Give the formula and describe the structure of a noble gas which is isostructural with
(i). `Icl_4^(ɵ)`
(ii). `I Br_2^(ɵ)`
(iii). `BrO_3`

Answer

(i) `(##RES_INO_CHM_XII_V02_C03_E01_018_A01##)`
Structure of `ICI_(4)^(-)`
No. of electrons in the valence shell of the central `I` atom =`7`.
No. of electrons provided by four `CI` atoms =`4xx1=4`
Charge on the central atom =`1`
`therefore` Total no. of electrons around of the central atom `=7+4+1=12`
Total no. of electrons pairs around the central atom `=12//2=6`
But the no. of bond pairs =`4`
(`because` there are four `I-CI` bonds)
`therefore` No. of lone pairs =`6-4=2`
Thus, `I` in `ICl_(4)^(-)` has `4` bond pairs and `2` lone pairs. Therefore, according to `VSEPR` theory, it should be square planar.
Now a noble gas compound having `12` electrons in the valence shell of the central atom is `XeF_(4)(8+1xx4=12)`.Like `ICl_(4)^(-)`, it also has `4` bond pairs and `2` lone pairs. Therefore, like `ICl_(4)^(-), XeF_(4)` is also square planar.
Structure of `IBr_(2)^(-)`.
No. of electrons in the valence shell of the central `I` atom =`7`.
No. of electrons provided by two `Br` atoms =`2xx1=2`
Charge on the central `I` atom =`1`
`therefore` Total no. of electrons around of the central `I` atom `=7+2+1=10`
But the no. of bond pairs =`2` (`because` there are two `I-Br` bonds)
`therefore` No. of lone pairs =`5-2=3`
Thus, `I` in `IBr_(2)^(-)` has two bond pairs and `2` lone pairs. Therefore, according to `VSEPR` theory, it should be linear .
Now a noble gas compound having `10` electrons in the valence shell of the central atom is `XeF_(2)(8+1xx2=10)`.Like `IBr^(-)`, it also has `2` bond pairs and `3` lone pairs.
`(##RES_INO_CHM_XII_V02_C03_E01_018_A02##)`
(iii)Structure of `BrO_(3)^(-)`.
In `BrO_(3)^(-)`, since `O` is more electronegative than `Br`, therefore, `-ve` charge stays on the `O` atom.
`(##RES_INO_CHM_XII_V02_C03_E01_018_A03##)`
Therefore, in `BrO_(3)^(-)`, there are two `Br=O` bonds and one bond `Br=O^(-)` bond.
Now according to `VSEPR` theory, double bonds do not contribute any electron while single bonds contribute one electron towards the total number of the central atom.However, both double and single bonds contribute one bond pair. Thus, total number of electrons is the valence shell of the central `Br` atom =`7+2xx0+1xx1=8`
`therefore` No. of electron pairs around `Br` atom =`8//2=4`
But total number of bond pairs = `2xx1 (Br=0) +1 xx 1(Br-O^(-))=3` and lone pairs =`4-3=1`
Thus, `BrO_(3)^(-)` has `3` bond pairs and one lone pair. Therefore, according to `VSERP` theory, it should be pyramidal.Now a noble gas compound having `8` electrons in the valence shell of the central atom is `XeO_(3)(8xx1+3xx0=8)`. Like `BrO_(3)^(-)`, it also has `3` bond pairs and one lone pair. Therefore, like `BrO_(3)^(-), XeO_(3)` is also pyramidal.