D & f- BLOCK ELEMENTS practice · Mixed difficulty

Option D (The $5f$-orbitals extend further from the nucleus than the $4f$ -orbitals) is correct because the greater extension of 5f orbitals in actinoids allows for more effective shielding and participation in bonding, leading to variable oxidation states.

Option C (KCN) is correct because silver sulfide ($\mathrm{A}{\mathrm{g}_2}\mathrm{S}$) dissolves in potassium cyanide (KCN) solution to form a soluble complex.

Option C is correct because silver (Ag) reacts with cyanide in the presence of water and oxygen to form a complex ion [Ag(CN)2]- and hydroxide ions.

Option B (ZnO) is correct because the thermal decomposition of zinc nitrate produces zinc oxide, nitrogen dioxide, and oxygen.

Option D is correct because VOSO4 contains V4+ which has an unpaired electron (d1 configuration), making it both paramagnetic and colored due to d-d transitions.

Option D is correct because Co3+ has the electronic configuration [Ar]3d⁶ after losing 3 electrons from its neutral state.

Option B ($10 \mathrm{~mL}$ of $1 \mathrm{M} \mathrm{~KMnO}_4$ solution oxidises $10 \mathrm{~mL}$ of $5 \mathrm{~N} \mathrm{~FeSO}_4$ solution) is correct because the n-factor for $KMnO_4$ in acidic medium is 5, so 1 M $KMnO_4$ is equivalent to 5 N $KMnO_4$.

Option D (Its colour is yellow) is correct because the colour of ${K}_2 {Cr}_2 {O}_7$ is orange, not yellow.

Option A (Strong oxidising agent) is correct because potassium permanganate ($\mathrm{KMn}{\mathrm{O}_4}$) acts as a strong oxidizing agent in both acidic and basic media.

Option A (Copper(I) Disproportionates into Cu and Cu(II) in aqueous solution) is not true because Copper(I) disproportionates into Cu and Cu(II) in aqueous solution is a true statement. Copper(I) oxide is actually a red powder.

Option D (The imperfect shielding on outer electrons by $4f$-electrons from the nuclear charge) is correct because the lanthanide contraction is caused by the imperfect shielding of the nuclear charge by the 4f electrons.

Option B (Co) is correct because Cobalt (Co) is a key component in alloys used for making permanent magnets.

Option C ($Cr > Mn > V > Ti$) is correct because the second ionization enthalpy is affected by the stability of the resulting electronic configurations.

Option B ($\mathrm{Mn}{\mathrm{O}_2}$) is correct because in a basic medium, $KMnO_4$ is reduced to $MnO_2$.

Option B is correct because in the highest oxidation state, transition metals tend to form acidic oxides and oxoanions, not basic oxides and cationic complexes.

Option D ($3{d^5}4{s^2}$) exhibits the largest number of oxidation states because it has the maximum number of electrons in the d and s orbitals available for bonding.

Option C (Cr > Mn > V > Ti) is correct because the second ionization enthalpy depends on electronic configuration and stability of the resulting ion.

Option D (Both (2) and (3)) is correct because both La3+ and Zn2+ have completely filled electronic configurations, making them diamagnetic.

The magnetic moment of $\mathrm{Z}{\mathrm{n}^{2 + }}$ ion is zero because $\mathrm{Z}{\mathrm{n}^{2 + }}$ has a $3d^{10}$ configuration, meaning all d-orbitals are filled and there are no unpaired electrons.

Option B is correct because La3+ has the electronic configuration [Xe], which means it has no unpaired electrons and is therefore diamagnetic.

Transition elements exhibit variable oxidation states because of the involvement of (n-1)d and ns electrons in bonding.

Option D ($+ 3$) is correct because the +3 oxidation state is the most stable and common oxidation state for lanthanides due to the electronic configuration they attain.

Option A (Yellow) is correct because Dy3+ and Sm3+ are complementary and exhibit similar colors due to similar electronic transitions.

Option A ($\mathrm{CuC}{\mathrm{l}_2}$) is correct because Cu(II) has an unpaired electron, making it paramagnetic.

Option B ($0.0075$) is correct because it represents the number of moles of $I_2$ formed based on the stoichiometry of the reaction between $KMnO_4$ and $KI$ in a weakly basic medium.

Option B (Cupric oxide) is correct because copper nitrate decomposes upon strong heating to form cupric oxide, nitrogen dioxide, and oxygen.

Transition metals form both ionic and covalent bonds because they have variable oxidation states and relatively high electronegativity.

Option B is correct because copper (Z=29) has 15 electrons with one spin and 14 with the opposite spin, considering its electronic configuration [Ar]3d¹⁰ 4s¹.

Option A (Electronic configuration) is correct because both lanthanoids and actinoids involve the filling of f-orbitals in their electronic configurations.

Option B ($\mathrm{A}{\mathrm{g}^ + }$) is correct because $Ag^+$ forms a linear complex $[Ag(CN)_2]^-$ with a coordination number of 2.