Explore topic-wise InterviewSolutions in .

(a) Nitrogen – Tetragens

(i) – (b)

(ii) – (d) 

(iii) – (a) 

(iv) – (c)

(d) Fluorine in OF₂ – +4

(d) Halogens – ns²np⁶

(i) On the basis of electronic configuration 2s² 2p² C and Si occur in tetravalent form but Ge, Sn, Pb show divalency due to inert pair effect.

(ii) Due to inert pair effect, Pb²⁺ is more stable than pb⁴⁺ whereas Sn⁴⁺ is more stable than Sn²⁺ .So tin acts as a reducing agent.

1. Radon is radioactive and used as a source of gamma rays 

2. Radon gas is sealed as small capsules and implanted in the body to destroy malignant i.e. cancer growth.

1. Krypton is used in fluorescent bulbs, flash bulbs etc… 

2. Lamps filed with krypton are used in airports as approaching lights as they can penetrate through dense fog

Argon prevents the oxidation of hot filament and prolongs the life in filament bulbs.

In H₃PO₂, two H atoms are bonded directly to P atom which imparts reducing character to the acid.

Thermal stability of hydrides, decreases from nitrogen to bismuth: NH₃ > PH₃ > AsH₃.

In NH₃,the nitrogen atom forms hydrogen bond because of the following reasons :

• Small size of nitrogen 
• High electronegativity (3.0) of nitrogen

N-H bond is polar forming hydrogen bond. 

P-H bond is almost purely covalent due to larger size and lesser electronegativity.

Nitrogen atom in NH₃ has one lone pair of electrons which is available for donation Therefore, it acts as a Lewis base.

(a) Nessler’s reagent

2. Cl₂, KClO₃

A halogen which is used in the preparation of TEL, an anti-knock compound in petroleum is Cl₂.

H – N – H bond angle in NH₃ is 107°

(c) Pyramidal

(a) Ostwald’s process

(c) Nitronium ion

1. Phosphine(PH₃).

2. By heating white phosphorus with concentrated NaOH solution in an inert atmosphere of CO₂.

P₄ + 3NaoH + 3H₂O \(\longrightarrow\) PH₃ + 3NaH₂PO₂ ( Sodium hypophosphite).

1. No. electronegativity of p-block elements increases along a period and decreases down a group.

2. Most of the p-block elements are nonmetallic in nature. Among p-block elements the metallic character decreases along a period and increases down a group.

3. F₂ > CI₂ > Br₂ > I₂ , because the standard electrode potential values decreases in the same order.

a) No.
b) From top to bottom in the group the bond angle of group 15 hydrides decreases. As the electronegativity of the central atom decreases on moving down the group, the bond pair-bond pair repulsion decreases. Hence the bond angle decreases ¡n the order NH₃ > PH₃ > AsH₃.

C) In all hydrides the central atom is sp³ hybridised. The molecules assume trigonal pyramidal geometry with a lone pair on the central atom.

d) Thermal stability of the group 15 hydrides increases BiH₃ < AsH₃ < PH₃ < NH₃ This is due to the fact that moving up the group the EH bond dissociation enthalpy (‘E’ is a group 15 element) increases due to a decrease in size of the central atom and the molecules will decompose only at higher temperatures.

1. Due to the presence of lone pair of electrons on nitrogen atom of NH₃. But in BiH₃,due to the large size of Bi the electron density decreases and hence is less basic.

2. The allotropic forms of P are White, Red and Black phosphorus. White phosphorus consists of tetrahedral P molecules. Red phosphorus is polymeric in structure consisting of chains of P tetrahedra linked together. Black phosphorus has a layer type structure and has α and β forms.

The bond enthalpy is highest for H₂.

Magnetic moment of an atom with atomic no. 24 in aqeous solution is 4.90 B.M.

(i) Al₂O₃. 2H₂O + 2NaOH → 2NaAlO₂ + 3H₂O

(ii) F₃B+:NH₃ → F₃B ←:NH₃

(i) 3Ga⁺ → 2Ga + Ga³⁺

This is because Ga³⁺ is more stable than Ga⁺.

(ii) Sum of three ionization of both the elements are very high. Thus, they have no tendency to lose electrons to form ionic compounds Instead, they form covalent compounds.

Similarities:

(i) Both have same number of valence electrons.

(ii) Both have similar electronic configuration.

Dissimilarties:

(i) B is a non-metal as Al is metal

(ii) B forms acidic oxide whereas Al forms amphoteric oxides.

B − Cl bond has dipole moment because of polarity. In BCl₃ since the molecule is symmetrical thus the polarities cancel out.

Because it is not able to release H⁺ ions on its own. It receives OH⁻ ions from water molecule to complete its octet and in turn releases H⁺ ions.

A thin protective layer of oxide is formed on the surface of Al and further reaction does not proceed. Such Al becomes passive.

(c) Calcium carbide + calcium phosphide

Borazole is B₃N₃H₆ It is known as inorganic benzene as it has a ring structure with alternate BH and NH groups.

HI is strongest acid

Both Al and Ga form amphoteric hydroxides. 

F₂ is the strongest oxidizing agent.

Noble gases are chemically inert because they have their octet complete except Helium, i.e. they have a stable electronic configurations.

H₂O molecules are associated with intermolecular Hbonding, H₂S is not because oxygen is more electronegative and smaller in size than sulphur. That is why H₂O is a liquid and H₂S is a gas.

NeF₂ compounds is not formed

(b) HF > HCl > HBr > HI

This is based on their crystal structure. Gold has a coordination number of 12 while boron has 6 or less than 6.

Boron atom being small in size is unable to accommodate large sized halogens around it. While in AlCl₃, Al atom has large size. They make use of vacant p-orbitals by coordinate bond i.e., metal atoms complete their octet by forming dimers, Al₂Cl₆.

Due to non-availability of d orbitals, boron is unable to expand its octet. Therefore, the maximum covalence of boron cannot exceed 4.

(i) it is due inert pair effect.

(ii) Molten AlBr₃ is poor conductor of electricity because it is a covalent compound. Therefore, it does not contain free Al³⁺ or B^r− ions to conduct electricity.

The dissimilarities between carbon and silicon in their compounds are as follow: 

(i) Carbon atoms are small, and this allows two carbon atoms to come closer, so that electrons in p-orbitals can overlap forming multiple bonds. While the larger size of the Si atom means that the overlap of the Si p-orbitals is poor, and hence that Si p-p bonding is poor. 

(ii) Both carbon and silicon have the same common electronic configuration as 2s² , 2p² . But in silicon, the electrons are spread into the 3rd energy level, whereas in carbon, it is only to the 2nd energy level. 

This difference occurs because of carbon in the 2nd period, but silicon in the 3d period.

(i) TiCl is more stable than TiCl₃ due to inert pair effect. Due to this effect, Ti will prefer to from TiCl rather than TiCl₃. It prefers to show an oxidation state of +1.

(ii) Because boron can absorb neutrons.

(i) Apply Fajan's rule, greater the charge on cation more is the polarising power and hence more will be covalent character. Polarising power of Pb⁴⁺ > Pb²⁻ thus PbCl₂ is more covalent than PbCl₄, so it is more stable than PbCl₄.

(ii) The size of fluorine is small therefore can be easily arranged around silicon to form [SiF₆ ]²⁻ while the size of chlorine is large enough which create steric hindrance to each other therefore cannot be arranged around silicon in excess.

(i) 2Al(s) + 2NaOH(aq) + 6H₂O(l) → 2Na⁺[Al(OH)₄]⁻(aq) + 3H₂ (g)

(ii) Na₂B₄O₇. 10H₂O \(\overset{\Delta}{\rightarrow}\) Na₂B₄O₇ \(\overset{\Delta}{\rightarrow}\) 2NaBO₂ + B₂O₃

(i) CaCO₃ (s) + 2HCl(aq) → CaCl₂ (aq) + CO₂ (g) + H₂O(l)

(ii) SiO₂ + 2NaOH(g) → Na₂SiO₃ + H₂O

(iii) Fe₂O₃ (s) + 3CO(g) \(\overset{\Delta}{\rightarrow}\) 2Fe(s) + 3CO₂(g)

(iv) C(s) + H₂O(g) \(\overset{473-1273K}{\rightarrow}\) CO (g) + H₂(g)

(v) 2C(s) + O₂(g) + 4N₂(g) \(\overset{1273K}{\rightarrow}\) 2CO(g) + 4N₂(g)

PCl₅ + D₂O → POCl₃ + DCl₂

POCl₃ + 3D₂O → D₃PO₄ + 3DCl

Therefore, the net reaction can be written as, PCl₅ + 4D₂O → D₃PO₄ + 5DCl