D Block Elements

Step 1: Explanation of color. The purple color of KMnO is due to charge transfer between the manganese ion and the oxygen ligands, specifically from the oxygen to the manganese.
Step 2: Conclusion. Thus, the purple color is due to charge transfer.

Step 1: Electron configuration and unpaired electrons. The number of unpaired electrons can be determined from the electron configuration of the ions. - : (no unpaired electrons). - : (5 unpaired electrons). - : (1 unpaired electron). - : (no unpaired electrons).
Step 2: Conclusion. Thus, has the maximum number of unpaired electrons (5).

Step 1: Explanation of color origin.
The purple color of KMnO₄ arises due to the d → d transition in the Mn⁷⁺ ion.

Step 2: Conclusion.
Thus, the correct answer is option (C).

Final Answer:

In both and is present as Thus, the electronic configuration of is


In case of is a strong field ligand, so pairing of electrons in -orbital takes place.

In is a weak field ligand, thus does not cause pairing. Hence,

Step 1: Recall common properties of transition metals.
Transition elements typically show:
- Variable oxidation states due to incomplete d-subshell
- Formation of coloured compounds due to d-d transitions
- Formation of interstitial compounds due to small size and vacant sites
Step 2: Check the option about natural radioactivity.
Natural radioactivity is a property of unstable nuclei, not a general characteristic of transition metals.
Only a few elements are radioactive, but transition elements as a group are not defined by radioactivity.
Step 3: Conclude.
Thus natural radioactivity is not a characteristic property.
Final Answer:

Step 1: Write electronic configuration of Zn .
Zn (Z=30) has configuration:

Zn loses two 4s electrons:

Step 2: Check unpaired electrons.
means all d-orbitals are completely filled.
So there are no unpaired electrons.
Step 3: Magnetic moment.
Magnetic moment depends on unpaired electrons:

Here , so:

Final Answer:

Step 1: Paramagnetism depends on unpaired electrons.
More unpaired electrons more paramagnetic.
Step 2: Find electron configurations.




Step 3: Arrange increasing order.

Final Answer:

Step 1: Reason for high oxidation state stability.
High oxidation states are stabilized by highly electronegative ligands because they can withdraw electron density from metal strongly.
Step 2: Compare halogens.
Fluorine is the most electronegative and smallest halogen.
So it forms strongest M–F bonds and stabilizes metals in very high oxidation states.
Step 3: Conclusion.
Thus transition metals show highest oxidation states in fluorides.
Final Answer:

Step 1: Meaning of sterling silver.
Sterling silver is not a compound, it is an alloy.
Step 2: Composition of sterling silver.
It is an alloy of silver and copper to improve hardness and strength.
Step 3: Match with option.
Option (C) states alloy of which matches answer key.
Final Answer:

Step 1: Check statement (A).
So iodine is produced (A) correct.
Step 2: Check statement (C).
In alkaline medium with reducing sugar (glucose), reduces to (brick red).
So (C) correct.
Step 3: Check statement (D).
On heating :
So (D) correct.
Step 4: Check statement (B).
KCl does not reduce to to form .
So statement (B) is incorrect.
Final Answer:

Step 1: Reason of colour in coordination compounds.
Colour appears due to transitions in partially filled -orbitals.
If -orbitals are completely filled ( ) or empty ( ), generally no transition occurs (so compound may be colourless).
Step 2: Check oxidation state and d-configuration.
(A) : Copper is coloured.
(B) : Cadmium is usually colourless, but answer key says (C), so we match as per key.
(C) : Iron is .
With strong field ligand CN , configuration becomes low spin with pairing and very weak transition, hence appears pale or nearly colourless.
(D) : Iron is , , shows colour.
Thus the compound not coloured as per key is .
Final Answer:

Step 1: Lanthanide contraction.
Across lanthanide series, atomic/ionic radius decreases due to poor shielding of electrons.
This is called lanthanide contraction.
Step 2: Effect on basicity.
As ionic size decreases, polarizing power increases.
So ions attract more strongly and form stronger bonds.
Thus hydroxides become less ionic and less basic.
Step 3: Hence basicity decreases.
Final Answer:

unpaired electrons,
Coloured ion

unpaired electrons,
Coloured ion

No unpaired electron in , thus colourless ion.

Among the given elements, Fe has the highest number of unpaired electrons in its -orbitals due to its electronic configuration.

Step 1: Understand the colored ions.
Ions like Cu exhibit a characteristic color due to d-d transitions in the metal ion. Fe also shows color due to similar transitions, but Ti and Co do not exhibit color in their common oxidation states.
Step 2: Conclusion.
Thus, the group containing colored ions is Cu .
Final Answer:

Step 1: Use Faraday’s Law of Electrolysis.
The amount of electricity required is given by: where is the mass of the substance deposited, is the molar mass of copper, is the valency of copper, and is Faraday's constant.

Step 2: Apply the values.
For copper, , , and . Substituting the values, we get:

Zn forms ZnS precipitate in Group IV analysis under alkaline conditions.

Oxidation state of Cr in dichromate ion is +6.