Ncert Solutions Chemistry Class 11th

Boric acid contains planar BO₃^3- ions which are linked together through hydrogen bonding shown in the fig.

Excess of CO₂ absorbs heat radiated by the earth. Some of it is dissipated into the atmosphere while the remaining part is radiated back to the earth. As a result, temperature of the earth increases. This is the cause of global warming.

1.7. 1 mole of CuSO₄ contains 1 mole (1 g atom) of Cu

Molar mass of CuSO₄= 63.5 + 32 + (4 x 16) = 159.5 g mol^-1

Thus, Cu that can be obtained from 159.5 g of CuSO₄ = 63.5 g

∴ Cu that can be obtained from 100 g of CuSO₄ =63.5/159.5 * 100  = 39.81g

The highly poisonous nature of CO arises because of its ability to form a complex with haemoglobin, which is about 300 times more stable than the oxygen-haemoglobin complex. This prevents haemoglobin in the red blood corpuscles from carrying oxygen round the body and ultimately resulting in death.

Since, anhydrous HF is a covalent compound and weak acid due to high bond dissociation energy. AlF₃does not dissolve in HF.
Whereas NaF is ionic compound.
3NaF + AlF₃ → Na₃ [AlF₆]
Na₃ [AlF₆] + 3BF₃ (g)→ AlF₃ + 3Na+ [BF]^–

B-Cl bond has dipole moment because of polarity. In BCl_3,  since the molecule is symmetrical (planar), the polarities cancel out and hence the dipole moment is zero

1.6. A mass percent of 69% means that 100 g of nitric acid solution contains 69 g of nitric acid by mass.
Molar mass of nitric acid HNO₃= 1 + 14 + (3x16) = 63 gmol^-1

The difference in bond length is due to the difference in the state of hybridisation. In BF₃ 'B' is sp² hybridised and in BF₄^– 'B' is sp³ hybridised.

1.5

Since the molar mass of sodium acetate is 82.0245 g mol^-1.

⇒ 1 mol of CH₃COONa has a mass = 82.0245 g

∴Mass of sodium acetate (CH₃COONa) required to make 500 ml of 0.375 molar aqueous solution

= 0.375 x 0.5 mol x 82.0245 g mol^-1

= 15.3795 g = 15.380 g

• PbCl₂ + Cl₂→ PbCl₄.
  This is because Pb can show +2 oxidation state more easily than +4 due to inert pair effect.
• PbCl₄→ PbCl₂ + Cl₂
  Because Pb²⁺ is more stable than Pb⁴⁺ due to inert pair effect.
• PbI₄does not exist because I^- ion being a powerful reducing agent reduces Pb⁴⁺ ion to Pb²⁺ ion in solution.

Pb⁴⁺ + 2I^– → Pb²⁺ + I₂