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`

Explanation

(i)Structure of `IC l_4^-`
No. of electrons in the valence shell of the central I atom =7
No. of electrons provided by four Cl atoms = 4 x 1 =4
Charge on the central atom =1
`therefore` Total no. of electrons around the central atom = 7+4+1=12
Total no. of electron pairs around the central atom = 12/2 =6
But the no. of bond pairs = 4 (`because` there are four I-Cl bonds)
`therefore` No of lone pairs =6-4=2
Thus, I in `IC l_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+1 x 4 =12) . Like `IC l_4^-` it also has 4 bond pairs and 2 lone pairs . Therefore , like , `XeF_4` is also square planar.
(ii)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 = 2 x 1 =2
Charge on the central I atom is =1
`therefore` Total no. of electrons around the central I atom = 7+ 2 +1 =1
But the no. of bonds pairs = 2 (`because` there are two I-Br bonds)
`therefore` No. of lone pairs = 5-2=3
Thus, I in `IBr_2^-` has two bonds pairs and three 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+1 x 2 =10) . Like `IBr^-` , it also has 2 bond pairs and 3 lone pairs.
(iii)Structure of `BrO_3^-`
In `BrO_3^-` , since O is more electronegative than Br, therefore , -ve charge stays on the O atom.
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+2 x 0 +1 x 1 =8
`therefore` No. of electron pairs around Br atom = 8/2=4
But total number of bond pairs = 2 x 1 (Br=0) + 1 x 1 (Br - `O^-`) =3 and lone pairs = 4-3=1
Thus, `BrO_3^-` has 3 bond pairs and one lone pair. Therefore, according to VSEPR theory, it should be pyramidal.
Now a noble gas compound having 8 electrons in the valence shell of the central atom is `XeO_3` (8 x 1 + 3 x 0 =8). Like `BrO_3^-` it also has 3 bond pairs and one lone pair. Therefore , like `BrO_3^(-), XeO_3` is also pyramidal.