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Question

The dipole moment of NaCl molecule is 8.5 D and its bond length is 2 oA. The percentage ionic character in this molecule will be:
  1. 50.25%
  2. 60.30%
  3. 88.50%
  4. 100%


Solution

The correct option is C 88.50%
μtheo=e×d=(4.8×1010)×(2×108)esu cm       =9.6×1018esu×cm       =9.6 D% Ionic character =μexpμtheo×100                                =8.59.6×100                                =88.5%

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