316928
Fluorine is not prepared by general methods because
1 \({\rm{HF}}\) can be easily oxidised.
2 \({\rm{HF}}\) cannot be easily oxidised.
3 \({\rm{HF}}\) is highly poisonous.
4 \({\rm{HF}}\) is a good conductor of electricity.
Explanation:
Fluorine is not prepared by general methods as used for the preparation of \(\mathrm{Cl}_{2}\) and \(\mathrm{Br}_{2}\) because \(\mathrm{HF}\) cannot be easily oxidised to \(\mathrm{F}_{2}\). (Fluorine is the most electronegative element)
CHXI11:THE P-BLOCK ELEMENTS
316929
Identify the statement which is not correct regarding \(\mathrm{CuSO}_{4}\).
1 It reacts with \(\mathrm{KI}\) to give iodine.
2 It reacts with \(\mathrm{KCl}\) to give \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\).
3 It reacts with \(\mathrm{NaOH}\) and glucose to give \(\mathrm{Cu}_{2} \mathrm{O}\).
4 It gives \(\mathrm{CuO}\) on strong heating in air.
Explanation:
\(\mathrm{CuSO}_{4}\) reacts with \(\mathrm{KI}\) to give white precipitate of \(\mathrm{Cu}_{2} \mathrm{I}_{2}\) and evolve \(\mathrm{I}_{2}\). But it does not react with \(\mathrm{KCl}\) in the same way.
CHXI11:THE P-BLOCK ELEMENTS
316930
Iodine is formed when potassium iodide reacts with a solution of
316928
Fluorine is not prepared by general methods because
1 \({\rm{HF}}\) can be easily oxidised.
2 \({\rm{HF}}\) cannot be easily oxidised.
3 \({\rm{HF}}\) is highly poisonous.
4 \({\rm{HF}}\) is a good conductor of electricity.
Explanation:
Fluorine is not prepared by general methods as used for the preparation of \(\mathrm{Cl}_{2}\) and \(\mathrm{Br}_{2}\) because \(\mathrm{HF}\) cannot be easily oxidised to \(\mathrm{F}_{2}\). (Fluorine is the most electronegative element)
CHXI11:THE P-BLOCK ELEMENTS
316929
Identify the statement which is not correct regarding \(\mathrm{CuSO}_{4}\).
1 It reacts with \(\mathrm{KI}\) to give iodine.
2 It reacts with \(\mathrm{KCl}\) to give \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\).
3 It reacts with \(\mathrm{NaOH}\) and glucose to give \(\mathrm{Cu}_{2} \mathrm{O}\).
4 It gives \(\mathrm{CuO}\) on strong heating in air.
Explanation:
\(\mathrm{CuSO}_{4}\) reacts with \(\mathrm{KI}\) to give white precipitate of \(\mathrm{Cu}_{2} \mathrm{I}_{2}\) and evolve \(\mathrm{I}_{2}\). But it does not react with \(\mathrm{KCl}\) in the same way.
CHXI11:THE P-BLOCK ELEMENTS
316930
Iodine is formed when potassium iodide reacts with a solution of
316928
Fluorine is not prepared by general methods because
1 \({\rm{HF}}\) can be easily oxidised.
2 \({\rm{HF}}\) cannot be easily oxidised.
3 \({\rm{HF}}\) is highly poisonous.
4 \({\rm{HF}}\) is a good conductor of electricity.
Explanation:
Fluorine is not prepared by general methods as used for the preparation of \(\mathrm{Cl}_{2}\) and \(\mathrm{Br}_{2}\) because \(\mathrm{HF}\) cannot be easily oxidised to \(\mathrm{F}_{2}\). (Fluorine is the most electronegative element)
CHXI11:THE P-BLOCK ELEMENTS
316929
Identify the statement which is not correct regarding \(\mathrm{CuSO}_{4}\).
1 It reacts with \(\mathrm{KI}\) to give iodine.
2 It reacts with \(\mathrm{KCl}\) to give \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\).
3 It reacts with \(\mathrm{NaOH}\) and glucose to give \(\mathrm{Cu}_{2} \mathrm{O}\).
4 It gives \(\mathrm{CuO}\) on strong heating in air.
Explanation:
\(\mathrm{CuSO}_{4}\) reacts with \(\mathrm{KI}\) to give white precipitate of \(\mathrm{Cu}_{2} \mathrm{I}_{2}\) and evolve \(\mathrm{I}_{2}\). But it does not react with \(\mathrm{KCl}\) in the same way.
CHXI11:THE P-BLOCK ELEMENTS
316930
Iodine is formed when potassium iodide reacts with a solution of
316928
Fluorine is not prepared by general methods because
1 \({\rm{HF}}\) can be easily oxidised.
2 \({\rm{HF}}\) cannot be easily oxidised.
3 \({\rm{HF}}\) is highly poisonous.
4 \({\rm{HF}}\) is a good conductor of electricity.
Explanation:
Fluorine is not prepared by general methods as used for the preparation of \(\mathrm{Cl}_{2}\) and \(\mathrm{Br}_{2}\) because \(\mathrm{HF}\) cannot be easily oxidised to \(\mathrm{F}_{2}\). (Fluorine is the most electronegative element)
CHXI11:THE P-BLOCK ELEMENTS
316929
Identify the statement which is not correct regarding \(\mathrm{CuSO}_{4}\).
1 It reacts with \(\mathrm{KI}\) to give iodine.
2 It reacts with \(\mathrm{KCl}\) to give \(\mathrm{Cu}_{2} \mathrm{Cl}_{2}\).
3 It reacts with \(\mathrm{NaOH}\) and glucose to give \(\mathrm{Cu}_{2} \mathrm{O}\).
4 It gives \(\mathrm{CuO}\) on strong heating in air.
Explanation:
\(\mathrm{CuSO}_{4}\) reacts with \(\mathrm{KI}\) to give white precipitate of \(\mathrm{Cu}_{2} \mathrm{I}_{2}\) and evolve \(\mathrm{I}_{2}\). But it does not react with \(\mathrm{KCl}\) in the same way.
CHXI11:THE P-BLOCK ELEMENTS
316930
Iodine is formed when potassium iodide reacts with a solution of
