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The hybridisation in $I{F_7}$ molecule is :A. $s{p^3}$B. $s{p^3}{d^2}$C. $s{p^3}d$D. $s{p^3}{d^3}$

Last updated date: 10th Sep 2024
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> In $I{F_7}$ molecule ,central metal atom is $I$. The electronic configuration of iodine atom is given below ;
${I_{53}} = 1{s^2},2{s^2},2{p^6},3{s^2},3{p^6},4{s^2},3{d^{10}},4{p^6},5{s^2},4{d^{10}},5{p^5}$ so here is seven atomic orbital that is one s orbitals, three p orbitals and three d orbitals which hybridizes to form seven hybrid orbitals. The geometry of $I{F_7}$ is pentagonal bipyramidal. Seven hybrid $s{p^3}{d^3}$ orbitals are directed towards the corners of a pentagonal bipyramid. Five of the hybrid orbitals are directed towards the corners of a regular pentagon whereas the remaining two are directed above and below the plane. So here option D is correct.
Note : We know that the number of valence electron in central metal atom Iodine is seven and in fluorine it is also seven so total number of valence electron in molecule is $7 + 7 \times 7 = 56$.Hence it need seven atomic orbital so possess $s{p^3}{d^3}$ hybridization. The bond angle is ${72^0}$ and ${90^0}$.