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Among the following which one is a wrong statement?
A
$SeF_4$ and $CH_4$ have same shape
B
$I_3^+$ has bent geometry
C
$PH_5$ and $BiCl_5$ do not exist
D
$p\pi$-$d\pi$ bonds are present in $SO_2$
Explanation
$SeF_4$ is see-saw; $CH_4$ is tetrahedral.
Detailed Solution
$SeF_4$: $sp^3d$, lp = 1, shape = see-saw; $CH_4$: $sp^3$, lp = 0, shape = tetrahedral. So they do not have the same shape.
$I_3^+$: $sp^3$, lp = 2, shape = bent/angular
$PH_5$ does not exist (d-orbital contraction absent); $BiCl_5$ does not exist due to inert pair effect ($Bi^{5+}$ acts as oxidising agent, $Cl^-$ as reducing agent)
$SO_2$ (O=S=O) has both $p\pi$–$d\pi$ and $p\pi$–$p\pi$ bonds
$I_3^+$: $sp^3$, lp = 2, shape = bent/angular
$PH_5$ does not exist (d-orbital contraction absent); $BiCl_5$ does not exist due to inert pair effect ($Bi^{5+}$ acts as oxidising agent, $Cl^-$ as reducing agent)
$SO_2$ (O=S=O) has both $p\pi$–$d\pi$ and $p\pi$–$p\pi$ bonds