The number of lone pair and bond pair electrons present in central iodine atom of 90.71kJ ion are respectively

Question

The number of lone pair and bond pair electrons present in central iodine atom of 90.71kJ ion are respectively (A)  122.55kJ  (B)  Hg2++2Cl-HgCl2,  (C)  k=1.65× 103  (D)  Hg2+

Tardigrade Admin · Accepted Answer

The structure of three lone pairs and two bond pairs is trigonal bipyramidal and the geometry of 25oC is linear as shown in the figure

Q. The number of lone pair and bond pair electrons present in central iodine atom of $90.71 k J$ ion are respectively

$122.55 k J$

$H g^{2 +} + 2 C l^{-} H g C l_{2},$

$k = 1.65 \times 10^{3}$

$H g^{2 +}$

The structure of three lone pairs and two bond pairs is trigonal bipyramidal and the geometry of $25^{o} C$ is linear as shown in the figure

Q. The number of lone pair and bond pair electrons present in central iodine atom of 90.71kJ ion are respectively

122.55kJ

Hg2++2Cl−HgCl2​,

k=1.65×103

Hg2+