Calculate The Percentage Ionic Character Of The HCl Class ... - Vedantu

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AnswerQuestion Answers for Class 12Class 12 BiologyClass 12 ChemistryClass 12 EnglishClass 12 MathsClass 12 PhysicsClass 12 Social ScienceClass 12 Business StudiesClass 12 EconomicsQuestion Answers for Class 11Class 11 EconomicsClass 11 Computer ScienceClass 11 BiologyClass 11 ChemistryClass 11 EnglishClass 11 MathsClass 11 PhysicsClass 11 Social ScienceClass 11 AccountancyClass 11 Business StudiesQuestion Answers for Class 10Class 10 ScienceClass 10 EnglishClass 10 MathsClass 10 Social ScienceClass 10 General KnowledgeQuestion Answers for Class 9Class 9 General KnowledgeClass 9 ScienceClass 9 EnglishClass 9 MathsClass 9 Social ScienceQuestion Answers for Class 8Class 8 ScienceClass 8 EnglishClass 8 MathsClass 8 Social ScienceQuestion Answers for Class 7Class 7 ScienceClass 7 EnglishClass 7 MathsClass 7 Social ScienceQuestion Answers for Class 6Class 6 ScienceClass 6 EnglishClass 6 MathsClass 6 Social ScienceQuestion Answers for Class 5Class 5 ScienceClass 5 EnglishClass 5 MathsClass 5 Social ScienceQuestion Answers for Class 4Class 4 ScienceClass 4 EnglishClass 4 MathsCalculate the percentage ionic character of the HCl molecule. The bond length is 1.275$A{}^\circ $ and the observed dipole moment is 1.03 D.AnswerVerified526.9k+ viewsHint: For calculation of ionic character in a molecule, we need to determine the dipole moment within the molecule. The dipole moment is the physical property which is a measure of the polarity of a chemical bond between two atoms in a molecule. The higher the electronegativity difference between the two atoms, the more will be the ionic character to the bond and vice versa.Complete answer: - The dipole moment is the mathematical product of the total amount of positive charge or negative charge and the distance between the center of the charge distribution.-For calculating the percent ionic character in a molecule, we need to follow the steps below-(a) We start by calculating the dipole moment by taking the product of the magnitude of charge and separation between charges, i.e \[\mu =q\times r\] where $\mu $ is the dipole moment q is the separated charge r is the distance between them Here, q= $1.6\times {{10}^{-19}}C$ ; r= \[1.275A{}^\circ \times {{10}^{-10}}m\]So, dipole moment ($\mu $ ) $=1.6\times {{10}^{-19}}\times 1.275\times {{10}^{-10}}m$ $=2.04\times {{10}^{-29}}cm$ (b) Now, we have to convert the centimeter into the standard unit of dipole moment, i.e Debye.We know that, $1D=3.335\times {{10}^{-30}}cm$ \[=\dfrac{2.04\times {{10}^{-29}}}{(3.335}\] \[=\dfrac{2.04\times {{10}^{-29}}}{\begin{align} & 3.335\times {{10}^{-30}} \\ & =\dfrac{20.4}{3.335} \\ & =6.1169D \\ \end{align}}\] (c) Now, percentage ionic character \[=\dfrac{\exp value}{theoretical value}\times 100\] \[=\dfrac{1.03D}{6.1169D}\times 100\] $=16.83%$ Note: For calculating the dipole moment you need to remember the charge of an electron i.e,$1.6\times {{10}^{-19}}$ . For the calculation of percentage ionic character, remember the conversion units i.e,$1D=3.335\times {{10}^{-30}}cm$ .Recently Updated PagesMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Economics: Engaging Questions & Answers for SuccessMaster Class 11 English: Engaging Questions & Answers for SuccessMaster Class 11 Maths: Engaging Questions & Answers for SuccessMaster Class 11 Biology: Engaging Questions & Answers for SuccessMaster Class 11 Computer Science: Engaging Questions & Answers for SuccessMaster Class 11 Business Studies: Engaging Questions & Answers for SuccessMaster Class 11 Economics: Engaging Questions & Answers for SuccessMaster Class 11 English: Engaging Questions & Answers for SuccessMaster Class 11 Maths: Engaging Questions & Answers for SuccessMaster Class 11 Biology: Engaging Questions & Answers for Success

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