You visited us 1 times! Enjoying our articles? Unlock Full Access!

Solid Angle

The arrangeme...

Question

The arrangement of $X^{-}$ ions around $A^{+}$ ion in solid $A X$ is given in the figure (not drawn to scale). If the radius of $X^{-}$ is $250$ pm, the radius (in pm) of $A^{+}$ will be:

104

Right on! Give the BNAT exam to get a 100% scholarship for BYJUS courses

125

No worries! We‘ve got your back. Try BYJU‘S free classes today!

183

No worries! We‘ve got your back. Try BYJU‘S free classes today!

57

No worries! We‘ve got your back. Try BYJU‘S free classes today!

Open in App

Solution

The correct option is A $104$
According to the given figure, $A^{+}$ is present in the octahedral void of $X^{-}$ . The limiting radius in an octahedral void is related to the radius of the sphere as:
$r_{v o i d} = 0.414 r_{s p h e r e}$
$r_{A^{+}} = 0.414 r_{X^{-}} = 0.414 \times 250 = 103.5$ $\approx 104$ pm
Hence, option A is correct.

Suggest Corrections

Similar questions

Q. The arrangement of

X^{-}

ions around

A^{+}

ion in solid

A X

is given in the figure (not drawn to scale). If the radius of

X^{-}

is 250 pm, the radius of

A^{+}

is:

The arrangement of $X^{-}$ ions around $A^{+}$ ion in solid AX is given in the figure (not drawn to scale). If the radius of $X^{-}$ is 250 pm, the ideal radius of $A^{+}$ is

Q. The arrangement of

X^{-}

ions around

A^{+}

ions in solid

A X

is given in the figure (not drawn to scale).
If the radius of

X^{-}

250 pm

, the radius of

A^{+}

is:

The arrangement of

X^{-}

ions around

A^{+}

ions in solid

A X

is given in the figure (not drawn to scale ). If the radius of

X^{-}

250 pm

, then the maxium possible radius of

A^{+}

is:

The arrangement of $X^{-}$ ions around $A^{+}$ ion in soild $A X$ is given in the figure(not drawn to scale). If the radius of $X^{-}$ is 250 pm, the radius of $A^{+}$ is:

Join BYJU'S Learning Program

The arrangement of X−ions around A+ ion in solid AX is given in the figure (not drawn to scale). If the radius of X− is 250 pm, the radius (in pm) of A+ will be:

The correct option is A 104According to the given figure, A+is present in the octahedral void of X−. The limiting radius in an octahedral void is related to the radius of the sphere as:rvoid=0.414rsphererA+=0.414rX−=0.414×250=103.5≈104 pmHence, option A is correct.

The arrangement of $X^{-}$ ions around $A^{+}$ ion in solid $A X$ is given in the figure (not drawn to scale). If the radius of $X^{-}$ is $250$ pm, the radius (in pm) of $A^{+}$ will be: