Question

e : ed3114#10001 14 01 25 Review | Constants Periodic Table Part A Predict the cation-anion distance using the values of ionic radii given in the figure. Enter your answers numerically separated by commas. VO ΑΣΦ de Na e ак Br db Submit Request Answer rson El lector de l a Polis Bennigsins Sacs 11:33 AM 11/21/2019

Answer 1

Part.A. Sol :-

(a). LiF :-

Ionic radius of Li⁺ cation = 0.68 Å

Ionic radius of F^- anion = 1.33 Å

Therefore,

Distance between cation and anion in LiF = d_Li⁺F^- =  Ionic radius of Li⁺ cation + Ionic radius of F^- anion

= 0.68 Å + 1.33 Å

= 2.09 Å

Hence, dLi+F- = 2.09 Å

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(b). NaCl :-

Ionic radius of Na⁺ cation = 1.16 Å

Ionic radius of Cl^- anion = 1.67 Å

Therefore,

Distance between cation and anion in NaCl = d_Na⁺Cl^-=  Ionic radius of Na⁺ cation + Ionic radius of Cl^- anion

= 1.16 Å + 1.67 Å

= 2.83 Å

Hence, dNa+Cl- = 2.83 Å

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(c). KBr :-

Ionic radius of K⁺ cation = 1.52 Å

Ionic radius of Br^- anion = 1.82 Å

Therefore,

Distance between cation and anion in KBr = d_K⁺Br^-=  Ionic radius of K⁺ cation + Ionic radius of Br^- anion

= 1.52 Å + 1.82 Å

= 3.34 Å

Hence, dK+Br- = 3.34 Å

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(d). RbI :-

Ionic radius of Rb⁺ cation = 1.66 Å

Ionic radius of I^- anion = 2.06 Å

Therefore,

Distance between cation and anion in RbI= d_Rb⁺I^-=  Ionic radius of Rb⁺ cation + Ionic radius of I^- anion

= 1.66 Å + 2.06 Å

= 3.72 Å

Hence, dRb+I- = 3.72 Å

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