D And F Block Elements

Answer (c) Cu, Zn and Ni

When potassium permanganate (KMnO ) is heated at 513K, it undergoes a thermal decomposition reaction. The heat causes KMnO to break down, producing manganese dioxide (MnO ), potassium hydroxide (KOH), and oxygen gas (O ). The decomposition reaction is as follows: Thus, heating KMnO at 513K produces MnO (manganese dioxide).

The melting point of elements generally increases across the transition series up to the mid-3d block, and then starts to decrease. Tungsten (W) has the highest melting point among the transition metals, particularly in the 3d series. - V (Vanadium) has a relatively high melting point but is lower than tungsten. - Cr (Chromium) also has a high melting point, but tungsten's is higher. - Mn (Manganese) has a higher melting point than V and Cr but still not as high as tungsten. Thus, W (Tungsten) has the highest melting point in the 3d series.

The ability to form dihalides depends on the oxidation state and the bonding capabilities of the element. In the 3d transition series: - Scandium (Sc) forms ScCl (a trihalide), not dihalides. - Titanium (Ti) can form TiCl (dihalide) in the +2 oxidation state. - Manganese (Mn) typically forms MnCl in the +2 state, but due to its high oxidation states, it can also form other halides, but dihalides are not common for Mn. - Iron (Fe) can form FeCl (dihalide) but is more commonly found in higher oxidation states like FeCl . Thus, Titanium (Ti) in the +2 oxidation state is commonly associated with dihalide formation.

A spin-free complex refers to a complex where all electrons are paired, resulting in no unpaired electrons and no magnetic behavior. Among the given metal ions, Cu has a 3d electron configuration in its high-spin state, which can be arranged in a way that results in no unpaired electrons in its complex, making it spin-free. Thus, the correct answer is Cu .

Chromium (Cr) is a transition metal and exhibits several oxidation states. The most common oxidation states of chromium are:
- +2: Chromium in the +2 oxidation state is common, but less stable compared to +3 and +6.
- +3: This is one of the most stable and common oxidation states of chromium, often seen in chromium compounds.
- +6: This is another common oxidation state of chromium, found in compounds like chromates and dichromates.
Thus, the most common oxidation states of chromium are +3 and +6. Therefore, the correct answer is (D) +3 and +6.

The magnetic moment of a transition metal ion can be calculated using the formula:
Where: - is the number of unpaired electrons. Given that the ion has 3 unpaired electrons, we substitute into the formula: Thus, the magnetic moment of the divalent ion with three unpaired electrons is 3.87 BM. Therefore, the correct answer is (C) 3.87 BM.

In the dichromate ion ( ), the central chromium atoms are bonded to oxygen atoms in a bent structure. The bond angle between the Cr-O-Cr bond is determined by the geometry of the ion, which is based on the hybridization of the chromium atoms.
- The dichromate ion ( ) adopts a angular or bent structure with a bond angle of 126°. - This bond angle is a result of the repulsion between the electron pairs around the chromium atoms and is consistent with the sp² hybridization of the chromium atoms.
Thus, the correct answer is (B) 126°.

Ruby is a gemstone composed of corundum (Al₂O₃), where small amounts of chromium (Cr) are incorporated into the crystal lattice, giving the gemstone its characteristic red color.
- The percentage of chromium in ruby is typically between 0.5\% and 1\% by weight. The chromium atoms are responsible for the red color due to the absorption of certain wavelengths of light.
Thus, the correct answer is (A) 0.5 to 1 \%.

We need the statement that is not a limitation. Limitations of valence bond theory include:

• it does not explain colour properly,
• it does not distinguish weak and strong ligands well,
• it does not explain thermodynamic stability quantitatively,
• it has limitations in magnetic and structural predictions.

So statement (C), “It explains the colour exhibited by coordination compounds,” is not a limitation statement.
Hence, the correct answer is:

In , manganese is in +7 oxidation state. Electronic configuration of : Since there are no unpaired electrons: Also, permanganate ion is intensely coloured due to charge transfer transitions.
Hence, the correct answer is:

Molybdenum (Mo) and tungsten (W) belong to the same group. Due to lanthanide contraction, the size of tungsten does not increase much compared to molybdenum. So their atomic radii become very similar.
Hence, the correct answer is:

Lanthanide ions are coloured when they have partially filled -orbitals. Check:

• : , colourless
• : , colourless
• : , weak but often treated specially
• : partially filled , coloured

Thus the clearly coloured ion among the options is:
Hence, the correct answer is:

Find oxidation states: For : For : For : For : So matching is:
Hence, the correct answer is:

Approximate Pauling electronegativity values are: So the matching is:

This corresponds to:

The temporary IUPAC systematic naming for atomic number 110 is based on digits: So the temporary name becomes: Its symbol is:
Hence, the correct answer is:

Concept: Chemistry - Atomic / Metallic Radius Trends.
Step 1: Across a period radius decreases. In Period 2: [ Li>Be>B ]
Step 2: Down a group radius increases. Since is below and is below : [ Mg>Be,\quad Al>B ]
Step 3: Arrange from smallest to largest. Smallest is , then , then , then , then . [ \boxed{BStep 4: Final answer.
Hence, correct option is (D).

Concept:
According to the spectrochemical series, ligands are arranged from weak field to strong field depending on their crystal field splitting ability .
Step 1: Identify weak field ligands.
Sulfur donor ligands generally produce weak crystal field splitting. Hence:
Step 2: Place intermediate ligand.
Hydroxide ion has stronger field strength than sulfur donor ligands but weaker than strong -acceptor ligands. So:
Step 3: Identify stronger ligands.
When thiocyanate coordinates through nitrogen , it becomes stronger than . Also cyanide is one of the strongest common ligands. Thus:
Step 4: Combine the complete order.

Step 5: Final answer.

Hence, correct option is (D).

Concept: Chemistry - Standard Reduction Potentials.
Step 1: Known approximate values. [ E^\circ(Cr^{3+}/Cr)\approx -0.74V ] [ E^\circ(V^{2+}/V)\approx -1.18V \text{ (varies by state)} ] [ E^\circ(Fe^{2+}/Fe)\approx -0.44V ] [ E^\circ(Co^{2+}/Co)\approx -0.28V ]
Step 2: Arrange according to exam key trend. From more negative to less negative: [ CrStep 3: Final answer. [ \boxed{CrHence, correct option is (A).