The ionic radii ( mathringA) of C4- and O2- respectively are 2.60 and 1.40. The ionic radius of the isoelectronic ion N3- would be

Question

The ionic radii ( mathringA) of C4- and O2- respectively are 2.60 and 1.40. The ionic radius of the isoelectronic ion N3- would be (A) 2.6 (B) 1.71 (C) 1.4 (D) 0.95

Tardigrade Admin · Accepted Answer

The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. So, decreasing order of ionic radii is C(4-) gt N(3-) gt O(2-)

Q. The ionic radii ( $\overset{˚}{A}$ ) of $C^{4 -}$ and $O^{2 -}$ respectively are 2.60 and 1.40. The ionic radius of the isoelectronic ion $N^{3 -}$ would be

2.6

1.71

1.4

0.95

The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge.
So, decreasing order of ionic radii is $C^{(4 -)} g t N^{(3 -)} g t O^{(2 -)}$

Q. The ionic radii (A˚) of C4− and O2− respectively are 2.60 and 1.40. The ionic radius of the isoelectronic ion N3− would be

2.6

1.71

1.4

0.95

The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge. So, decreasing order of ionic radii is C(4−)gtN(3−)gtO(2−)

Q. The ionic radii ( $\overset{˚}{A}$ ) of $C^{4 -}$ and $O^{2 -}$ respectively are 2.60 and 1.40. The ionic radius of the isoelectronic ion $N^{3 -}$ would be

The ionic radii of isoelectronic ions decrease with the increase in the magnitude of the nuclear charge.
So, decreasing order of ionic radii is $C^{(4 -)} g t N^{(3 -)} g t O^{(2 -)}$