Two ions $A^{+}$ and $B^{-}$ have radii 88pm and 200pm, respectively. In the close-packed crystal of the compound AB, predict the coordination number of $A^{+}$

Q. Two ions

A^{+}

and

B^{-}

have radii 88 and 200 pm respectively. In the closed packed crystal of compound predict the co-ordination number of

A^{+}

Q. In

A^{+} B^{-}

ionic compound, radii of

A^{+}

and

B^{-}

ions are

180 p m

and

187 p m

respectively. The crystal structure of this compound will be:

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In a compound AB, the ionic radii of A+ and B− are 88 pm and 200 pm respectively the coordination number of A+ is:

The correct option is A 6Let consider the problem.r+r−=88200=0.44Therefore it lies in the range of 0.414−0.732Hence the coordination number of A+=6.

In a compound $A B$ , the ionic radii of $A^{+}$ and $B^{-}$ are $88$ pm and $200$ pm respectively the coordination number of $A^{+}$ is:

The correct option is A $6$
Let consider the problem.
$\frac{r^{+}}{r^{-}} = \frac{88}{200} = 0.44$
Therefore it lies in the range of $0.414 - 0.732$
Hence the coordination number of $A^{+} = 6.$