D & f- BLOCK ELEMENTS practice · Mixed difficulty

Option B (26) is correct because the atomic number is calculated by adding the number of electrons in Argon, the 3d orbital, and the oxidation state.

The question is incomplete. It asks why Zr and Y have about the same radius, but provides options related to reactions of iron compounds. Assuming the question meant to ask which reaction is correct, then Option B (Ferric sulphate reacts with ${\mathrm{K}_4}[{\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_6}} ]$) is correct because it forms Prussian blue.

The correct answer is B ($\mathrm{KMn}{\mathrm{O}_4}$) because manganese is already in its highest oxidation state (+7) in $\mathrm{KMn}{\mathrm{O}_4}$, so it cannot be further oxidized by ozone.

The correct answer is 1.73. The magnetic moment is calculated using the number of unpaired electrons in the $Cu^{2+}$ ion.

Option A (Gives hardness to steel and can remove oxygen and sulphur) is correct because manganese improves steel's hardness and acts as a deoxidizer and desulfurizer.

Option C (Mercuric chloride) is correct because the chemical name of corrosive sublimate is Mercuric chloride ($\mathrm{HgC}{\mathrm{l}_2}$).

Option A (Mn) is correct because Mn has the maximum number of unpaired electrons (5) in its 3d orbitals, leading to maximum paramagnetism.

Option B ($\mathrm{N}{\mathrm{a}_2}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_5}\mathrm{NO}} \right]$) is correct because the formula for sodium nitroprusside is $\mathrm{N}{\mathrm{a}_2}\left[ {\mathrm{Fe}{{\left( {\mathrm{CN}} \right)}_5}\mathrm{NO}} \right]$.

Option A ($\mathrm{M}{\mathrm{n}^{2 + }}$) is correct because it has a stable half-filled d-orbital configuration ($3d^5$).

Option B (Transition elements) is correct because transition elements often exhibit variable oxidation states and crystal defects, leading to the formation of non-stoichiometric compounds.

Option A ($\mathrm{C}{\mathrm{u}_2}\mathrm{O}$) is the correct answer because $\mathrm{C}{\mathrm{u}_2}\mathrm{O}$ is known as ruby copper.

Option A (Invar) is correct because Invar is a nickel-iron alloy, making it a ferrous alloy.

Option C ($\mathrm{Yb}^{3+}$) is correct because $\mathrm{Yb}^{3+}$ has a filled or nearly filled f-orbital configuration, resulting in fewer electronic transitions and thus less color.

Option B ($\mathrm{H}{\mathrm{g}_2}\mathrm{C}{\mathrm{l}_2}$) is correct because Mercurous chloride ($\mathrm{H}{\mathrm{g}_2}\mathrm{C}{\mathrm{l}_2}$), also known as calomel, is used as a purgative in medicine.

Option C is correct because transition metals have high melting points due to the involvement of (n-1)d and ns electrons in metallic bonding, but the reason that they have completely filled d-orbitals is false.

Option D (Four series) is correct because the d-block elements are arranged in four series corresponding to the filling of the 3d, 4d, 5d, and 6d orbitals.

The correct answer is C (Gd) because Gd³⁺ has a half-filled 4f subshell (4f⁷).

The correct answer is C because the order of polarizing power is incorrect. Polarizing power increases with charge and decreases with size.

Option D (Sulphide) is correct because the gas produced that turns K2Cr2O7 paper green and lead acetate paper black is hydrogen sulfide (H2S), which is produced from a sulfide.

Option D (0.83) is correct because it is calculated by equating the gram equivalents of $\mathrm{Fe(NO_2)_2}$ and $\mathrm{Cr_2O_7^{2-}}$.

Option D (Mercury) is correct because cinnabar is the principal ore from which mercury is extracted, and its chemical formula is HgS.

Option A (Heating steel in an atmosphere of ammonia) is correct because nitriding involves heating steel in an atmosphere of ammonia to introduce nitrogen into the surface.

Option A ($\mathrm{Fe}{\mathrm{S}_2} + {\mathrm{H}_2}\mathrm{O} + {\mathrm{O}_2}$) is correct because green vitriol (ferrous sulfate) is formed by the oxidation of iron pyrite in the presence of water and oxygen.

Option D (${Mn}^{2+}, {Ti}^{3+}$) is correct because both ions have unpaired electrons, allowing for d-d transitions and thus color in aqueous solution.

Option B is correct. KI reacts with CuSO4 to produce Cu2I2, I2, and K2SO4.

Option D (6) is correct because the oxidation number of Mn in K2MnO4, the product of alkaline oxidative fusion of MnO2, is +6.

Option D is correct because strong heating of ${\mathrm{K}_2}\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_7}$ yields ${\mathrm{K}_2}\mathrm{Cr}{\mathrm{O}_4}$, $\mathrm{C}{\mathrm{r}_2}{\mathrm{O}_3}$, and ${\mathrm{O}_2}$.

The correct answer is A ($\mathrm{Tb}(\mathrm{IV})$) because $\mathrm{Tb}^{4+}$ has unpaired electrons in its electronic configuration, making it paramagnetic.

Option A ($Cr{O_2}$) is correct because Chromium(IV) oxide is a metallic and ferromagnetic compound used in magnetic tapes.

Option B (${K}_2 {MnO}_4$) is correct because when $MnO_2$ is fused with potassium hydroxide in the presence of $KNO_3$, potassium manganate ($K_2MnO_4$), a deep green compound, is formed.