Statement I: The halogen that makes the longest bond with hydrogen in HX, has the smallest covalent radius in its group. [2026]

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Q.
Given below are two statements: [2026]

Statement I: The halogen that makes the longest bond with hydrogen in HX, has the smallest covalent radius in its group.

Statement II: A group 15 element’s hydride ${EH}_{3}$ has the lowest boiling point among corresponding hydrides of other group 15 elements. The maximum covalency of that element E is 4.

In the light of the above statements, choose the correct answer from the options given below.

1 Both Statement I and Statement II are true

2 Statement I is true but Statement II is false

3 Both Statement I and Statement II are false

4 Statement I is false but Statement II is true

Ans.
(3)

$HF < HCl < HBr < HI (bond length order)$

$F < Cl < Br < I (radius order)$

${PH}_{3} < {AsH}_{3} < {NH}_{3} < {SbH}_{3} < {BiH}_{3} (Boiling point order)$

$Maximum possible covalency of phosphorus is 6$

Solution

1 Both Statement I and Statement II are true

2 Statement I is true but Statement II is false

3 Both Statement I and Statement II are false

4 Statement I is false but Statement II is true

Ans.
(3)

$HF < HCl < HBr < HI (bond length order)$

$F < Cl < Br < I (radius order)$

${PH}_{3} < {AsH}_{3} < {NH}_{3} < {SbH}_{3} < {BiH}_{3} (Boiling point order)$

$Maximum possible covalency of phosphorus is 6$

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Solution

1 Both Statement I and Statement II are true

2 Statement I is true but Statement II is false

3 Both Statement I and Statement II are false

4 Statement I is false but Statement II is true

Ans. (3) HF<HCl<HBr<HI (bond length order) F<Cl<Br<I (radius order) PH3<AsH3<NH3<SbH3<BiH3 (Boiling point order) Maximum possible covalency of phosphorus is 6

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Ans.
(3)

$HF < HCl < HBr < HI (bond length order)$

$F < Cl < Br < I (radius order)$

${PH}_{3} < {AsH}_{3} < {NH}_{3} < {SbH}_{3} < {BiH}_{3} (Boiling point order)$

$Maximum possible covalency of phosphorus is 6$