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ADNAN MUHAMMED

Grade 12,

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​Q. ARRANGE THE FOLLOWING IN INCREASING ORDER OF PROPERTIES AS INDICATED IN BRACKET AGAINST- N 2 , O 2 , F 2 , Cl 2 ( BOND ENERGY) PH 3 , AsH 3 , SbH 3 , NH 3 ( BOND ANGLE)

​Q. ARRANGE THE FOLLOWING IN INCREASING ORDER OF PROPERTIES AS INDICATED IN BRACKET AGAINST-
 
  1. N2  , O2 , F2 , Cl2 ( BOND ENERGY)
  2. PH3 , AsH3 , SbH3 , NH3 ( BOND ANGLE)

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1 Answers

Sunil Kumar FP
askIITians Faculty 183 Points
9 years ago
(1)INCREASING ORDER OF BOND ENERGY
N2>O2>Cl2>F2
N2 has triple bond>O2(Double bond)Between F2 and Cl2
as we go down the group, the radius becomes bigger, and the bonding electrons is further away from the nuclei, causing the covalent bonds to get weaker and weaker down the group. however, flourine is a special case. due to the small size of the atoms, the nuclei and bonding electrons are so close that the experience serious repulsion. weakening the covalent bonds.
(2)INCREASING ORDER OF BOMD ANGLE
Bond angle: NH3 (107.8°) > PH3 (99.5°) > AsH3 (91.8°) ≈ SbH3
(91.3°) > BiH3 (90°)
Electronegativity of N is highest. Therefore, the lone pairs will be
towards nitrogen and hence more repulsion between bond pairs.
Therefore bond angleof NH3 is the highest.

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