Question 1

Given below are two statements: Statement I: The number of pairs among [ Ti ^ 4+ , V ^ 2+ ], [ V ^ 2+ , Mn ^ 2+ ], [ Mn ^ 2+ , Fe ^ 3+ ] and [ V ^ 2+ , Cr ^ 2+ ] in which both ions are coloured is 3.

Accepted Answer

For Statement I: The electronic configurations of the given transition metal ions are: Ti ^ 4+ : 3d^0 (colourless due to absence of unpaired electrons) V ^ 2+ : 3d^3 (coloured) Mn ^ 2+ : 3d^5 (coloured) Fe ^ 3+ : 3d^5 (coloured) Cr ^ 2+ : 3d^4 (coloured) The pairs in which both ions are coloured are [ V ^ 2+ , Mn ^ 2+ ], [ Mn ^ 2+ , Fe ^ 3+ ], and [ V ^ 2+ , Cr ^ 2+ ]. The pair [ Ti ^ 4+ , V ^ 2+ ] contains Ti ^ 4+ which is colourless. Thus, there are exactly 3 pairs where both ions are coloured

Question 2

Given below are two statements for catalytic properties of transition metals. Statement I: First row transition metals which act as catalyst utilise their 3d electrons only for formation of bonds betw

Accepted Answer

Statement I is incorrect because first row transition metals utilise both 3d and 4s electrons for the formation of bonds between reactant molecules and atoms on the surface of the catalyst. Statement II is incorrect because the increase in the concentration of reactants on the surface of the catalyst weakens the bonds in the reacting molecules, which lowers the activation energy. Therefore, both Statement I and Statement II are incorrect. Answer: Both Statement I and Statement II are incorrect

Question 3

The correct order of first ( _i H_1) and second ( _i H_2) ionisation enthalpy values of Cr and Mn are : A. _i H_1 : Cr > Mn B. _i H_2 : Cr > Mn C. _i H_1 : Mn > Cr D. _i H_2 : Mn > Cr Choose the corre

Accepted Answer

The electronic configuration of Cr is [Ar] 3d^5 4s^1 and Mn is [Ar] 3d^5 4s^2. For the first ionization enthalpy, the electron is removed from the 4s orbital. Mn has a fully filled 4s^2 configuration and a higher effective nuclear charge compared to Cr. Thus, the first ionization enthalpy of Mn is greater than that of Cr ( _i H_1 : Mn > Cr). This makes statement C correct. After the first ionization, the electronic configurations become: Cr^+ : [Ar] 3d^5 Mn^+ : [Ar] 3d^5 4s^1 For the second ioni

Question 4

The correct statements among the following are. A. Basic vanadium oxide is used in the manufacture of H _2 SO _4. B. The spin-only magnetic moment value of the transition metal halide employed in Zieg

Accepted Answer

Statement A: Vanadium pentoxide ( V _2 O _5) is used as a catalyst in the manufacture of H _2 SO _4 by the Contact process. V _2 O _5 is an amphoteric oxide, not a basic oxide. Thus, statement A is incorrect. Statement B: The transition metal halide used in Ziegler-Natta polymerization is TiCl _4. In TiCl _4, titanium is in the +4 oxidation state with a 3 d ^0 electronic configuration. The number of unpaired electrons is zero, so its spin-only magnetic moment is 0 BM. Thus, statement B is incorr

Question 5

Given below are two statements : Statement I: Presence of large number of unpaired electrons in transition metal atoms results in higher enthalpies of their atomisation. Statement II: d_ xy = d_ xz =

Accepted Answer

Statement I: The enthalpy of atomisation of transition metals depends on the strength of metallic bonding. A greater number of unpaired electrons in the d-orbitals leads to stronger metallic bonding, which results in a higher enthalpy of atomisation. Thus, Statement I is correct. Statement II: [Fe(H_2O)_6]^ 3+ is an octahedral complex. In an octahedral crystal field, the d-orbitals split such that the t_ 2g set of orbitals (d_ xy , d_ xz , d_ yz ) has lower energy than the e_g set of orbitals (d

Question 6

A paper dipped in a dil. H_2SO_4 solution of 'X' upon treatment with SO_2 gas turns into green. The compound 'X' is :

Accepted Answer

SO_2 gas acts as a reducing agent. When a paper dipped in an acidified solution of potassium dichromate (K_2Cr_2O_7) is exposed to SO_2 gas, the orange-colored dichromate ion (Cr_2O_7^ 2- ) is reduced to the green-colored chromium(III) ion (Cr^ 3+ ). The chemical reaction is: K_2Cr_2O_7 + H_2SO_4 + 3SO_2 K_2SO_4 + Cr_2(SO_4)_3 + H_2O The formation of chromium(III) sulfate, Cr_2(SO_4)_3, is responsible for the green color. Thus, the compound 'X' is K_2Cr_2O_7. Answer: K_2Cr_2O_7

Question 7

Which of the following is NOT a physical or chemical characteristics of interstitial compounds?

Accepted Answer

Interstitial compounds are formed when small atoms like H, C, or N are trapped inside the crystal lattices of transition metals. The principal physical and chemical characteristics of interstitial compounds are: (i) They have high melting points, higher than those of pure metals. (ii) They are very hard, and some borides approach diamond in hardness. (iii) They retain metallic conductivity. (iv) They are chemically inert. (v) They are usually non-stoichiometric. Therefore, the statement that the

Question 8

Pairs of elements with the same number of electrons in their respective 4f orbital are [Atomic number. Eu-63, Gd-64, Dy-66, Ho-67, Tm-69, Yb-70, Lu-71, Hf-72] A. (Eu and Gd) B. (Dy and Ho) C. (Yb and

Accepted Answer

The electronic configurations of the given elements are: Eu (63): [Xe] 4f^7 6s^2 Gd (64): [Xe] 4f^7 5d^1 6s^2 Dy (66): [Xe] 4f^ 10 6s^2 Ho (67): [Xe] 4f^ 11 6s^2 Tm (69): [Xe] 4f^ 13 6s^2 Yb (70): [Xe] 4f^ 14 6s^2 Lu (71): [Xe] 4f^ 14 5d^1 6s^2 Hf (72): [Xe] 4f^ 14 5d^2 6s^2 From the above configurations: Eu and Gd both have 7 electrons in their 4f orbitals. Yb and Hf both have 14 electrons in their 4f orbitals. Dy and Ho have 10 and 11 electrons respectively in their 4f orbitals. Lu and Tm have

Question 9

Consider |x| is the difference in oxidation states of Mn in highest manganese fluoride and highest manganese oxide. The ions with |x| number of unpaired electrons from the following are: A. Sc^ 3+ B.

Accepted Answer

The highest manganese fluoride is MnF_4, where the oxidation state of Mn is +4. The highest manganese oxide is Mn_2O_7, where the oxidation state of Mn is +7. The difference in oxidation states is |x| = |7 - 4| = 3. The number of unpaired electrons in the given ions are: Sc^ 3+ : [Ar] 3d^0 0 unpaired electrons Zn^ 2+ : [Ar] 3d^ 10 0 unpaired electrons V^ 2+ : [Ar] 3d^3 3 unpaired electrons Fe^ 2+ : [Ar] 3d^6 4 unpaired electrons Co^ 2+ : [Ar] 3d^7 3 unpaired electrons Thus, V^ 2+ (C) and Co^ 2+

Question 10

The correct statements among the following are, A. Mo(VI) and W(VI) are less stable than Cr(VI). B. Ce^ 4+ and Tb^ 4+ are oxidant while Eu^ 2+ and Yb^ 2+ are reductant. C. Cm and Am have seven unpaire

Accepted Answer

In d-block elements, the stability of higher oxidation states increases down a group. Therefore, Mo(VI) and W(VI) are more stable than Cr(VI). Statement A is incorrect. The most stable oxidation state for lanthanoids is +3. Ions like Ce^ 4+ and Tb^ 4+ tend to gain an electron to achieve the +3 state, acting as oxidants. Conversely, Eu^ 2+ and Yb^ 2+ tend to lose an electron to achieve the +3 state, acting as reductants. Statement B is correct. The electronic configuration of Am is [Rn] 5f^ 7 7s^

Question 11

Correct statements from the following are A. Potassium dichromate is an oxidising agent and it oxidises FeSO_4 to Fe_2(SO_4)_3 in acidic medium. B. Sodium dichromate can be used as primary standard in

Accepted Answer

Statement A is correct. Potassium dichromate acts as a strong oxidising agent in acidic medium and oxidises Fe^ 2+ to Fe^ 3+ , converting FeSO_4 to Fe_2(SO_4)_3. Statement B is incorrect. Sodium dichromate is deliquescent (absorbs moisture from the atmosphere), which prevents it from being weighed accurately. Therefore, it cannot be used as a primary standard in volumetric estimation. Potassium dichromate is used instead. Statement C is correct. Chromate (CrO_4^ 2- ) and dichromate (Cr_2O_7^ 2-

Question 12

Number of paramagnetic ions among the following d- and f-block metal ions is _______. Mn^ 2+ , Cu^ 2+ , Zn^ 2+ , Yb^ 2+ , Sc^ 3+ , La^ 3+ , Gd^ 3+ , Lu^ 3+ , Ti^ 4+ , Ce^ 4+ (Atomic number of Mn = 25,

Accepted Answer

The magnetic nature of an ion depends on the presence of unpaired electrons. Ions with at least one unpaired electron are paramagnetic, while those with all paired electrons are diamagnetic. Let us write the electronic configurations of the given ions: Mn^ 2+ (Z=25) : [Ar] 3d^5 5 unpaired electrons (Paramagnetic) Cu^ 2+ (Z=29) : [Ar] 3d^9 1 unpaired electron (Paramagnetic) Zn^ 2+ (Z=30) : [Ar] 3d^ 10 0 unpaired electrons (Diamagnetic) Yb^ 2+ (Z=70) : [Xe] 4f^ 14 0 unpaired electrons (Diamagnetic

Question 13

Which of the following sets includes all the species that will change the orange colour of K_2Cr_2O_7 in acidic medium?

Accepted Answer

K_2Cr_2O_7 in acidic medium acts as a strong oxidizing agent. It changes its colour from orange (Cr_2O_7^ 2- ) to green (Cr^ 3+ ) when it reacts with a reducing agent. We need to identify the set in which all species can act as reducing agents (i.e., they can be oxidized to a higher oxidation state). In option (A), Fe^ 2+ oxidizes to Fe^ 3+ , Sn^ 2+ oxidizes to Sn^ 4+ , I^- oxidizes to I_2, and S^ 2- oxidizes to S. All of these are reducing agents. In option (B), Fe^ 3+ is already in its maximum

Question 14

Consider the following statements about manganate and permanganate ions. Identify the correct statements. A. The geometry of both manganate and permanganate ions is tetrahedral. B. The oxidation state

Accepted Answer

Manganate ion → MnO _4^ 2- Permanganate ion → MnO _4^ - (A) Both are tetrahedral (d^3s hybridisation) TRUE (B) MnO _4^ - has Mn in +7 oxidation state; MnO _4^ 2- has Mn in +6 oxidation state. The statement says +7 for manganate and +6 for permanganate, which is reversed. FALSE (C) Oxidation of Mn ^ 2+ by peroxodisulphate ( S _2 O _8^ 2- ) gives permanganate ion ( MnO _4^ - ), not manganate ion. FALSE (D) MnO _4^ - is diamagnetic; MnO _4^ 2- is paramagnetic. TRUE (E) Acidified permanganate ion ox

Question 15

Given below are two statements: Statement I : The number of pairs, from the following, in which both the ions are coloured in aqueous solution is 3. [ Sc ^ 3+ , Ti ^ 3+ ], [ Mn ^ 2+ , Cr ^ 2+ ], [ Cu

Accepted Answer

[Sc³⁺, Ti³⁺] has Sc³⁺ colorless (no d-electrons) and Ti³⁺ colored; [Mn²⁺, Cr²⁺] has Mn²⁺ colorless (d⁵ half-filled, no d-d transitions) and Cr²⁺ colored; [Cu²⁺, Zn²⁺] has Cu²⁺ colored and Zn²⁺ colorless (d¹⁰ filled); [Ni²⁺, Ti⁴⁺] has Ni²⁺ colored and Ti⁴⁺ colorless (no d-electrons). Zero pairs have both ions colored. Statement I is FALSE. Th⁴⁺ is not a reducing agent. Statement II is FALSE. Both statements are false.

Question 16

" X " is an oxoanion of the lightest element of group 7 (in the periodic table). The metal is in +6 oxidation state in " X ". The color of the potassium salt of X is

Accepted Answer

The lightest element of group 7 (considering transition metals) is manganese (Mn). An oxoanion with Mn in +6 oxidation state is manganate, MnO₄²⁻. In manganate, Mn exhibits +6 oxidation state (where oxygen is -2, so Mn + 4(-2) = -2, giving Mn = +6). The potassium salt K₂MnO₄ (potassium manganate) is green in color. This is distinct from permanganate KMnO₄ (where Mn is +7) which is purple. The characteristic green color of manganate solutions and salts makes this the answer.

Question 17

The oxidation state of chromium in the final product formed in the reaction between KI and acidified K _ 2 Cr _ 2 O _ 7 solution is:

Accepted Answer

In the reaction between KI and acidified K_2Cr_2O_7, Cr in the dichromate ion exists in +6 oxidation state. Iodide ions are oxidized to I_2 while chromium is reduced. The reduction product is Cr^ 3+ ions, which form chromium(III) salts like Cr_2(SO_4)_3 or coordination complexes. Therefore, the oxidation state of chromium in the final product is +3.

Question 18

Given below are two statements : Statement I : The first ionization enthalpy of Cr is lower than that of Mn . Statement II : The second and third ionization enthalpies of Cr are higher than those of M

Accepted Answer

Statement I: The electronic configuration of Cr is [Ar] 3d^5 4s^1 and Mn is [Ar] 3d^5 4s^2. For Cr, the first electron is removed from 4s^1, while for Mn, it is removed from a stable fully filled 4s^2 subshell. Additionally, Mn has a higher nuclear charge. Thus, the first ionization enthalpy (IE_1) of Cr (653 kJ/mol ) is lower than that of Mn (717 kJ/mol ). Statement I is true. Statement II: For the second ionization enthalpy (IE_2), Cr^+ has configuration [Ar] 3d^5 (stable half-filled) and Mn^+

Question 19

Among the following oxides of 3 d elements, the number of mixed oxides are \_\_\_\_。 Ti _ 2 O _ 3 , ~V _ 2 O _ 4 , Cr _ 2 O _ 3 , Mn _ 3 O _ 4 , Fe _ 3 O _ 4 , Fe _ 2 O _ 3 , Co _ 3 O _ 4

Accepted Answer

Mixed oxides contain the same element in two different oxidation states. Analyzing each oxide: Ti₂O₃ contains Ti(+3) only - simple oxide. V₂O₄ contains V(+4) only - simple oxide. Cr₂O₃ contains Cr(+3) only - simple oxide. Mn₃O₄ can be written as MnO·Mn₂O₃, containing both Mn(+2) and Mn(+3) - mixed oxide. Fe₃O₄ can be written as FeO·Fe₂O₃, containing both Fe(+2) and Fe(+3) - mixed oxide. Fe₂O₃ contains Fe(+3) only - simple oxide. Co₃O₄ can be written as CoO·Co₂O₃, containing both Co(+2) and Co(+3

Question 20

A first row transition metal ( M ) does not liberate H _ 2 gas from dilute HCl.1 mol of aqueous solution of MSO _ 4 is treated with excess of aqueous KCN and then H _ 2 ~S ( ~g ) is passed through the

Accepted Answer

The first row transition metal (M) that does not liberate H_2 gas from dilute HCl is Copper (Cu), as it has a positive reduction potential (E^ _ Cu^ 2+ /Cu = +0.34 V ). When 1 mol of CuSO_4 reacts with excess aqueous KCN, it first forms Cu(CN)_2, which is unstable and decomposes to CuCN and cyanogen gas (CN)_2. 2CuSO_4 + 4KCN 2Cu(CN)_2 + 2K_2SO_4 2Cu(CN)_2 2CuCN + (CN)_2 The CuCN then dissolves in excess KCN to form a highly stable soluble complex, potassium tetracyanocuprate(I): CuCN + 3KCN K_3

d and f Block Elements — Previous Year Questions & Solutions