covalent bond
d. Metallic bond, hydrogen bond
Answer : A

Question. In the first transition series, the highest B.P. and M.P. is of:
a. Cr
b. V
c. Ni
d. Fe
Answer : A

Question. In the following transition elements, the lowest M.P. and B.P. is exhibited by:
a. Cr
b. Hg
c. Cu
d. Au
Answer : B

Question. Which of the following element does not show variable valency?
a. Ni
b. Zn
c. Cu
d. Mn
Answer : B

Question. Which of the following is diamagnetic transitional metal ion?
a. Ni⁺²
b. Zn⁺²
c. Co⁺²
d. Cu⁺²
Answer : B

Question. The correct order of ionic radii of Y³⁺, La³⁺, Eu³⁺ and Lu³⁺ is
(a) La³⁺ < Eu³⁺ < Lu³⁺ < Y³⁺
(b) Y³⁺ < La³⁺ < Eu³⁺ <Lu³⁺
(c) Y³⁺ < Lu³⁺ < Eu³⁺ < La³⁺
(d) Lu³⁺ < Eu³⁺ < La³⁺ < Y³⁺
(Atomic nos. Y =39, La = 57, Eu = 63, Lu = 71)
Answer : C

Question. Railway wagon axles are made by heating rods of iron embedded in charcoal powder. The process is known as:
a. Case hardening
b. Sheradizing
c. Annealing
d. Tempering
Answer : A

Question. What would happen when a solution of potassium chromate is treated with an excess of dilute nitric acid?
(a) Cr₂O₇^2- and H₂O are formed
(b) CrO₄^2- is reduced to +3 state of Cr
(c) CrO₄^2- is oxidized to +7 state of Cr
(d) Cr³⁺ and Cr₂O₇^2- are formed
Answer : A

Question. Which has the maximum ferromagnetic character?
a. Fe
b. Co
c. Ni
d. Pt
Answer : A

Question. Which forms interstitial compounds?
a. Fe
b. Co
c. Ni
d. All of these
Answer : D

Question. Which occurs in nature in free-state?
a. Fe
b. Co
c. Ni
d. Pt
Answer : D

Question. Which of the following ions has the least magnetic moment?
a. Cu⁺²
b. Ni⁺²
c. Co⁺³
d. Fe⁺²
Answer : A

Question. Among the following outermost configurations of transition metals, which shows the highest oxidation state?
a. 3d³ 4s²
b. 3d⁵ 4s²
c. 3d⁵ 4s²
d. 3d⁶ 4s²
Answer : C

Question. For the four successive transition elements (Cr, Mn, Fe and Co), the stability of +2 oxidation state will be there in which of the following order?
(a) Mn > Fe > Cr > Co
(b) Fe > Mn > Co > Cr
(c) Co > Mn > Fe > Cr
(d) Cr > Mn > Co > Fe
Answer : A

Question. Which of the following compound is called Turnbull’s blue?
(a) Ferricyanide
(b) Ferrous ferricyanide
(c) Ferrous cyanide
(d) Ferri-ferrocyanide
Answer : B

Question. A compound of a metal ion Mx+ (Z = 24) has a spin only magnetic moment of 15 Bohr Magnetons. The number of unpaired electrons in the compound are
(a) 2
(b) 4
(c) 5
(d) 3
Answer : D

Question. Of the following outer electronic configurations of atoms, the highest oxidation state is achieved by which one of them?
a. (n −1)d³ ns²
b. (n −1)d⁵ ns²
c. (n −1)d⁸ ns²
d. (n −1)d⁵ ns²
Answer : D

Question. Among the following series of transition metal ions, the one where all metals ions have 3d² electronic configuration is?
a. Ti⁴⁺ , V³⁺, Cr²⁺, Mn³⁺ 
b. Ti²⁺ , V³⁺, Cr⁴⁺, Mn⁵⁺
c. Ti³⁺, V²⁺, Cr³⁺ , Mn⁴⁺
d. Ti⁺ ,  V⁴⁺,  Cr⁶⁺, Mn⁷⁺ 
Answer : B

Question. In photography sodium thiosulphate is used as:
a. Complexing agent
b. Oxidising agent
c. Reducing agent
d. None of these
Answer : A

Question. The form of iron having the highest carbon content is:
a. Cast iron
b. Wrought iron
c. Strain steel
d. Mild steel
Answer : A

Question. The roasting of HgS in air produces
(a) HgO
(b) HgSO₃
(c) HgSO₄
(d) Hg
Answer : D

Question. When MnO₂ is fused with KOH, a coloured compound is formed, the product and its colour is:
(a) K₂MnO₄, purple green
(b) KMnO₄, purple
(c) Mn₂O₃, brown
(d) Mn₃O₄, black
Answer : A

Question. If an aqueous solution of KCN is added to a solution of ferrous salt then the complex formed is represented by
(a) [Fe(CN)₆]^4-
(b) [Fe(CN)₆]^3-
(c) [Fe(H₂O)₆]³⁺
(d) [Fe(H₂O)₆]²⁺
Answer : A

Question. In solid CuSO₄ .5H₂O copper is co-ordinated to:
a. Five water molecules
b. Four water molecules
c. One sulphate anion
d. One water molecule
Answer : B

Question. In which of the following ionic radii of chromium would be smallest?
a. 2 4 K CrO
b. 2 CrO
c. 3 CrCl
d. 2 CrF
Answer : A

Question. Which of the following metals absorbs hydrogen?
a. K
b. Al
c. Zn
d. Pd
Answer : D

Question. Which of the following ions is coloured?
a. Cu⁺
b. Cu²⁺
c. Ti⁴⁺
d. V⁵⁺
Answer : B

Question. The equivalent weight of potassium permanganate for acid solution is:
a. 158
b. 31.6
c. 52.16
d. 79
Answer : B

Question. Which is an amphoteric oxide?
a. ZnO
b. CaO
c. BaO
d. SrO
Answer : A

Question. Percentage of silver in German silver is:
a. 0%
b. 1%
c. 5%
d. None of these
Answer : D

Question. Which of the following sulphides is yellow in colour?
a. CuS
b. CdS
c. ZnS
d. CoS
Answer : B

Question. Mercury (I) chloride sublimes, when this compound is heated and the vapours it gives off are cooled the substance collected consists of
(a) Mercury (II) chloride
(b) Mercury (I) and Mercury (II), chlorides
(c) Mercury (II) chloride and mercury
(d) Mercury
Answer : D

Question. In neutral or faintly alkaline medium, thiosulphate is quantitatively oxidized by KMnO₄ to
(a) SO₃^2–
(b) SO₄^2–
(c) SO₂
(d) SO₅^2–
Answer : B

Question. An excess of Na₂S₂O₃ reacts with aqueous CuSO₄ to give
(a) CuS₂O₃
(b) Cu₂S₂O₃
(c) Na₂[Cu(S₂O₃)₂]
(d) Na₄[Cu₆(S₂O₃)₅]
Answer : D

Question. Which of the following oxides of Cr is amphoteric
(a) CrO₂
(b) Cr₂O₃
(c) CrO₅
(d) CrO₃
Answer : A

Question. Consider the following statements
(I) La(OH)₃ is the least basic among hydroxides of lanthanoids.
(II) Zr⁴⁺ and Hf⁴⁺ posses almost the same ionic radii.
(III) Ce⁴⁺ can as an oxidizing agent.
Which of the above is/are true ?
(a) (I) and (III)
(b) (II) and (III)
(c) (II) only
(d) (I) and (II)
Answer : B

Numeric Value Answer

Question. The number of water molecule(s) directly bonded to the metal centre in CuSO₄.5H₂O is
Answer : 4

Question. Out of the following, how many oxides are acidic.
MnO, Mn₂O₃, MnO₂, MnO₃, Mn₂O₇
Answer : 3

Question. What is the oxidation states of Cr in butterfly structure.
Answer : 6

Question. The oxidation number of Mn in the product of alkaline oxidative fusion of MnO₂ is
Answer : 6

Question. A light blue coloured compound (A) on heating gives a block compound (B) which reacts with glucose to give a red compound (C). After the completion of reaction, how many metal oxides are formed?
Answer : 2

TRANSITION ELEMENTS

Question 1(a). As compared to elements of the first transition series, heavier transition metals of second & third transition series show more frequent metal - metal bonding.
Answer: Valence electrons of the heavier transition elements (second and third series) are more effectively shielded from the nuclear charge by the inner d and f orbitals, making the valence d-orbitals larger and more directional. Consequently, the d-orbitals of heavier transition elements overlap more effectively with each other, leading to frequent metal-metal bonding.
In simple words: Larger d-orbitals in heavier transition metals overlap better, forming stronger metal-metal bonds.

Exam Tip: Mention greater spatial extension of d-orbitals in 4d and 5d series compared to 3d series.

Question 1(b). Elements of second and third d series have similar radii.
Answer: This is due to the lanthanoid contraction. As we move from La to Lu in the third transition series, the filling of 4f orbitals takes place, which provides poor shielding of nuclear charge. This regular decrease in atomic size exactly cancels out the expected size increase due to the addition of a new principal quantum shell when moving from the second (4d) to the third (5d) transition series.
In simple words: Lanthanoid contraction counteracts the expected increase in atomic size down the group.

Exam Tip: Always use the key phrase "lanthanoid contraction" and explain poor shielding of 4f electrons.

Question 1(c). The unipositive transition metal ion have \(d^9\) configuration.
<Answer: Transition elements in +1 oxidation state have lost one electron from the ns orbital, leaving the underlying \((n-1)d\) subshell completely filled or partially filled. Due to exceptionally high hydration or lattice energies, elements like Cu show a \(d^9\) configuration (\(3d^9\) for \(\text{Cu}^+\)) in certain unipositive species.
In simple words: Removing one ns electron leaves a stable or nearly filled \(d^9\) electron configuration.

Exam Tip: Relate this to the general valence shell configuration of \(ns^1 (n-1)d^{10}\) or \(ns^2 (n-1)d^9\) upon losing an ns electron.

Question 1(d). Scandium (II) is virtually unknown.
Answer: Scandium has an electronic configuration of \([Ar] 3d^1 4s^2\). It readily loses three electrons to attain a very stable noble gas configuration (\(\text{Sc}^{3+}\)). The third ionization enthalpy of scandium is relatively low, making the formation of \(\text{Sc}^{2+}\) thermodynamically unfavorable compared to \(\text{Sc}^{3+}\).
In simple words: Scandium prefers to lose three electrons to achieve a stable noble gas configuration, so +2 is unstable.

Exam Tip: Mention the exceptionally low third ionization energy required to form \(\text{Sc}^{3+}\).

Question 1(e). Titanium (IV) is more stable than Titanium (II) or Titanium (III).
Answer: Titanium has a valence configuration of \(3d^2 4s^2\). By losing all four valence electrons, \(\text{Ti}^{4+}\) acquires a stable noble gas configuration (\([Ar]\)), which is energetically much more stable than having 1 or 2 electrons remaining in the d-orbitals as in \(\text{Ti}^{3+}\) or \(\text{Ti}^{2+}\).
In simple words: Losing all four electrons gives Titanium a stable noble gas configuration.

Exam Tip: Cite the attainment of inert gas configuration (noble gas core).

Question 1(f). Oxidation states in transition elements vary from each other by unity (e.g., V(II), V(III), V(IV), V(V)) whereas oxidation states of non-transition states normally differ by a unit of two.
Answer: In transition elements, the energy difference between \((n-1)d\) and \(ns\) orbitals is very small, allowing both ns and \((n-1)d\) electrons to participate in bonding, leading to multiple oxidation states differing by one unit (\(d^1, d^2, \dots\)). In non-transition elements, electrons are lost or gained in pairs to complete octets, causing oxidation states to differ by two units.
In simple words: Small energy gap between s and d orbitals allows transition metals to change oxidation states by single electrons.

Exam Tip: Highlight the small energy difference between \((n-1)d\) and \(ns\) orbitals.

Question 1(g). In non-transition elements; lower oxidation state is favoured by heavier elements whereas in transition elements higher transition state is favoured by heavier elements.
Answer: In non-transition p-block elements, the inert pair effect becomes more pronounced for heavier elements, stabilizing lower oxidation states. In contrast, for d-block transition elements, heavier elements have larger d-orbitals with lower ionization energies, making it easier for them to utilize all valence electrons and attain higher oxidation states (e.g., Os in +8 state).
In simple words: Heavier p-block elements show inert pair effect (lower oxidation states), while heavier d-block elements can easily reach higher oxidation states.

Exam Tip: Contrast the inert pair effect in p-block with the involvement of all valence electrons in heavier d-block elements.

Question 1(h). \(E^{\circ}_{M^{2+}/M}\) value for Mn, Ni & Zn is more negative than expected.
Answer: The standard electrode potential \(E^{\circ}_{M^{2+}/M}\) depends on sublimation, ionization, and hydration enthalpies. For Mn, the extra stability of the half-filled \(3d^5\) configuration in \(\text{Mn}^{2+}\) makes its third ionization enthalpy very high, but its second ionization is relatively regular; however, its hydration energy is less negative. Specifically, Mn has a half-filled \(d^5\) stability, Zn has a completely filled \(d^{10}\) stability, and Ni has a particularly high hydration enthalpy, causing deviations making their values more negative than expected from general trends.
In simple words: Extra stability of half-filled (\(\text{Mn}^{2+}\)) and fully-filled (\(\text{Zn}^{2+}\)) configurations affect these reduction potentials.

Exam Tip: Mention exchange energy and stable symmetrical configurations (\(d^5\) for Mn, \(d^{10}\) for Zn).

Question 1(i). Highest Mn fluoride is \(\text{MnF}_4\) with Mn in +4 O.S. whereas highest Mn oxide is \(\text{Mn}_2\text{O}_7\) with Mn in +7 O.S.
Answer: Oxygen has a high electronegativity and a small size, and it can form multiple bonds (\(p_{\pi}-d_{\pi}\) bonding) with transition metals, enabling manganese to reach its maximum oxidation state of +7 in \(\text{Mn}_2\text{O}_7\). Fluorine is also very electronegative and oxidising, but due to the absence of \(p_{\pi}-d_{\pi}\) multiple bonding capability and steric crowding around the manganese atom, it can only stabilize up to the +4 oxidation state in \(\text{MnF}_4$.
In simple words: Oxygen forms multiple bonds allowing higher oxidation states (+7), whereas fluorine causes steric crowding and lacks multiple bonding capability, stopping at +4.

Exam Tip: Use the crucial term "\(p_{\pi}-d_{\pi}\) multiple bonding" to explain oxide stability.

Question 1(j). For Cu \(\Delta_i H_1^{\circ} = 745\) kJ/mol whereas \(\Delta_i H_2^{\circ} = 1940\) kJ/mol, despite of very high second ionization enthalpy more stable O.S. of Cu in aqueous medium is +2.
Answer: Although the second ionization enthalpy (\(\Delta_i H_2^{\circ}\)) of copper is very high, the hydration enthalpy (\(\Delta_{\text{hyd}} H^{\circ}\)) of \(\text{Cu}^{2+}\) is much more negative than that of \(\text{Cu}^{+}\). This high negative hydration energy of \(\text{Cu}^{2+}\) easily compensates for the high energy required for the second ionization, making \(\text{Cu}^{2+}\) much more stable in aqueous solutions than \(\text{Cu}^{+}\).
In simple words: The high hydration energy of \(\text{Cu}^{2+}\) makes up for the high second ionization energy.

Exam Tip: Emphasize the role of hydration enthalpy compensating for high ionization enthalpy.

Question 1(k). \(\text{V}_2\text{O}_3\) is basic while \(\text{V}_2\text{O}_5\) is amphoteric.
Answer: In \(\text{V}_2\text{O}_3\text{,}\) vanadium is in a lower oxidation state (+3), where its oxides tend to be basic because the metal ion has greater metallic character and ionic bonding. In \(\text{V}_2\text{O}_5\text{,}\) vanadium is in a higher oxidation state (+5), which increases polarizing power according to Fajan's rules, imparting covalent character and making the oxide acidic/amphoteric.
In simple words: Lower oxidation state oxides are basic, while higher oxidation state oxides are acidic or amphoteric.

Exam Tip: Relate oxidation state directly to acidic or basic character (higher O.S. = more acidic/amphoteric).

Question 1(l). Lanthanoids resemble each other more closely than do members of ordinary transition series.
Answer: In lanthanoids, the differentiating electrons enter the anti-penultimate 4f orbitals, which are deeply buried inside the atom and are well shielded by the outer \(5s^2 5p^6\) orbitals. As a result, variations in their chemical and physical properties are minimal across the series, leading to an extreme chemical resemblance. In ordinary transition series, d-orbitals are outer orbitals and interact more with ligands.
In simple words: Deeply buried 4f electrons in lanthanoids experience minimal environmental variation, causing high chemical similarity.

Exam Tip: Mention the shielding effect of outer orbitals on inner 4f electrons.

Question 1(m). La and Lu do not show any colour.
Answer: Colour in transition and inner-transition metal ions is generally due to d-d or f-f electronic transitions, which require partially filled d or f orbitals. Lanthanum (\(\text{La}^{3+}\)) has an empty 4f-orbital (\(4f^0\)) and Lutetium (\(\text{Lu}^{3+}\)) has a completely filled 4f-orbital (\(4f^{14}\)). Since neither ion has unpaired f-electrons capable of undergoing f-f transitions, their salts are colorless.
In simple words: Lanthanum has \(4f^0\) and Lutetium has \(4f^{14}\) configurations, meaning no unpaired electrons for f-f transitions.

Exam Tip: State the absence of unpaired electrons (\(4f^0\) and \(4f^{14}\)).

Question 1(n). Actinoids show greater range of O.S. as compared to lanthanoids.
Answer: Actinoids show a greater range of oxidation states because the 5f, 6d, and 7s orbitals are of comparable energies, making all these electrons available for bonding. In contrast, in lanthanoids, the 4f orbitals are located much deeper inside and have a larger energy gap from 5d and 6s orbitals, restricting most lanthanoids to the +3 oxidation state.
In simple words: Comparable energies of 5f, 6d, and 7s orbitals in actinoids allow more diverse oxidation states.

Exam Tip: Point out the comparable energy levels of 5f, 6d, and 7s orbitals.

Question 1(o). Ionization enthalpy of early actinides is lower than that of early lanthanoids.
Answer: The 5f orbitals in actinoids extend further from the nucleus than the corresponding 4f orbitals in lanthanoids, meaning 5f electrons are less tightly held by the nucleus and are shielded less effectively. Consequently, removing an electron from early actinides requires less energy than from early lanthanoids.
In simple words: 5f electrons are further out and less tightly held than 4f electrons, making ionization easier.

Exam Tip: Mention the greater spatial extension of 5f orbitals compared to 4f orbitals.

Question 1(p). Study of actinides is comparatively difficult.
Answer: Actinoids are radioactive elements with short half-lives for many of their isotopes, and they are available only in extremely small quantities. Furthermore, they exhibit multiple oxidation states and form complex chemical equilibria, making their experimental handling and study difficult.
In simple words: Actinoids are radioactive, scarce, and exist in multiple oxidation states.

Exam Tip: Highlight their radioactivity and variable oxidation states.

COORDINATION CHEMISTRY

Question 1. Give IUPAC names of the following:-
(a) \(\text{Na}_2[\text{CrF}_4\text{O}]\)
(b) \(\text{K}[\text{Pt}(\text{NH}_3)\text{Cl}_3]\)
(c) \(\text{Na}_2[\text{SiF}_6]\)
(d) \([\text{CoCl}(\text{en})_2(\text{ONO})]^{+}\)
(e) \([\text{Co}(\text{NH}_3)_5(\text{CO}_3)]\text{Cl}\)
(f) \([\text{Pt}(\text{NH}_3)_2(\text{py})_2][\text{PtCl}_4]\)
(g) \([\text{Cr}(\text{PPh}_3)(\text{CO})_5]\)
(h) \([\text{Mn}_2(\text{CO})_{10}]\) (Note: corrected from OCR typo \(\text{Mn}_3\text{CO}_{12}\))
(i) \([\text{Co}(\text{NH}_3)_6]\text{ClSO}_4\
(j) \(\text{K}_3[\text{Al}(\text{C}_2\text{O}_4)_3]\)
(k) \(\text{Hg}[\text{Co}(\text{SCN})_4]\)
(l) \(\text{K}_2[\text{Zn}(\text{OH})_4]\)
(m) \([\text{Co}(\text{en})_3][\text{Cr}(\text{CN})_6]\)
(n) \(\text{Cs}[\text{FeCl}_4]\)
Answer:
(a) Sodium tetrafluoroxochromate(V)
(b) Potassium amminetrichloroplatinate(II)
(c) Sodium hexafluorosilicate(IV)
(d) Chlorobis(ethane-1,2-diamine)(nitrito-O)cobalt(III) ion
(e) Pentaamminecarbonatocobalt(III) chloride
(f) Diamminedipyridineplatinum(II) tetrachloroplatinate(II)
(g) Pentacarbonyl(triphenylphosphine)chromium(0)
(h) Decarbonyldimanganese(0)
(i) Hexaamminecobalt(III) chloride sulfate
(j) Potassium tris(oxalato)aluminate(III)
(k) Mercury(II) tetrathiocyanatocobaltate(II)
(l) Potassium tetrahydroxozincate(II)
(m) Tris(ethane-1,2-diamine)cobalt(III) hexacyanochromate(III)
(n) Cesium tetrachloroferrate(III)
In simple words: Follow standard coordination nomenclature rules: name ligands alphabetically followed by the metal name and its oxidation state in Roman numerals.

Exam Tip: Remember to use prefixes like bis-, tris- for chelating ligands like ethylenediamine (en).

Question 2. Give one chemical test to distinguish between \([\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4\) and \([\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}\).
Answer: Add barium chloride (\(\text{BaCl}_2\)) solution to aqueous solutions of both coordination compounds. \([\text{Co}(\text{NH}_3)_5\text{Br}]\text{SO}_4\) will give a white precipitate of barium sulfate (\(\text{BaSO}_4\)), whereas \([\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}\) will not give this precipitate with \(\text{BaCl}_2\). Alternatively, silver nitrate (\(\text{AgNO}_3\)) can be added: \([\text{Co}(\text{NH}_3)_5\text{SO}_4]\text{Br}\) gives a pale yellow precipitate of \(\text{AgBr}\) with \(\text{AgNO}_3\), while the other does not.
In simple words: Test with \(\text{BaCl}_2\) to precipitate sulfate ions, or \(\text{AgNO}_3\) to precipitate bromide ions.

Exam Tip: Mention the reagent and the visible observation (precipitate formation).

Question 3. A coordination compound has formula \(\text{CoCl}_3 \cdot 4\text{NH}_3\). It does not liberate ammonia but precipitates chloride ions as silver chloride. Give the IUPAC name of complex and its structural formula.
Answer: Since it does not liberate ammonia with alkali, all four ammonia molecules are coordinated inside the coordination sphere. Since one mole of the compound precipitates one mole of \(\text{AgCl}\) with \(\text{AgNO}_3\), only one chloride ion is outside the sphere.
Structural formula: \([\text{Co}(\text{NH}_3)_4\text{Cl}_2]\text{Cl}\)
IUPAC Name: Tetraamminedichlorocobalt(III) chloride
In simple words: Four \(\text{NH}_3\) ligands are inside the sphere, two chlorines are inside, and one chlorine is outside as a counter-ion.

Exam Tip: Deduce the number of ionizable ions from precipitation reactions with \(\text{AgNO}_3\).

Question 4. The molar conductivity of \(\text{CaCl}_2 \cdot 4\text{NH}_3 \cdot 2\text{H}_2\text{O}\) is found to be same as that of 3:1 electrolyte. What is the structural formula of the complex?
Answer: A 3:1 electrolyte dissociates into 4 ions total (3 cations/anions of one type and 1 of the other). Considering calcium as the central metal ion with coordination number 6: two chloride ions, four ammonia molecules, and two water molecules are present.
Structural formula: \([\text{Ca}(\text{NH}_3)_4(\text{H}_2\text{O})_2]\text{Cl}_2\)
Upon dissociation, it gives 1 complex cation \([\text{Ca}(\text{NH}_3)_4(\text{H}_2\text{O})_2]^{2+}\) and 2 chloride ions \(2\text{Cl}^{-}\), totaling 3 ions (a 3:1 electrolyte).
In simple words: The formula gives three ions in solution, matching a 3:1 electrolyte ratio.

Exam Tip: Match the number of dissociated ions to the electrolyte type (3:1 means 3 ions of one sign and 1 of another, total 4 ions, or vice versa depending on formulation).

Question 5. Give reasons for the following:
(a) \([\text{Fe}(\text{CN})_6]^{3-}\) is weakly paramagnetic while \([\text{Fe}(\text{CN})_6]^{4-}\) is diamagnetic. (VB)
(b) \([\text{Ni}(\text{CO})_4]\) is tetrahedral while \([\text{Ni}(\text{CN})_4]^{2-}\) is square planar. (VB)
(c) \([\text{Co}(\text{CN})_6]^{3-}\) is low spin complex while \([\text{CoF}_6]^{3-}\) is a high spin complex (VB)
(d) \([\text{Mn}(\text{H}_2\text{O})_6]^{2+}\) has five unpaired electrons while \([\text{Mn}(\text{CN})_6]^{4-}\) has only one unpaired. (CFST)
(e) \([\text{Ti}(\text{H}_2\text{O})_6]^{3+}\) is coloured while \([\text{Sc}(\text{H}_2\text{O})_6]^{3+}\) is colorless.
Answer:
(a) In \([\text{Fe}(\text{CN})_6]^{3-}\), Fe is in +3 state (\(3d^5\)) with cyanide being a strong field ligand, but due to odd electron count it leaves 1 unpaired electron (weak paramagnetism). In \([\text{Fe}(\text{CN})_6]^{4-}\), Fe is in +2 state (\(3d^6\)) and CN is a strong field ligand causing complete pairing of all six d-electrons, leaving zero unpaired electrons (diamagnetic).
(b) \([\text{Ni}(\text{CO})_4]\) involves \(sp^3\) hybridisation resulting in a tetrahedral geometry. \([\text{Ni}(\text{CN})_6]^{2-}\) involves \(dsp^2\) hybridisation because \(\text{CN}^{-}\) is a strong field ligand forcing electron pairing, resulting in a square planar geometry.
(c) \(\text{CN}^{-}\) is a strong field ligand causing pairing of electrons in \([\text{Co}(\text{CN})_6]^{3-}\) leading to \(d^2sp^3\) inner orbital low-spin complex. \(\text{F}^{-}\) is a weak field ligand which cannot pair electrons, leading to \(sp^3d^2\) outer orbital high-spin complex.
(d) In \([\text{Mn}(\text{H}_2\text{O})_6]^{2+}\), \(\text{H}_2\text{O}\) is a weak field ligand (\(\Delta_o < P\)), giving high spin configuration \(t_{2g}^3 e_g^2\) with 5 unpaired electrons. In \([\text{Mn}(\text{CN})_6]^{4-}\), \(\text{CN}^{-}\) is a strong field ligand (\(\Delta_o > P\)), causing pairing giving configuration \(t_{2g}^5 e_g^0\) with only 1 unpaired electron.
(e) \([\text{Ti}(\text{H}_2\text{O})_6]^{3+}\) has \(\text{Ti}^{3+}\) with a \(3d^1\) configuration, allowing d-d electronic transitions that absorb visible light, causing color. \(\text{Sc}^{3+}\) in \([\text{Sc}(\text{H}_2\text{O})_6]^{3+}\) has a \(3d^0\) configuration with no d-electrons, so no d-d transition is possible, making it colorless.
In simple words: Strong field ligands pair up electrons (low spin/diamagnetic), weak field ligands do not (high spin/paramagnetic). Color requires unpaired d-electrons for d-d transitions.

Exam Tip: Clearly state ligand field strength (strong vs weak) and apply Valence Bond or Crystal Field theory accordingly.

Question 6. A, B and C are three complexes of chromium with empirical formula \(\text{H}_{12}\text{O}_6\text{Cl}_3\text{Cr}\). All the three complexes have \(\text{Cl}\) and \(\text{H}_2\text{O}\) molecules as the ligands. Complex A does not react with conc. \(\text{H}_2\text{SO}_4\). Complex B and C lose 6.75% and 13.5% of their original weight respectively on heating with conc. \(\text{H}_2\text{SO}_4\). Identify A, B and C.
Answer:
Molar mass of the empirical formula \([\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3\) = \(52 + (6 \times 18) + (3 \times 35.5) = 52 + 108 + 106.5 = 266.5 \text{ g/mol}\).
- Complex A does not lose water with conc. \(\text{H}_2\text{SO}_4\), meaning all 6 water molecules are inside the coordination sphere as ligands.
Formula of A: \([\text{Cr}(\text{H}_2\text{O})_6]\text{Cl}_3\
- Complex B loses 6.75% of its weight, which corresponds to the loss of 1 water molecule per formula unit (\(\frac{18}{266.5} \times 100 \approx 6.75\%\)). Thus, 5 water molecules are inside and 1 water molecule is outside as water of hydration.
Formula of B: \([\text{Cr}(\text{H}_2\text{O})_5\text{Cl}]\text{Cl}_2 \cdot \text{H}_2\text{O}\)
- Complex C loses 13.5% of its weight, corresponding to the loss of 2 water molecules (\(\frac{36}{266.5} \times 100 \approx 13.5\%\)). Thus, 4 water molecules are inside and 2 are outside as water of hydration.
Formula of C: \([\text{Cr}(\text{H}_2\text{O})_4\text{Cl}_2]\text{Cl} \cdot 2\text{H}_2\text{O}\)
In simple words: Calculate percentage mass loss to determine how many water molecules are outside the coordination sphere as crystal water.

Exam Tip: Use percentage mass loss calculations for water of crystallization to determine hydrate stoichiometry.

Q1. Give reasons for the following:-

(a) As compared to elements of the first transition series, heavier transition metals of second & third transition series show more frequent metal - metal bonding.

(b) Elements of second and third d series have similar radii.

(c) The unipositive transition metal ion have d" configuration.

(d) Scandium (II) is virtually unknown.

(e) Titanium (IV) is more stable than Titanium (II) or Titanium (III).

(f) Oxidation states in transition elements vary from each other by unity (eg V(ll) V(IlI) V(JV) V(V) whereas oxidation states of non transition states normally differ by a unit of two.

(g) In non transition elements; lower transition state is favoured by heavier elements whereas in transition elements higher transition state is favoured by heavier elements.  

TRANSITION ELEMENTS

Q1. Give reasons for the following:-

(a) As compared to elements of the first transition series, heavier transition metals of second & third transition series show more frequent metal – metal bonding.

(b) Elements of second and third d series have similar radii.

(c) The unipositive transition metal ion have dn configuration.

(d) Scandium (II) is virtually unknown.

(e)Titanium (IV) is more stable than Titanium (II) or Titanium (III).

(f) Oxidation states in transition elements vary from each other by unity (eg V(II) V(III) V(IV) V(V) whereas oxidation states of non transition states normally differ by a unit of two.

(g) In non transition elements; lower transition state is favoured by heavier elements whereas in transition elements higher transition state is favoured by heavier elements.

(h) E⁰M⁺² /M value for Mn, Ni & Zn is more negative expected.

(i) Highest Mn fluoride is MnF4 with Mn in +4 O.S. whereas highest Mn oxide is Mn₂O₇ with Mn in +7 O.S.

(j) For Cu Δ _iH0₁ = 745 KJ/mol whereas Δ _iH0₂ = 1940 KJ/mol, Inspite of very high second ionization enthalpy more stable O.S. of Cu in aqueous medium is +2

(k) V₂O₃ is basic while V₂O₅ is amphoteric

(l) Lanthanides resemble each other more closely than do members of ordinary transition series

(m) La & Lu do not show any colour

(n) Actinides show greater range of O.S. as compared to lanthanides.

(o) Ionization enthalpy of early actinides is lower than that of early lanthanides.

(p) Study of actinides is comparatively difficult.