D & f- BLOCK ELEMENTS practice · Mixed difficulty

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 B (Deep blue) is correct because cobalt(II) compounds typically impart a deep blue color to glass due to the electronic transitions of Co2+ ions.

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 D is correct because Assertion (A) is false, as the spin-only magnetic moment of Sc3+ is 0 BM, and Reason (R) is true, as it correctly states the formula for calculating spin-only magnetic moment.

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 B (Mn) is correct because manganese is a crucial component in the water-splitting complex of photosynthesis, which oxidizes water to produce oxygen.

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 D (${\mathrm{K}_2}\mathrm{S}{\mathrm{O}_4}.\mathrm{C}{\mathrm{r}_2}{\left( {\mathrm{S}{\mathrm{O}_4}} \right)_3}.24{\mathrm{H}_2}\mathrm{O}$) is correct because it represents chrome alum, which is widely used in the tanning of leather.

Option C (2) is correct because the number of electrons present in the outermost orbit of Fe atom is 2.

Option C (Ag) is correct because silver nitrate decomposes upon strong heating to produce silver metal, nitrogen dioxide, and oxygen gas.

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 B (Spitting of silver) is correct because the phenomenon of oxygen being absorbed by molten silver and then released upon cooling, causing the silver particles to scatter, is known as spitting of silver.

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 A (Silver bromide) is correct because silver bromide is a light-sensitive compound that undergoes chemical changes upon exposure to light, making it essential for photographic films.

Option D ($\mathrm{AgBr}$) is correct because Silver Bromide is the photosensitive material primarily used in photographic films and papers.

Option D (German silver) is correct because German silver is an alloy that contains nickel, along with copper and zinc.

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 $Cu^+$.