D & f- BLOCK ELEMENTS practice · Medium

Option B (Fe) is correct because iron is used in the Bosch process to convert steam to hydrogen.

Option A (${MnO}$) is correct because Manganese(II) oxide (MnO) is a well-known antiferromagnetic substance.

Option B is correct because $\mathrm{FeC}{\mathrm{l}_3}$ is an oxidizing agent itself and cannot be decolorized by acidified $\mathrm{KMn}{\mathrm{O}_4}$ solution.

Option B ($\mathrm{Ru}{\left( {\mathrm{CO}} \right)_5}$) is correct because Ruthenium forms pentacarbonyl with the formula $\mathrm{Ru}{\left( {\mathrm{CO}} \right)_5}$

The correct answer is D because the color of K2Cr2O7 is orange, not yellow.

Option A (Manganese) is correct because manganese steel is used in making armoured steel due to its high strength and wear resistance.

Option C (Tempering) is correct because tempering involves heating steel to a temperature below redness and then slowly cooling it to reduce brittleness.

Option D (Silver) is correct because silver thiosulphate is white but decomposes to form black silver sulfide in the presence of moisture.

Option B ($\sqrt{15} \mathrm{BM}$) is correct because the spin-only magnetic moment is calculated using the formula $\sqrt{n(n+2)}$, where n is the number of unpaired electrons.

Option A (5, 9) is correct because chromate ions ($CrO_4^{2-}$) convert to dichromate ions ($Cr_2O_7^{2-}$) in acidic conditions (low pH, around 5), and the reverse occurs in basic conditions (high pH, around 9).

Option C (${X}=$ black, ${MnO}_2, {Y}=$ green, ${K}_2 {MnO}_4, {Z}={KMnO}_4$) is correct because $MnO_2$ reacts with $K_2CO_3$ in the presence of air to form green $K_2MnO_4$, which then reacts with $Cl_2$ to form pink $KMnO_4$.

Option B ($\mathrm{AgBr}$) is correct because silver bromide is highly sensitive to light, making it essential in photographic films and papers.

Option A (exhibit +2 and +3 oxidation states) is correct because both Samarium and Europium readily exhibit +2 and +3 oxidation states due to their electronic configurations and stability of half-filled/fully-filled f-orbitals.

Option A (Statement I and statement II are correct and statement II is the correct explanation of statement I) is correct because IIB group elements are d-block elements but not transition elements because their d-orbitals are completely filled, which is explained by the incomplete filling of (n-1) shells in transition elements.

The correct answer is C (Insoluble chlorides) because silver, mercury, and lead are placed in the same group of qualitative analysis because they form insoluble chlorides that precipitate out with dilute HCl.

Option C (6) is correct because in $\mathrm{B}_2 \mathrm{H}_6$, there are four terminal hydrogen atoms and two boron atoms that lie in one plane.

Option B is correct because the smaller energy difference between 5f and 6d orbitals in actinides allows for more unpaired electrons and thus more oxidation states.

Option C ($\mathrm{Mn}< \mathrm{Fe},$ $\mathrm{Tc}< \mathrm{Ru}$ and $\mathrm{Os}< \mathrm{Re}$) is correct because melting points generally increase with the number of unpaired electrons, but decrease when the d-orbitals are nearly half-filled or completely filled.

Option C (Mo) is correct because molybdenum is known to form heteropoly acids, such as ${\mathrm{H}_2}\mathrm{Mo}{\mathrm{O}_4}$ and ${\mathrm{H}_6}\mathrm{M}{\mathrm{o}_7}{\mathrm{O}_{24}}$.

Option A (Cu) is correct because copper is commonly added to gold to increase its hardness and durability for jewelry making.

Option D (ZnO) is correct because Zinc Oxide is known to exhibit thermochromism, appearing yellow when hot and white when cold.

Option C (Np) is correct because Neptunium (Np) can exhibit a +7 oxidation state due to its electronic configuration and the availability of its f-electrons for bonding.

Option A (Formation of $Fe{\left( {CO} \right)_5}$ from $Fe$) is correct because the oxidation state of iron remains 0 in both the reactant and the product, indicating no oxidation has occurred.

Option C (${\mathrm{K}_2}\mathrm{Hg}{\mathrm{I}_4} + \mathrm{NaOH}$) is correct because Nessler's reagent is a solution of potassium tetraiodomercurate(II) in potassium hydroxide.

Option D (Spiegeleisen) is correct because Spiegeleisen, an alloy of iron, manganese, and carbon, is added to molten steel to remove dissolved oxygen and prevent blowholes.

Option D ($d - d$ transition) is correct because the color in transition metal compounds arises from the absorption of light during $d - d$ electronic transitions.

The correct answer is $\mathrm{CrO}_2 \mathrm{Cl}_2, \mathrm{MnO}_4^{-}$ because Cr has a +6 oxidation state and Mn has a +7 oxidation state, which are their highest possible oxidation states.

Option A (${(N{H_4})_2}C{r_2}{O_7}$) is correct because it decomposes to give nitrogen gas instead of oxygen gas upon heating.

If both Assertion & Reason are false. $Cu^{2+}$ is more stable than $Cu^+$.

Option A (Assertion is correct, reason is correct; reason is a correct explanation for assertion.) is correct because actinoids exhibit more oxidation states due to the smaller energy difference between 5f and 6d orbitals, allowing for greater participation of electrons in bonding.