d and f Block Elements

To identify the incorrect statement, let's analyze each option :

Option A : "In manganate and permanganate ions, the π-bonding takes place by overlap of p-orbitals of oxygen and d-orbitals of manganese." This statement is true. In both manganate ${MnO}_4^{2-} $ and permanganate ${MnO}_4^- $ ions, the manganese atom is in a high oxidation state (+6 and +7, respectively) and forms bonds with oxygen atoms involving the overlap of p-orbitals of oxygen and d-orbitals of manganese.

Option B : "Manganate ion is green in colour and permanganate ion in purple in colour." This statement is true. Manganate ions are indeed green and permanganate ions are purple in aqueous solution.

Option C : "Manganate and permanganate ions are paramagnetic." This statement is incorrect. Both manganate and permanganate ions are diamagnetic, not paramagnetic. In these ions, manganese is in +6 and +7 oxidation states, respectively, and has no unpaired electrons.

Option D : "Manganate and permanganate ions are tetrahedral." This statement is true. Both manganate and permanganate ions have a tetrahedral geometry, with the manganese atom at the center and the four oxygen atoms surrounding it.

Therefore, the incorrect statement is option C.

The highest possible oxidation states of uranium and plutonium can be identified by considering the electronic configurations of these elements and noting the number of valence electrons available for bonding.

Uranium (U, atomic number 92) has the electronic configuration: $ [Rn] 5f^3 6d^1 7s^2 $

Plutonium (Pu, atomic number 94) has the electronic configuration: $ [Rn] 5f^6 7s^2 $

To find the highest oxidation state, we sum the number of $d$, $f$, and $s$ electrons in the valence shell.

For uranium :

$ 3 (f) + 1 (d) + 2 (s) = 6 $

So the highest oxidation state of uranium is +6.

For plutonium :

$ 6 (f) + 2 (s) = 8 $

So the highest oxidation state of plutonium is +8. However, in practice, plutonium is known to exhibit a highest oxidation state of +7.

Comparing with the given options, the correct answer is :

Option D : 6 and 7.

Cu²⁺ and Ni²⁺ ions belong to Group II and Group IV respectively.

Group II sulphides precipitated out even under low concentration of sulphides, while Group IV sulphides require higher concentration of sulphides. To fulfill this reaction condition, dil. HCl is chosen in Group II reagents and NH₄OH is chosen in Group IV reagents.

In order to determine which pair of compounds have metals in their highest oxidation states, we need to consider the common oxidation states of the metals involved, as well as the rules for determining oxidation states in compounds and complex ions. Let's evaluate each option:

Option A: MnO₂ and CrO₂Cl₂

In MnO₂, manganese (Mn) has an oxidation state of +4. This is not the highest oxidation state manganese can achieve; it can go up to +7, as seen in KMnO₄ (potassium permanganate).

In CrO₂Cl₂, chromium (Cr) has an oxidation state of +6, which is indeed its highest oxidation state.

Thus, this option contains one metal in its highest oxidation state (Cr) but not the other (Mn).

Option B: [NiCl₄]^2- and [CoCl₄]^2-

For both [NiCl₄]^2- and [CoCl₄]^2-, the nickel (Ni) and cobalt (Co) are in a +2 oxidation state. Neither of these represents the highest oxidation state for these metals. Nickel can have a +3 state, and cobalt can go up to +4 in very rare cases.

Option C: [Fe(CN)₆]^3- and [Cu(CN)₄]^2-

In [Fe(CN)₆]^3-, the iron (Fe) has an oxidation state of +3, which is a common high oxidation state for iron, though iron can also exist in a +2 state commonly and +6 in very rare conditions. Therefore, +3 is considered a high oxidation state but not the absolute highest.

In [Cu(CN)₄]^2-, copper (Cu) has an oxidation state of +2, which is the highest stable oxidation state for copper in most of its compounds.

Option D: [FeCl₄]^- and Co₂O₃

In [FeCl₄]^-, iron (Fe) has an oxidation state of +3. As mentioned, +3 is a high state for iron, but not the absolute highest oxidation state it can achieve.

In Co₂O₃, cobalt (Co) has an oxidation state of +3, as evidenced by the formula, where two Co atoms interact with three O atoms (each oxygen providing a -2 charge, for a total of -6, requiring each Co to be +3 to balance the charge). +3 is indeed among the highest common oxidation states for cobalt, although, as mentioned, Co can technically reach +4.

Conclusion: Based on above analysis, while none of the options perfectly fits the criteria of both elements being in their absolute highest oxidation states (especially considering the rare or less common states), Option A and Option D come closest, with chromium and cobalt being in their common high oxidation states in the provided compounds. However, the question asks for the highest oxidation state, and strictly speaking, none of the options perfectly satisfy the condition for both elements included. Among the choices, Option A is the closest because chromium is in its highest oxidation state of +6 in CrO₂Cl₂, and manganese in MnO₂ is in a +4 state, which while not its absolute highest, represents a commonly encountered high oxidation state. Given the context and conventional exams' focus on common oxidation states, Option A would be the best fit despite the slight discrepancy with the question's phrasing.

On addition of impurities to the colourless gemstones, brilliant colours are produced. On addition of chromium to colourless corundum (Al₂O₃), a red ruby is formed. Red colour ruby is obtained when Cr³⁺ ions replace Al³⁺ ions in the octahedral sites of corundum. A green emerald emerges when chromium is added to colourless beryl (Be₃Al₂(SiO₃)₆). Green colour emerald is obtained when Cr³⁺ ions replace Al³⁺ in the octahedral site of beryl.

MnO₄²⁻ disproportionate as (in neutral or acidic solution)

3MnO₄²⁻ + 4H⁺ → 2MnO₄²⁻ + MnO₂ + 2H₂O

To match the catalysts to the correct processes :

1. $TiCl_3$ is well-known as a catalyst in Ziegler-Natta polymerization which is used for the production of polymers of olefins. So, $(A) - (ii)$.




2. $PdCl_2$ is used in the Wacker process, which is used to convert ethene to acetaldehyde. Hence, $(B) - (i)$.




3. $V_2O_5$ is used in the Contact process to manufacture sulfuric acid by catalyzing the oxidation of sulfur dioxide to sulfur trioxide. Therefore, $(D) - (iii)$.




4. $CuCl_2$ is used in Deacon's process for the oxidation of hydrochloric acid to chlorine. Hence, $(C) - (iv)$.

Comparing these pairings with the options given, we find that the correct option is :

$ \left( A \right) - \left( {ii} \right),\,\,\left( B \right) - \left( {i} \right),\,\,\left( C \right) - \left( {iv} \right),\,\,\left( D \right) - \left( {iii} \right) $