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the trend for bond dissociation energy of halogens is Cl2>Br2>F2>I2... But as we move down in group strength of intermolecular forces increase( in case of halogens LDF) because of increasing polarizability so there should be more energy needed to break I-I bond(in addition to strong LDF it is in solid state also) but it doesn’t happens actually bcz in given trend Cl-Cl bond has greatest dissociation energy.....now plzzzz explain this all why not this trend I2>Br2>Cl2>F2?????

the trend for bond dissociation energy of halogens is Cl2>Br2>F2>I2... But as we move down in group strength of intermolecular forces increase( in case of halogens LDF) because of increasing polarizability so there should be more energy needed to break I-I bond(in addition to strong LDF it is in solid state also) but it doesn’t happens actually bcz in given trend Cl-Cl bond has greatest dissociation energy.....now plzzzz explain this all why not this trend I2>Br2>Cl2>F2?????

Grade:12th pass

2 Answers

Arun
25750 Points
5 years ago

The Bond dissociation energy of halogens decreases down the group as the size of the halogen is increases .

The bond dissociation energy of fluorine is however, lower than chlorine and bromine due to its small size its inter electronic repulsion is very high .

That's why the the bond dissociation enthalpy follows the order Cl2 >Br2>F2>I2.

Dibya jyoti Newar
15 Points
3 years ago
𝘾𝙄-𝘾𝙄>𝘽𝙧-𝘽𝙧>𝙁-𝙁>𝙡-𝙡
𝙍𝙚𝙖𝙨𝙤𝙣:
•𝙄𝙩 𝙙𝙪𝙚 𝙩𝙤, 𝙩𝙝𝙚 𝙗𝙤𝙣𝙙 𝙚𝙣𝙚𝙧𝙜𝙮 𝙤𝙛 𝙝𝙖𝙡𝙤𝙜𝙚𝙣𝙨 𝙙𝙚𝙘𝙧𝙚𝙖𝙨𝙚 𝙜𝙤𝙞𝙣𝙜 𝙙𝙤𝙬𝙣 𝙩𝙝𝙚 𝙜𝙧𝙤𝙪𝙥 𝙙𝙪𝙚 𝙩𝙤 𝙞𝙣𝙘𝙧𝙚𝙖𝙨𝙚 𝙖𝙩𝙤𝙢𝙞𝙘 𝙨𝙞𝙯𝙚 𝙤𝙛 𝙚𝙡𝙚𝙢𝙚𝙣𝙩. 
• 𝘽𝙤𝙣𝙙 𝙙𝙞𝙨𝙨𝙤𝙘𝙞𝙖𝙩𝙞𝙤𝙣 𝙤𝙛 𝙛𝙡𝙤𝙪𝙧𝙞𝙣𝙚 𝙞𝙨 𝙡𝙤𝙬𝙚𝙧 𝙩𝙝𝙖𝙣 𝙘𝙝𝙡𝙤𝙧𝙞𝙣𝙚 𝙖𝙣𝙙 𝙖𝙡𝙨𝙤 𝙗𝙧𝙤𝙢𝙞𝙣𝙚 𝙙𝙪𝙚 𝙩𝙤 𝙨𝙢𝙖𝙡𝙡 𝙨𝙞𝙯𝙚 𝙞𝙩'𝙨 𝙞𝙣𝙩𝙚𝙧 𝙚𝙡𝙚𝙘𝙩𝙧𝙤𝙣𝙞𝙘 𝙧𝙚𝙥𝙪𝙡𝙨𝙞𝙤𝙣 𝙞𝙨 𝙫𝙚𝙧𝙮 𝙝𝙞𝙜𝙝. 
•𝙎𝙤 𝙩𝙝𝙖𝙩 𝙗𝙤𝙣𝙙 𝙙𝙞𝙨𝙨𝙤𝙘𝙞𝙖𝙩𝙞𝙤𝙣 𝙚𝙣𝙩𝙝𝙖𝙡𝙥𝙮 𝙞𝙣 𝙜𝙞𝙫𝙚𝙣 𝙗𝙚𝙡𝙤𝙬 𝙩𝙝𝙞𝙨 𝙤𝙧𝙙𝙚𝙧
𝘾𝙡2>𝘽𝙧2>𝙁2>𝙡2>

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