Chemistry NCERT Exemplar Solutions Class 12th Chapter Seven

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iii) H₃PO₄

When it reacts with a metal atom, it can produce three different forms of salts, referred known as the three series of salts.

H₃PO₄ + Na→NaH₂PO₄

NaH₂PO₄ + Na→Na₂HPO₄

Na₂HPO₄ + Na→Na₃PO₄

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iii)

When heated with concentrated NaOH in an inert CO₂ environment. It formsPH₃

P₄ + 3NaOH + 3H₂O→PH₃ + 3NaH₂PO₂

As we progress from the top to the bottom in a group, the basic character of hydrides reduces PH₃ is basic than NH₃.

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iv)

The E-H bond dissociation energy of SbH₃ is the lowest among the aforementioned compounds, it easily releases H, making it a strong reducing agent. The greater the lowering agent, the weaker the E-H bond.

This is a multiple choice answer as classified in NCERT Exemplar

Correct option (i)

Fluorine is the most reactive element, forming hydrogen fluoride when it combines with hydrogen gas.

Due to the shorter bond length between the two atoms, it takes a lot of energy to break the link between hydrogen and fluorine.

As a result, as we move down the group, the bond dissociation energy drops.

This is a multiple choice answer as classified in NCERT Exemplar

Correct option (i)

The compound has the following number of electrons:

NO₃^-  =  32e^- 

CO₃^2-  =  32e^- 

ClO₃^-  =  42e^- 

SO₃ ^-  =  42e^- 

Hence,  NO₃^- and CO₃^2-  are isoelectronic

This is a multiple choice answer as classified in NCERT Exemplar

Correct option (iii)

If an electron pair is donated to a vacant orbital on one atom and a lone pair of electrons on the other, the bonding is designated pπ-dπ depending on the orbital to which the electron pair is donated and the orbital from which the electron pair is donated.

Because their valence shells lack d-orbitals, nitrogen, carbon, and boron cannot form a pi bond. Phosphorus, on the other hand, can produce pi bonds.

This is a multiple choice answer as classified in NCERT Exemplar

Correct option (ii)

When H₂S is passed through an aqueous solution of salt acidified with dil. HCl in qualitative analysis, a black precipitate is produced. When the precipitate is boiled with dil. HNO3, a blue solution results. When an excess of ammonia aqueous solution is added to this solution, it produces a deep blue  [Cu (NH₃)₄]²⁺  solution.

This is a multiple choice answer as classified in NCERT Exemplar

Correct option is (iii)

Hydrogen iodide is a more powerful reducing agent than sulphuric acid, it reduces the amount of iodine in the solution from H₂SO₄ to SO₂ and HI to I₂

When chloride salts are treated with sulfuric acid,  HCl gas is formed, which produces a colourless gas.

NaCl + H₂SO₄→HCl + Na₂SO₄

Violet fumes are produced during the reaction due to the creation of iodine (I₂) gas.

2NaI + H₂SO₄→Na₂SO₄ + 2HI 2H₂O + SO₂ + I₂