D & f- BLOCK ELEMENTS practice · Easy

Option B is correct. Zinc and mercury have completely filled d-orbitals, preventing them from easily losing or sharing d-electrons to exhibit variable valency.

Option A (Rare earths) is correct because the 4f subshell is filled in the lanthanide series, also known as rare earth elements.

Option D ($({n}-1) {d}$ ns orbitals) is correct because transition elements lose electrons from both the (n-1)d and ns orbitals during ionization.

Option D (Fe) is correct because mercury does not form amalgams with iron.

Option B ($\mathrm{Ni}^{2+}$) is correct because, for isoelectronic species, ionic radii decrease with increasing nuclear charge (atomic number).

Option B (${Mn}$) is correct because manganese has an anomalous electronic configuration that results in a lower melting point compared to its neighbors in the periodic table.

Option A (LnC) is correct because lanthanides form carbides of the types Ln3C, Ln2C3, and LnC2, but not LnC.

Option D ($+5$) is correct because vanadium's most stable and common oxidation state is +5, corresponding to the loss of all its valence electrons.

Option D (Transition elements) is correct because transition elements exhibit similar properties both vertically (within a group) and horizontally (across a period) due to their electronic configurations and metallic bonding.

Option D is correct because $\mathrm{Cu}^{2+}$ is unstable with $\mathrm{I}^{-}$ due to the formation of $\mathrm{CuI}$ and $\mathrm{I}_2$.

Option C (2) is correct because the outermost orbit of an Fe atom ($4s^2$) contains 2 electrons.

Option D (All of the above) is correct because all the statements about Hg are factual: it's the only liquid metal at room temperature, Hg2+ salts are more stable than Hg22+ salts, and it doesn't form amalgams with iron and platinum.

Option B (${K_2}Mn{O_4}$) is correct because the decomposition of $KMn{O_4}$ produces ${K_2}Mn{O_4}$, $Mn{O_2}$, and ${O_2}$.

Option B (90) is correct because 21.6 carat gold is 21.6/24 pure, which equals 90%.

Option A (Fe + C + Mn) is correct because manganese steel is an alloy of iron, carbon, and manganese, known for its high tensile strength and wear resistance.

Option B ($Ti < Zr = Hf$) is correct because of the Lanthanoid contraction, which causes Zr and Hf to have similar atomic sizes.

Option B (20) is the correct answer because molten Ag absorbs approximately 20 times its volume of ${\mathrm{O}_2}$.

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

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 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.

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

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