D And F Block Elements

Step 1: Existence of Similar Complexes

• (hexafluorosilicate) is well-known as a stable species.
• is also known, due to the relatively larger size of Ge allowing stable complex formation with six chlorines.
• (stannate(II)) also exists, with tin in a high coordination environment.

Step 2: Why Does Not Exist

• Silicon does not form a stable complex with six chloride ligands.
• The large charge density and possible multiple-bond character with chlorine prevent a viable ion.

Hence, does not exist.

- The correct statement is that noble gases have positive electron gain enthalpies. This is true because noble gases are very stable and do not easily gain electrons. - The other statements are incorrect. For example, in the contact process, high temperature and high pressure are used to manufacture sulfuric acid.

This is a redox titration problem. We need to write the balanced chemical equation for the reaction between permanganate ions (MnO ) and oxalate ions (C O ) in an acidic medium.
Step 1: Write the half-reactions.
Reduction half-reaction (permanganate):
In acidic medium, MnO is reduced to Mn . The oxidation state of Mn changes from +7 to +2.
The change in oxidation state is 5. So, the n-factor for MnO is 5.
MnO + 8H + 5e Mn + 4H O.
Oxidation half-reaction (oxalate):
Oxalate ion (C O ) is oxidized to carbon dioxide (CO ). The oxidation state of carbon changes from +3 (in C O ) to +4 (in CO ).
For the two carbon atoms in one oxalate ion, the total change in oxidation state is .
So, the n-factor for C O is 2.
C O 2CO + 2e .
Step 2: Balance the electrons and combine the half-reactions.
To balance the electrons, we multiply the reduction half-reaction by 2 and the oxidation half-reaction by 5.
MnO + 8H + 5e Mn + 4H O)
C O 2CO + 2e )
This gives:
2MnO + 16H + 10e 2Mn + 8H O
5C O 10CO + 10e
Adding them together gives the overall balanced equation:
2MnO + 5C O + 16H 2Mn + 10CO + 8H O.
Step 3: Determine the molar ratio.
From the balanced equation, we can see that 2 moles of permanganate ions react with 5 moles of oxalate ions.
The question asks for the number of moles of oxalate ions oxidized by one mole of permanganate ions.
Moles of oxalate = 1 mole MnO moles.
Alternatively, using the concept of equivalents:
Equivalents of MnO = Equivalents of C O Moles -factor = Moles -factor Moles .

Step 1: Determine the number of electrons and -orbital configuration for each ion.
An ion is paramagnetic if it has unpaired electrons. Both ions in the pair must have the same number of electrons (isoelectronic) and both must have unpaired electrons.
Lanthanides range from to with general configuration . The ions generally lose and electrons.

Step 2: Analyze the electron configuration for option (A).
, so ( , diamagnetic).
, so ( , diamagnetic).
Isoelectronic (68 ), but both are diamagnetic. Incorrect.

Step 3: Analyze the electron configuration for option (B).
, so ( , 6 unpaired , paramagnetic).
, so ( , 5 unpaired , paramagnetic).
Not isoelectronic (60 vs 59 ). Incorrect.

Step 4: Analyze the electron configuration for option (C).
, so ( , 7 unpaired , paramagnetic).
, so ( , 7 unpaired , paramagnetic).
Both are isoelectronic ( ) and paramagnetic ( unpaired ). Correct.

Step 5: Analyze the electron configuration for option (D).
, so ( , diamagnetic).
, so ( , diamagnetic).
Isoelectronic ( ), but both are diamagnetic. Incorrect.

Step 1: Analyze Reactions and Products:

• (A) White alkali: Produces Phosphine ( ) and Hypophosphite ( ). Hypophosphorous acid ( ) has one , two , one . No P-P bond.
• (B) : Produces Orthophosphorous acid ( ). Has one , one , two . No P-P bond.
• (C) : Produces Pyrophosphorous acid ( ). Structure has linkage. No direct P-P bond.
• (D) Red alkali: Oxidation of Red Phosphorus with alkali produces salts of Hypophosphoric acid ( ). The structure of Hypophosphoric acid contains a direct P-P bond ( ).

Step 2: Conclusion:
Hypophosphoric acid ( ), formed from Red P and alkali, contains a P-P bond.

Step 1: Understanding the Concept:
The question asks for the highest (most positive or least negative) standard electrode potential ( ) among the given first-row transition metals. The standard electrode potential reflects the tendency of a species to be reduced. A more positive value indicates a greater tendency for the M²⁺ ion to be reduced to the metal M.
Step 2: Analyzing the Trend and Standard Values:
The standard electrode potentials for the M²⁺/M couple in the first transition series generally show a trend of becoming less negative from Ti to Cu, but with some notable exceptions. Let's look at the standard values for the given elements:
- V (Vanadium): V
- Cr (Chromium): V
- Mn (Manganese): V
- Fe (Iron): V
Step 3: Comparing the Values:
Comparing the given values:
- V
- V
- V
- V
The highest value (least negative) among these is -0.44 V, which corresponds to Iron (Fe).
Step 4: Explaining the Anomalies:
- The value for Mn is more negative than expected from the general trend. This is due to the extra stability of the Mn²⁺ ion, which has a half-filled d-orbital configuration ( ). - The value for Fe is less negative (higher) than its neighbors. This can be attributed to a combination of factors including hydration enthalpy and ionization enthalpy.
Final Answer:
Comparing the standard reduction potentials, Iron (Fe) has the highest value (-0.44 V). Therefore, option (A) is correct.

1. i. P O , N O : Form acidic oxides (H PO , HNO ), correct.
2. ii. N O, NO: Neutral oxides (minimal reaction with water), correct.
3. iii. SeO , TeO : Acidic (form selenic, telluric acids), not basic, incorrect.
4. iv. As O , Sb O : Amphoteric (react with acids and bases), correct.
5. Thus, i, ii, iv are correct, so the answer is (1).

Acidity of HX in water depends on bond strength and solvation effects. HF has the strongest H–F bond → weakest acid in water, despite high bond dissociation enthalpy. Therefore, R is correct but does not explain A.