Ncert Solutions Chemistry Class 11th
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1.2. Calculate the mass percent of different elements present in sodium sulphate (Na2SO4).
1.2. Calculate the mass percent of different elements present in sodium sulphate (Na2SO4).
1.2 Molar mass of Na2SO4= (2 x Atomic mass of Sodium) + Atomic mass of Sulphur + (4 x Atomic mass of Oxygen)
= (2x 23) + 32 + (4x 16)
= 46 + 32 + 64
= 142 g/mol

11.7. Describe the shapes of BF3 and BH4–. Assign the hybridisation of boron in these species.
11.7. Describe the shapes of BF3 and BH4–. Assign the hybridisation of boron in these species.
In BF3, boron is sp2 hybridized.
? shape of BF3 = planar.
In [BH4]–, boron is sp3 hybridized, thus the shape is tetrahedral.

On heating boric acid above 370 K, it forms metaboric acid, HBO2 which on further heating yields boric oxide B2O3.
H3B2O3 → HBO2 → B2O3
Boric acid is a Lewis acid since it accepts electrons from hydroxyl ion of H2O molecule. It is not a protic acid.
B (OH)3 + 2HOH → [B (OH)4]– + H3O+
11.4. Consider the compounds, BCl3, and CCl4. How will they behave with water? Justify.
11.4. Consider the compounds, BCl3, and CCl4. How will they behave with water? Justify.
In BCl3, there is only six electrons in the valence shell of B atom. Thus, the octet is incomplete and it can accept a pair of electrons from water and hence BCl3 undergoes hydrolysis. Whereas, in CCl4, C atom has 8 electrons and its octet is complete. That's why it has no tendency to react wit
11.3. Why does borontrifluoride behave as a Lewis acid?
11.3. Why does borontrifluoride behave as a Lewis acid?
In BF3, central atom has only six electrons after sharing with the electrons of the F
atoms. It is an electron-deficient compound and thus behaves as a Lewis acid.
11.2. How can you explain higher stability of BCl3 as compared to TlCl3?
11.2. How can you explain higher stability of BCl3 as compared to TlCl3?
BCl3 is quite stable. Because there is absence of d- and f-electrons in boron three valence electrons (2s2 2px1) are there for bonding with chlorine atom. In Tl the valence s-electron (6s2) are experiencing maximum inert pair effect. Thus, only 6p1 electron is available for bonding. T
11.1. Discuss the pattern of variation in the oxidation states of
(i) B to Tl and (ii) C to Pb.
11.1. Discuss the pattern of variation in the oxidation states of
(i) B to Tl and (ii) C to Pb.
(i) B to Tl
Common oxidation states are +1 and +3. The stability of +3 oxidation state decreases from B to Tl while +1 oxidation state increases from B to Tl.
(ii) C to Pb
The common oxidation states are +4 and +2. Stability of +4 oxidation state decreases from C to Pb.
1.1. Calculate the molecular mass of the following:
(i) H2O (ii) CO2 (iii) CH4
1.1. Calculate the molecular mass of the following:
(i) H2O (ii) CO2 (iii) CH4
1.1. (i) Molecular mass of H2O = (2x Atomic mass of Hydrogen)+ Atomic mass of Oxygen
Atomic mass of Hydrogen = 1.008 amu
Atomic mass of Oxygen = 16.00 amu
So,
Molecular mass of H2O = 2 (1.008 amu) + 16.00 amu =18.016 amu
(ii) Molecular mass of CO2 = Atomic mass of Carbon + (2x Atomic mass of
10.32. Which one of the alkaline earth metal carbonates is thermally most stable?
(a) MgCO3 (b) CaCO3 (c) SrCO3 (d) BaCO3(Intermediate)
10.32. Which one of the alkaline earth metal carbonates is thermally most stable?
(a) MgCO3 (b) CaCO3 (c) SrCO3 (d) BaCO3(Intermediate)
(d) BaCO3
Thermal stability is directly proportional to the size of the cation i.e., larger the size of the atom, greater is its thermal stability. Ba being the biggest cation among the given compounds, has a greater thermal stability.