316249
Assertion : \(\mathrm{H}_{2} \mathrm{Se}\) is more acidic than \(\mathrm{H}_{2} \mathrm{~S}\). Reason : \(\mathrm{S}\) is less electronegative than \(\mathrm{Se}\).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Down the group, electronegativity decreases. The bond dissociation energy also decreases due to increased atomic size, and the release of \(\mathrm{H}^{+}\) ion becames easier. Thus \(\mathrm{H}_{2} \mathrm{Se}\) is more acidic than \(\mathrm{H}_{2} \mathrm{~S}\). S is more electronegative than Se. So option (3) is correct.
CHXI11:THE P-BLOCK ELEMENTS
316250
Which is the correct thermal stability order for\({{\rm{H}}_{\rm{2}}}{\rm{E}}{\mkern 1mu} {\rm{(E = O,}}{\mkern 1mu} {\rm{S,}}{\mkern 1mu} {\rm{Se,}}{\mkern 1mu} {\rm{Te}}\,\,{\rm{and}}\,\,{\rm{Po)}}\) ?
On going down the group thermal stability order for \(\mathrm{H}_{2} \mathrm{E}\) decreases because H-E bond energy decreases \(\therefore \quad\) Order of stability would be : \(\mathrm{H}_{2} \mathrm{Po} < \mathrm{H}_{2} \mathrm{Te} < \mathrm{H}_{2} \mathrm{Se} < \mathrm{H}_{2} \mathrm{~S} < \mathrm{H}_{2} \mathrm{O}\)
NEET - 2019
CHXI11:THE P-BLOCK ELEMENTS
316251
Bond dissociation energy of " \({\rm{E}} - {\rm{H}}\) " bond of the " \(\mathrm{H}_{2} \mathrm{E}\) " hydrides of group 16 elements (given below) follows order A. O B. S C. Se D. Te Choose the correct from the options given below.
1 A \(>\) B \(>\) D \(>\) C
2 B \(>\) A \(>\) C \(>\) D
3 D \(>\) C \(>\) B \(>\) A
4 A \(>\) B \(>\) C \(>\) D
Explanation:
Bond dissociation energy of group 16 elements has the following order :
Thus, option (4) is correct.
JEE - 2023
CHXI11:THE P-BLOCK ELEMENTS
316252
A chalcogen combines directly with hydrogen to form a hydride but with a great difficulty. On burning in air, this chalcogen forms a solid polymeric dioxide and it also has the highest electrical resistance amongst the metals. This chalcogen is
1 Sulphur
2 Selenium
3 Polonium
4 Tellurium
Explanation:
Selenium is a chalcogen combines directly with hydrogen to form a hydride but with a great difficulty. On burning in air, it forms a solid polymeric dioxide and it also possesses the highest electrical resistance amongst the metals.
CHXI11:THE P-BLOCK ELEMENTS
316253
Which of the following is a highly basic oxide?
1 \(\mathrm{SO}_{2}\)
2 \(\mathrm{SO}_{3}\)
3 \(\mathrm{TeO}_{2}\)
4 \(\mathrm{TeO}_{3}\)
Explanation:
In a group from top to bottom basic nature of oxides increases and the element with lower oxidation state is more basic in nature.
316249
Assertion : \(\mathrm{H}_{2} \mathrm{Se}\) is more acidic than \(\mathrm{H}_{2} \mathrm{~S}\). Reason : \(\mathrm{S}\) is less electronegative than \(\mathrm{Se}\).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Down the group, electronegativity decreases. The bond dissociation energy also decreases due to increased atomic size, and the release of \(\mathrm{H}^{+}\) ion becames easier. Thus \(\mathrm{H}_{2} \mathrm{Se}\) is more acidic than \(\mathrm{H}_{2} \mathrm{~S}\). S is more electronegative than Se. So option (3) is correct.
CHXI11:THE P-BLOCK ELEMENTS
316250
Which is the correct thermal stability order for\({{\rm{H}}_{\rm{2}}}{\rm{E}}{\mkern 1mu} {\rm{(E = O,}}{\mkern 1mu} {\rm{S,}}{\mkern 1mu} {\rm{Se,}}{\mkern 1mu} {\rm{Te}}\,\,{\rm{and}}\,\,{\rm{Po)}}\) ?
On going down the group thermal stability order for \(\mathrm{H}_{2} \mathrm{E}\) decreases because H-E bond energy decreases \(\therefore \quad\) Order of stability would be : \(\mathrm{H}_{2} \mathrm{Po} < \mathrm{H}_{2} \mathrm{Te} < \mathrm{H}_{2} \mathrm{Se} < \mathrm{H}_{2} \mathrm{~S} < \mathrm{H}_{2} \mathrm{O}\)
NEET - 2019
CHXI11:THE P-BLOCK ELEMENTS
316251
Bond dissociation energy of " \({\rm{E}} - {\rm{H}}\) " bond of the " \(\mathrm{H}_{2} \mathrm{E}\) " hydrides of group 16 elements (given below) follows order A. O B. S C. Se D. Te Choose the correct from the options given below.
1 A \(>\) B \(>\) D \(>\) C
2 B \(>\) A \(>\) C \(>\) D
3 D \(>\) C \(>\) B \(>\) A
4 A \(>\) B \(>\) C \(>\) D
Explanation:
Bond dissociation energy of group 16 elements has the following order :
Thus, option (4) is correct.
JEE - 2023
CHXI11:THE P-BLOCK ELEMENTS
316252
A chalcogen combines directly with hydrogen to form a hydride but with a great difficulty. On burning in air, this chalcogen forms a solid polymeric dioxide and it also has the highest electrical resistance amongst the metals. This chalcogen is
1 Sulphur
2 Selenium
3 Polonium
4 Tellurium
Explanation:
Selenium is a chalcogen combines directly with hydrogen to form a hydride but with a great difficulty. On burning in air, it forms a solid polymeric dioxide and it also possesses the highest electrical resistance amongst the metals.
CHXI11:THE P-BLOCK ELEMENTS
316253
Which of the following is a highly basic oxide?
1 \(\mathrm{SO}_{2}\)
2 \(\mathrm{SO}_{3}\)
3 \(\mathrm{TeO}_{2}\)
4 \(\mathrm{TeO}_{3}\)
Explanation:
In a group from top to bottom basic nature of oxides increases and the element with lower oxidation state is more basic in nature.
316249
Assertion : \(\mathrm{H}_{2} \mathrm{Se}\) is more acidic than \(\mathrm{H}_{2} \mathrm{~S}\). Reason : \(\mathrm{S}\) is less electronegative than \(\mathrm{Se}\).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Down the group, electronegativity decreases. The bond dissociation energy also decreases due to increased atomic size, and the release of \(\mathrm{H}^{+}\) ion becames easier. Thus \(\mathrm{H}_{2} \mathrm{Se}\) is more acidic than \(\mathrm{H}_{2} \mathrm{~S}\). S is more electronegative than Se. So option (3) is correct.
CHXI11:THE P-BLOCK ELEMENTS
316250
Which is the correct thermal stability order for\({{\rm{H}}_{\rm{2}}}{\rm{E}}{\mkern 1mu} {\rm{(E = O,}}{\mkern 1mu} {\rm{S,}}{\mkern 1mu} {\rm{Se,}}{\mkern 1mu} {\rm{Te}}\,\,{\rm{and}}\,\,{\rm{Po)}}\) ?
On going down the group thermal stability order for \(\mathrm{H}_{2} \mathrm{E}\) decreases because H-E bond energy decreases \(\therefore \quad\) Order of stability would be : \(\mathrm{H}_{2} \mathrm{Po} < \mathrm{H}_{2} \mathrm{Te} < \mathrm{H}_{2} \mathrm{Se} < \mathrm{H}_{2} \mathrm{~S} < \mathrm{H}_{2} \mathrm{O}\)
NEET - 2019
CHXI11:THE P-BLOCK ELEMENTS
316251
Bond dissociation energy of " \({\rm{E}} - {\rm{H}}\) " bond of the " \(\mathrm{H}_{2} \mathrm{E}\) " hydrides of group 16 elements (given below) follows order A. O B. S C. Se D. Te Choose the correct from the options given below.
1 A \(>\) B \(>\) D \(>\) C
2 B \(>\) A \(>\) C \(>\) D
3 D \(>\) C \(>\) B \(>\) A
4 A \(>\) B \(>\) C \(>\) D
Explanation:
Bond dissociation energy of group 16 elements has the following order :
Thus, option (4) is correct.
JEE - 2023
CHXI11:THE P-BLOCK ELEMENTS
316252
A chalcogen combines directly with hydrogen to form a hydride but with a great difficulty. On burning in air, this chalcogen forms a solid polymeric dioxide and it also has the highest electrical resistance amongst the metals. This chalcogen is
1 Sulphur
2 Selenium
3 Polonium
4 Tellurium
Explanation:
Selenium is a chalcogen combines directly with hydrogen to form a hydride but with a great difficulty. On burning in air, it forms a solid polymeric dioxide and it also possesses the highest electrical resistance amongst the metals.
CHXI11:THE P-BLOCK ELEMENTS
316253
Which of the following is a highly basic oxide?
1 \(\mathrm{SO}_{2}\)
2 \(\mathrm{SO}_{3}\)
3 \(\mathrm{TeO}_{2}\)
4 \(\mathrm{TeO}_{3}\)
Explanation:
In a group from top to bottom basic nature of oxides increases and the element with lower oxidation state is more basic in nature.
NEET Test Series from KOTA - 10 Papers In MS WORD
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CHXI11:THE P-BLOCK ELEMENTS
316249
Assertion : \(\mathrm{H}_{2} \mathrm{Se}\) is more acidic than \(\mathrm{H}_{2} \mathrm{~S}\). Reason : \(\mathrm{S}\) is less electronegative than \(\mathrm{Se}\).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Down the group, electronegativity decreases. The bond dissociation energy also decreases due to increased atomic size, and the release of \(\mathrm{H}^{+}\) ion becames easier. Thus \(\mathrm{H}_{2} \mathrm{Se}\) is more acidic than \(\mathrm{H}_{2} \mathrm{~S}\). S is more electronegative than Se. So option (3) is correct.
CHXI11:THE P-BLOCK ELEMENTS
316250
Which is the correct thermal stability order for\({{\rm{H}}_{\rm{2}}}{\rm{E}}{\mkern 1mu} {\rm{(E = O,}}{\mkern 1mu} {\rm{S,}}{\mkern 1mu} {\rm{Se,}}{\mkern 1mu} {\rm{Te}}\,\,{\rm{and}}\,\,{\rm{Po)}}\) ?
On going down the group thermal stability order for \(\mathrm{H}_{2} \mathrm{E}\) decreases because H-E bond energy decreases \(\therefore \quad\) Order of stability would be : \(\mathrm{H}_{2} \mathrm{Po} < \mathrm{H}_{2} \mathrm{Te} < \mathrm{H}_{2} \mathrm{Se} < \mathrm{H}_{2} \mathrm{~S} < \mathrm{H}_{2} \mathrm{O}\)
NEET - 2019
CHXI11:THE P-BLOCK ELEMENTS
316251
Bond dissociation energy of " \({\rm{E}} - {\rm{H}}\) " bond of the " \(\mathrm{H}_{2} \mathrm{E}\) " hydrides of group 16 elements (given below) follows order A. O B. S C. Se D. Te Choose the correct from the options given below.
1 A \(>\) B \(>\) D \(>\) C
2 B \(>\) A \(>\) C \(>\) D
3 D \(>\) C \(>\) B \(>\) A
4 A \(>\) B \(>\) C \(>\) D
Explanation:
Bond dissociation energy of group 16 elements has the following order :
Thus, option (4) is correct.
JEE - 2023
CHXI11:THE P-BLOCK ELEMENTS
316252
A chalcogen combines directly with hydrogen to form a hydride but with a great difficulty. On burning in air, this chalcogen forms a solid polymeric dioxide and it also has the highest electrical resistance amongst the metals. This chalcogen is
1 Sulphur
2 Selenium
3 Polonium
4 Tellurium
Explanation:
Selenium is a chalcogen combines directly with hydrogen to form a hydride but with a great difficulty. On burning in air, it forms a solid polymeric dioxide and it also possesses the highest electrical resistance amongst the metals.
CHXI11:THE P-BLOCK ELEMENTS
316253
Which of the following is a highly basic oxide?
1 \(\mathrm{SO}_{2}\)
2 \(\mathrm{SO}_{3}\)
3 \(\mathrm{TeO}_{2}\)
4 \(\mathrm{TeO}_{3}\)
Explanation:
In a group from top to bottom basic nature of oxides increases and the element with lower oxidation state is more basic in nature.
316249
Assertion : \(\mathrm{H}_{2} \mathrm{Se}\) is more acidic than \(\mathrm{H}_{2} \mathrm{~S}\). Reason : \(\mathrm{S}\) is less electronegative than \(\mathrm{Se}\).
1 Both Assertion and Reason are correct and Reason is the correct explanation of the Assertion.
2 Both Assertion and Reason are correct but Reason is not the correct explanation of the Assertion.
3 Assertion is correct but Reason is incorrect.
4 Assertion is incorrect but Reason is correct.
Explanation:
Down the group, electronegativity decreases. The bond dissociation energy also decreases due to increased atomic size, and the release of \(\mathrm{H}^{+}\) ion becames easier. Thus \(\mathrm{H}_{2} \mathrm{Se}\) is more acidic than \(\mathrm{H}_{2} \mathrm{~S}\). S is more electronegative than Se. So option (3) is correct.
CHXI11:THE P-BLOCK ELEMENTS
316250
Which is the correct thermal stability order for\({{\rm{H}}_{\rm{2}}}{\rm{E}}{\mkern 1mu} {\rm{(E = O,}}{\mkern 1mu} {\rm{S,}}{\mkern 1mu} {\rm{Se,}}{\mkern 1mu} {\rm{Te}}\,\,{\rm{and}}\,\,{\rm{Po)}}\) ?
On going down the group thermal stability order for \(\mathrm{H}_{2} \mathrm{E}\) decreases because H-E bond energy decreases \(\therefore \quad\) Order of stability would be : \(\mathrm{H}_{2} \mathrm{Po} < \mathrm{H}_{2} \mathrm{Te} < \mathrm{H}_{2} \mathrm{Se} < \mathrm{H}_{2} \mathrm{~S} < \mathrm{H}_{2} \mathrm{O}\)
NEET - 2019
CHXI11:THE P-BLOCK ELEMENTS
316251
Bond dissociation energy of " \({\rm{E}} - {\rm{H}}\) " bond of the " \(\mathrm{H}_{2} \mathrm{E}\) " hydrides of group 16 elements (given below) follows order A. O B. S C. Se D. Te Choose the correct from the options given below.
1 A \(>\) B \(>\) D \(>\) C
2 B \(>\) A \(>\) C \(>\) D
3 D \(>\) C \(>\) B \(>\) A
4 A \(>\) B \(>\) C \(>\) D
Explanation:
Bond dissociation energy of group 16 elements has the following order :
Thus, option (4) is correct.
JEE - 2023
CHXI11:THE P-BLOCK ELEMENTS
316252
A chalcogen combines directly with hydrogen to form a hydride but with a great difficulty. On burning in air, this chalcogen forms a solid polymeric dioxide and it also has the highest electrical resistance amongst the metals. This chalcogen is
1 Sulphur
2 Selenium
3 Polonium
4 Tellurium
Explanation:
Selenium is a chalcogen combines directly with hydrogen to form a hydride but with a great difficulty. On burning in air, it forms a solid polymeric dioxide and it also possesses the highest electrical resistance amongst the metals.
CHXI11:THE P-BLOCK ELEMENTS
316253
Which of the following is a highly basic oxide?
1 \(\mathrm{SO}_{2}\)
2 \(\mathrm{SO}_{3}\)
3 \(\mathrm{TeO}_{2}\)
4 \(\mathrm{TeO}_{3}\)
Explanation:
In a group from top to bottom basic nature of oxides increases and the element with lower oxidation state is more basic in nature.
