Home
Class 12
CHEMISTRY
The radius of Na(+) is 95 pm and that of...

The radius of `Na(+)` is 95 pm and that of `CI^(-)`ions is 181 pm. Predict whethr the coordination number of `Na^(+)` is 6 or4.

Text Solution

Verified by Experts

Radius of` Na^(+)`= 95 pm
Radius of` CI_(-)`=181 pm
Radius ratio, `r_(+)/r=(r(Na^(+)))/(r(CI^(-)))=95/181=0.524`
The radius ratio lies between 0.414-0.732. Hence, `Na^(+)` ions prefer to occupy octahefral holeshaving coordination number 6.
Promotional Banner

Topper's Solved these Questions

  • XII BOARDS

    XII BOARDS PREVIOUS YEAR|Exercise Outside Delhi : SET-II|10 Videos

  • XII BOARDS

    XII BOARDS PREVIOUS YEAR|Exercise Outside Delhi : SET-III|13 Videos

  • XII BOARDS

    XII BOARDS PREVIOUS YEAR|Exercise [SET-II]|10 Videos

  • SAMPLE PAPER 2019

    XII BOARDS PREVIOUS YEAR|Exercise SECTION: D|1 Videos

Similar Questions

Explore conceptually related problems

The radius of the Na^(2+) is 95p m and that of Cl^(-) ion is 181 p m . Predict the co - ordination number of Na^(+) :

The radius of the Na^(+) is 95 pm and that of CI ion is 181 pm Predict the coordination number of Na^(+) ?

If the radius of Na^(+) ion is 95 p m and that of Cl^(-) ion is 181 p m , then :

Coordinating number of Na^(+) in NaCl is

The radius of Ag^(+) ion is 126 pm and that of I^(-) ion is 216 pm. The coordination number of Ag^(+) ion is:

The radius of cation is 100 pm and that of anion is 188 pm, then the coordination number of cation is __________.