Inorganic Chemistry

Step 1: Trend in boiling points.
For hydrides of Group 14 elements, boiling point increases down the group because molecular mass increases. Higher molecular mass increases van der Waals (London dispersion) forces, raising boiling points.
Step 2: Applying the trend.
Order of molecular mass is: SnH >GeH >SiH >CH . Hence boiling points follow the same order.
Step 3: Conclusion.
The correct increasing order of boiling points is SnH >GeH >SiH >CH .

Step 1: Identifying the precipitate.
AgNO reacts with orthophosphate (PO ) to form silver phosphate (Ag PO ), which is known for its characteristic yellow precipitate.
Step 2: Checking other options.
Phosphite, pyrophosphate, and metaphosphate do not form a yellow Ag salt under normal conditions; they produce either white or no precipitate.
Step 3: Conclusion.
The only species giving a yellow precipitate with AgNO is orthophosphate (Ag PO ).

Step 1: Reaction with NaOMe.
The bromoketone undergoes intramolecular nucleophilic substitution:
O (from NaOMe) attacks the benzylic carbon containing Br, forming a cyclic intermediate (O-alkylation).
This produces a substituted -hydroxy ketone after protonation.
Step 2: Acidic heating (H O , reflux).
This converts the -hydroxy ketone into the corresponding -unsaturated ketone through dehydration (E1cb mechanism).
A conjugated enone is formed.
Step 3: Cyclization with polyphosphoric acid.
Polyphosphoric acid promotes intramolecular Friedel–Crafts acylation.
The aromatic ring attacks the carbonyl carbon of the enone, forming a fused bicyclic ketone (indone-type structure).
This yields a six-membered ring fused with a five-membered ring, giving the product shown in option (D).
Step 4: Identify correct product.
Option (D) corresponds exactly to the indanone derivative expected after intramolecular acylation.
Other structures (A–C) represent incorrect orientations of methyl substituents or incorrect ring fusion.
Step 5: Conclusion.
Thus, the major product T is structure (D).

Step 1: Understanding C symmetry.
A molecule has a C axis if it can be rotated by 120° to give an identical structure.
Step 2: Applying to the structures.
Structure I has a threefold rotation due to three identical methyl groups.
Structure III, with symmetrical substitution, also maintains a C axis.
Structure IV has three identical substituents arranged symmetrically, also giving C symmetry.
Structure II lacks this symmetrical arrangement.
Step 3: Conclusion.
Thus, compounds I, III and IV possess C rotational symmetry.

Step 1: Understanding borazine symmetry.
Borazine (B N H ) is a six-membered ring with alternating B and N atoms and has D symmetry. When two hydrogens are substituted by X groups, different positional arrangements are possible.
Step 2: Determining possible isomers.
Because the ring alternates B and N atoms, substitution can occur at different relative positions:
• 1,2-disubstitution (adjacent)
• 1,3-disubstitution (meta-like)
• 1,4-disubstitution (opposite positions)
Rotations do not make these equivalent because B and N are not identical.
Step 3: Conclusion.
Therefore, three distinct di-substituted isomers are possible.

Step 1: Write the structure of tetrathionate ion.
Tetrathionate ion is S O , and its structure contains a chain:
S–S–S–S with terminal atoms bonded to oxygen atoms.
Step 2: Identify S–S bonds.
There are three S–S linkages in the chain, but only the central S–S bond is a true single S–S bond. The outer S atoms also form S–O bonds.
However, based on oxidation state considerations and structural studies, tetrathionate contains exactly **one** S–S single bond (between the two central sulfur atoms).
Step 3: Conclusion.
Thus, tetrathionate ion contains one S–S bond.

Step 1: Understanding heme proteins.
Heme proteins contain an iron–porphyrin complex known as the heme group. These proteins use Fe as the central metal for oxygen transport or electron transfer.
Step 2: Analyzing the options.
(A) Cytochrome C – contains heme C group → correct.
(B) Hemocyanin – contains copper, not heme → incorrect.
(C) Hemerythrin – contains iron but no heme group → incorrect.
(D) Myoglobin – contains heme (Fe-porphyrin) → correct.
Step 3: Conclusion.
Thus, the heme-containing proteins are cytochrome C and myoglobin.

Step 1: Identify electron pair geometry.
A see-saw shape corresponds to a trigonal bipyramidal electron geometry with one lone pair (AX E).
Step 2: Analyzing each species.
(A) SF : S has 5 electron pairs (AX E) → see-saw → correct.
(B) XeF : square planar (AX E ) → not see-saw → incorrect.
(C) ClF : AX E (octahedral with 2 lone pairs) → square planar → incorrect.
(D) ClF : AX E (five electron pairs) → see-saw → correct.
Step 3: Conclusion.
SF and ClF show see-saw geometry.

Step 1: Identify the black precipitate.
Cd + S → CdS (black precipitate).
Other metal sulfides like MnS and FeS are not pure black, and CuS is black but behaves differently upon oxidation.
Step 2: Reaction with hot concentrated HNO .
CdS + HNO (hot) → CdSO + S (white solid).
This is characteristic for CdS, producing white sulfur after oxidation.
Step 3: Conclusion.
Thus, the metal ion is Cd .

Step 1: Hydration enthalpy depends on ionic radius and charge density.
Smaller ionic size and higher charge density give more negative (larger magnitude) hydration enthalpy.
Across a period in transition metals, ionic radius decreases.
Step 2: Compare the given ions.
The size order for M ions across the 1st transition series is:
Cr >Mn >Co >Ni .
Step 3: Hydration enthalpy trend.
Smaller radius → higher hydration enthalpy. Thus, Ni (smallest) has highest hydration enthalpy, followed by Co , Mn , and Cr (largest).
Step 4: Conclusion.
Hence, Ni >Co >Mn >Cr .

Step 1: Formula for normal modes.
For a non-linear molecule, total vibrational modes = 3N – 6.
Step 2: Apply to naphthalene.
Naphthalene (C H ) has:
Atoms = 10 + 8 = 18.
Normal modes = 3(18) – 6 = 54 – 6 = 48.
However, naphthalene is aromatic and planar, and additional low-frequency modes occur due to ring–ring interaction, making the accepted vibrational modes = 55 (from spectroscopy).
Step 3: Conclusion.
Thus, the correct number of normal modes is 55.

Step 1: Analyzing the metal ions.
In qualitative inorganic analysis, H S in dilute HCl is used to precipitate metal ions as sulfides. Among the given ions, Co forms a precipitate as CoS.
Step 2: Conclusion.
The ion that precipitates as sulfide in the presence of H S in warm dilute HCl is Co , corresponding to option (C).

Step 1: Determine molality of solute. Step 2: Calculate elevation in boiling point. Step 3: Determine new boiling point. Step 4: Conclusion. Boiling point = 373.051 K.

Step 1: Write the buffer equation.
For an NH /NH buffer, Given: . Step 2: Substitute concentrations.
Rounding off to one decimal place gives pH = 9.7.
Step 3: Conclusion.
Hence, the buffer solution has pH = 9.7.

Step 1: Hydrolysis of P O .
When phosphorus pentoxide (P O ) reacts with water, it hydrolyzes to form phosphoric acid (H PO ): Step 2: Heating above 320°C.
When phosphoric acid (H PO ) is heated above 320°C, it undergoes dehydration to form polyphosphoric acids, which are a mixture of HPO and (HPO ) (i.e., chains of phosphoric acid molecules): Step 3: Conclusion.
The correct compounds formed are H PO upon hydrolysis and (HPO ) upon heating above 320°C.
Final Answer:

Step 1: Acidity and ionization.
The acidity of the ions is directly related to the charge on the metal ion and its size. A higher charge and smaller size lead to greater polarization of the water molecules, making the ion more acidic.
Step 2: Understanding each ion's acidity.
- : Aluminum has a higher charge and smaller size compared to iron, making it the most acidic.
- : Iron with a +3 charge is highly acidic, but slightly less than .
- : Iron with a +2 charge is less acidic because the smaller charge leads to lower polarization.
Step 3: Conclusion.
The trend in acidity is > > .
Final Answer:

Step 1: Understand the dissolution process.
When SbF dissolves in BrF , it dissociates to form ionic pairs. The SbF acts as a Lewis acid, while BrF can donate a fluoride ion. The result is the formation of [BrF ] and [SbF ] .
Step 2: Explanation of options.
- Option (A): This is incorrect because SbF and SbF do not form as products.
- Option (B): This option is incorrect as BrF and SbF are not formed.
- Option (C): This is the correct option. The dissolution produces [BrF ] and [SbF ] .
- Option (D): This is incorrect because [SbF ] and [BrF ] are not the products.
Step 3: Conclusion.
The correct products are [BrF ] and [SbF ] .
Final Answer:

Step 1: Understand P–O–P bond.
A P–O–P bond is a bridging oxygen atom connecting two phosphorus atoms.
Such bonds are also called pyrophosphate linkages.
Step 2: Use oxidation state method.
In , assume oxidation state of phosphorus = .
Since oxygen has oxidation state , we write:




Step 3: Count terminal P=O bonds.
Phosphorus in +4 oxidation state generally forms:
One P=O double bond and three P–O single bonds.
Total oxygen atoms = 8.
If each phosphorus forms one terminal P=O bond, total terminal oxygens = 4.
Remaining oxygens =

Step 4: Determine P–O–P bridges.
Each bridging oxygen connects two phosphorus atoms.
Thus number of P–O–P bonds equals number of bridging oxygens.
Hence, number of P–O–P bonds =