Question

Structure of ${{\mathrm{ICl}}_2}^{-}$

• A. Trigonal
• B. Octahedral
• C. Square planar
• D. Distorted trigonal pyramidal

Answer 1

The correct option is D

Distorted trigonal pyramidal

Explanation for correct answer:-

Option (D) Distorted trigonal pyramidal:

Structure of ${{\mathbf{ICl}}_{\mathbf{2}}}^{\mathbf{-}}$:-

• Valence Shell Electron Pair Repulsion (VSEPR) theory is a theory that is used in order to predict the molecular shape of any compound, with the help of the electron pairs that are present around the central atom of the compound.
• The total number of bond pairs and the lone pair can be calculated with the help of the formula:$\frac{1}{2}[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}]$
• Here, V= No. of valance electron on the central atom, M= No. of monovalent atoms directly attached to central atom, C= Charge on Cation, A= Charge on Anion.
• For ${{\mathrm{ICl}}_2}^{-}$ if we use this formula we can find out the number of lone pairs and number of bond pairs and thus predict its shape.
• Here in ${{\mathrm{ICl}}_2}^{-}$, V= 7 as Iodine has 7 valence electrons, M= 2 as there are two monovalent chlorine atoms attached to Iodine, C= 0 and A= $-1$as the charge of the compound is $-1$
• There applying the formula we get,

$$\begin{gathered} \frac{1}{2}[\mathrm{V}+\mathrm{M}-\mathrm{C}+\mathrm{A}] \\ =\frac{1}{2}[7+2-0+1)] \\ =\frac{1}{2}\times 10 \\ =5 \end{gathered}$$

• From here we can see that there are 2 bond pairs $(\mathrm{I}-\mathrm{Cl})$and 3 lone pairs, and the hybridization of the compound is ${\mathrm{sp}}^3\mathrm{d}$.
• So according to this, the shape is Trigonal bipyramidal.
• But due to the presence of 3 lone pairs of the Iodine atom that are present in the equatorial position leaving the the Cl and I in a linear molecular shape.
• Thus the shape of the compound is Distorted trigonal pyramidal.

Explanation for incorrect answer:-

Option (A) Trigonal:

• Since from the VSEPR theory we have seen that the hybridization of ${{\mathrm{ICl}}_2}^{-}$is ${\mathrm{sp}}^3\mathrm{d}$as its VSEPR number is 5.
• Thus it cannot be trigonal.

Option (B) Octahedral

• Since from the VSEPR theory we have seen that the hybridization of ${{\mathrm{ICl}}_2}^{-}$is ${\mathrm{sp}}^3\mathrm{d}$as its VSEPR number is 5.
• For octahedral, the hybridization has to be ${\mathrm{sp}}^3{\mathrm{d}}^2$.

Option (C) Square planar

• From the VSEPR theory, we have seen that the hybridization of ${{\mathrm{ICl}}_2}^{-}$is ${\mathrm{sp}}^3\mathrm{d}$as its VSEPR number is 5.
• For square planar, the hybridization has to be ${\mathrm{sp}}^3{\mathrm{d}}^2$.

Therefore, the Structure of ${{\mathrm{ICl}}_2}^{-}$is an option (D) Distorted trigonal pyramidal.