p- Block Elements
MCQsNitrogen is relatively inert as compared to phosphorus. Why ?
Answer: Because P - P single bond is much weaker than N - N triple bond. The bond length of nitrogen is small and bond dissociation energy is very large which makes it inert and urtreactive. That's why phosphorus becomes more reactive.
Answer: Because P - P single bond is much weaker than N - N triple bond. The bond length of nitrogen is small and bond dissociation energy is very large which makes it inert and urtreactive. That's why phosphorus becomes more reactive.
Though nitrogen exhibits +5 oxidation state, it does not form pentahalide. Why ?
Answer: Due to non-availability of d-orbitals in its valence electronic configuration nitrogen does not form pentahalide.
Answer: Due to non-availability of d-orbitals in its valence electronic configuration nitrogen does not form pentahalide.
What is the basicity of H3PO2 and Why ?
Answer: H3PO2 has one replaceable H atom so it is monobasic.
Answer: H3PO2 has one replaceable H atom so it is monobasic.
Bond enthalpy of fluorine is lower than that of chlorine. Why ?
Answer: Because F2 is very small in size and its interelectronic repulsions between the lone pairs of electrons are very large.
Answer: Because F2 is very small in size and its interelectronic repulsions between the lone pairs of electrons are very large.
On adding NaOH to ammonium sulphate, a colourless gas with pungent odour is evolved which forms a blue coloured complex with Cu2+ ion. Identify the gas.
Answer: The gas with a pungent odour is NH3 and the blue coloured complex is Tetra-ammine copper (II) sulphate monohydrate.
Answer: The gas with a pungent odour is NH3 and the blue coloured complex is Tetra-ammine copper (II) sulphate monohydrate.
Why is BiH3 the strongest reducing agent amongst all the hydrides of group 15 elements ?
Answer: Reducing nature depends upon the stability of M - H bond. Stability of the bond decreases from N to Bi hydrides, So, BiH3 is the strongest reducing agent.
Answer: Reducing nature depends upon the stability of M - H bond. Stability of the bond decreases from N to Bi hydrides, So, BiH3 is the strongest reducing agent.
What is the covalency of nitrogen in N2O5 ?
Answer: The covalency of nitrogen in N2O5 is 4 because each nitrogen atom has four shared electrons pairs.
Answer: The covalency of nitrogen in N2O5 is 4 because each nitrogen atom has four shared electrons pairs.
Fluorine exhibits only -1 oxidation state whereas other halogens exhibit +1, +3, +5 and +7 oxidation states also. Why is it so ?
Answer: Because fluorine is the most electronegative element and it does not have vacant d-orbitals.
Answer: Because fluorine is the most electronegative element and it does not have vacant d-orbitals.
Why is F2 a stronger oxidising agent than Cl2 ?
Answer: Due to low bond dissociation enthalpy and high electronegativity of Fluorine, it has very strong tendency to accept electrons and get reduced.
F + e– → F–
That's why, F2 acts as strong oxidising agent than Cl2.
Answer: Due to low bond dissociation enthalpy and high electronegativity of Fluorine, it has very strong tendency to accept electrons and get reduced.
F + e– → F–
That's why, F2 acts as strong oxidising agent than Cl2.
NF3 is an exothermic compound but NCl3 is an endothermic compound. Explain.
Answer: Due to smaller size of F, the N – F bond is much stronger than N – Cl bond while bond dissociation energy of F2 is much lower than that of Cl2. Therefore, energy released during the formation of NF3 molecule is more than the energy needed to break N2 and F2 molecules into atoms.
In other words, formation of NF3 is an exothermic reaction. The energy released during the formation of NCl3 molecule is less than the energy needed to break N2 and Cl2 molecules into atoms. Thus formation of NCl3 is an endothermic reaction.
Answer: Due to smaller size of F, the N – F bond is much stronger than N – Cl bond while bond dissociation energy of F2 is much lower than that of Cl2. Therefore, energy released during the formation of NF3 molecule is more than the energy needed to break N2 and F2 molecules into atoms.
In other words, formation of NF3 is an exothermic reaction. The energy released during the formation of NCl3 molecule is less than the energy needed to break N2 and Cl2 molecules into atoms. Thus formation of NCl3 is an endothermic reaction.
Why is BiH3 the strongest reducing agent amongst all the hydrides of group 15 elements ?
Answer: Reducing nature depends upon the stability of M - H bond. Stability of the bond decreases from N to Bi hydrides, So, BiH3 is the strongest reducing agent.
Answer: Reducing nature depends upon the stability of M - H bond. Stability of the bond decreases from N to Bi hydrides, So, BiH3 is the strongest reducing agent.
Orthophosphoric acid (H3PO4) is non-reducing whereas hypophosphorus acid (H3PO2) is a strong reducing agent.” Explain and justify the above statement with suitable example.
Answer: Orthophosphoric acid is not a reducing agent because it doesn’t have any P-H bond whereas hypophosphorus acid is a strong reducing agent as it has two P-H bonds.
Answer: Orthophosphoric acid is not a reducing agent because it doesn’t have any P-H bond whereas hypophosphorus acid is a strong reducing agent as it has two P-H bonds.
ICl is more reactive than I2.
Answer: I-Cl bond is weaker than I-I bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence ICl is more reactive.
Answer: I-Cl bond is weaker than I-I bond as a result of which ICl breaks easily to form halogen atoms which readily bring about the reaction, hence ICl is more reactive.
NH3 is a stronger base than PH3.
Answer: Both N and P contain lone pairs of electrons but due to small size and high electronegativity of Nitrogen in NH3, the electron density is much higher than PH3, therefore it can easily donate electrons and acts as strong Lewis base than PH3.
Answer: Both N and P contain lone pairs of electrons but due to small size and high electronegativity of Nitrogen in NH3, the electron density is much higher than PH3, therefore it can easily donate electrons and acts as strong Lewis base than PH3.
SF4 is easily hydrolysed whereas SF6 is not easily hydrolysed.
Answer: In SF4, due to less steric hindrance by four F atoms, H2O molecules can attack easily while in SF6 the S atom is completely protected by six F atoms and does not allow H2O molecules to attack the S atom.
Answer: In SF4, due to less steric hindrance by four F atoms, H2O molecules can attack easily while in SF6 the S atom is completely protected by six F atoms and does not allow H2O molecules to attack the S atom.
PbCl4 is more covalent than PbCl2.
Answer: According to Fajan’s rule, high charged cations (Pb4+) can polarise the anion (Cl–) more effectively than low charged cations (Pb2+) and hence PbCl4 is more covalent than PbCl2.
Answer: According to Fajan’s rule, high charged cations (Pb4+) can polarise the anion (Cl–) more effectively than low charged cations (Pb2+) and hence PbCl4 is more covalent than PbCl2.
At room temperature, N2 is much less reactive.
Answer: Due to presence of triple bonds between two N atoms, their bond length decreases and hence bond dissociation energy increases which makes N2 lesser reactive at room temperature.
Answer: Due to presence of triple bonds between two N atoms, their bond length decreases and hence bond dissociation energy increases which makes N2 lesser reactive at room temperature.
Thermal stability decreases from H2O to H2Te.
Answer: Thermal stability decreases from H2O to H2Te due to weakening of bond between hydrogen and the atom from O to Te as size is increasing down the group.
Answer: Thermal stability decreases from H2O to H2Te due to weakening of bond between hydrogen and the atom from O to Te as size is increasing down the group.
Fluoride ion has higher hydration enthalpy than chloride ion.
Answer: Due to stronger attractions of smaller in size fluoride ion.
Answer: Due to stronger attractions of smaller in size fluoride ion.
Red phosphorus is less reactive than white phosphorus.
Answer: Red phosphorus is less reactive than white phosphorus because white phosphorus possess angle strain where long angles are only 60° making it more reactive.
Also, red phosphorus being polymeric is less reactive than white phosphorus which has discrete tetrahedral structure.
Answer: Red phosphorus is less reactive than white phosphorus because white phosphorus possess angle strain where long angles are only 60° making it more reactive.
Also, red phosphorus being polymeric is less reactive than white phosphorus which has discrete tetrahedral structure.
N2O5 is more acidic than N2O3.
Answer: N2O5 is more acidic than N2O3 because higher the oxidation state, higher will be acidic character. In N2O5 N has +5 oxidation state and In N2O3 N has +3 oxidation state.
Answer: N2O5 is more acidic than N2O3 because higher the oxidation state, higher will be acidic character. In N2O5 N has +5 oxidation state and In N2O3 N has +3 oxidation state.
How the supersonic jet aeroplanes are responsible for the depletion of ozone layer ?
Answer: The oxides of nitrogen released by the exhausts of supersonic jetplanes are causing the depletion of ozone layer.
Answer: The oxides of nitrogen released by the exhausts of supersonic jetplanes are causing the depletion of ozone layer.
HF is not stored in glass bottles but is kept in wax-coated bottles.
Answer: HF is highly corrosive and etches glass hence it is kept in wax-coated bottles.
Answer: HF is highly corrosive and etches glass hence it is kept in wax-coated bottles.
Bleaching of flowers by Cl2 is permanent while that of SO2 is temporary.
Answer: Chlorine bleaches the material by oxidation hence it is permanent while SO2 bleaches the material by reduction and as the material is exposed to air, it gets oxidised and the colour is restored, hence it is temporary.
Answer: Chlorine bleaches the material by oxidation hence it is permanent while SO2 bleaches the material by reduction and as the material is exposed to air, it gets oxidised and the colour is restored, hence it is temporary.
Bi(V) is a stronger oxidizing agent than Sb(V).
Answer: Bi(V) is a stronger oxidizing agent than Sb(V) due to inert pair effect as the stability of lower oxidation state (+3) increases down the group.
Answer: Bi(V) is a stronger oxidizing agent than Sb(V) due to inert pair effect as the stability of lower oxidation state (+3) increases down the group.
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