Among [Co(CN)_4]^{4-}, [Co(CO)_3(NO)], XeF_4, | vedclass

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(D) $1$. $[Co(CN)_4]^{4-}$: $Co^{0}$ $(d^9 s^2)$ in $CN^-$ field is tetrahedral.$2$. $[Co(CO)_3(NO)]$: $Co$ is $d^9 s^2$,$CO$ and $NO$ are strong fiel

Vedclass · Accepted Answer

$5$

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(D) $1$. $[Co(CN)_4]^{4-}$: $Co^{0}$ $(d^9 s^2)$ in $CN^-$ field is tetrahedral.$2$. $[Co(CO)_3(NO)]$: $Co$ is $d^9 s^2$,$CO$ and $NO$ are strong field ligands,geometry is tetrahedral.$3$. $XeF_4$: $sp^3d^2$ hybridization,square planar geometry.$4$. $[PCl_4]^{+}$: $sp^3$ hybridization,tetrahedral geometry.$5$. $[PdCl_4]^{2-}$: $Pd^{2+}$ $(d^8)$ is square planar.$6$. $[ICl_4]^{-}$: $sp^3d^2$ hybridization,square planar geometry.$7$. $[Cu(CN)_4]^{3-}$: $Cu^{+}$ $(d^{10})$ is tetrahedral.$8$. $P_4$: $P$ atoms are $sp^3$ hybridized,tetrahedral structure.Thus,the species with tetrahedral geometry are $[Co(CN)_4]^{4-}, [Co(CO)_3(NO)], [PCl_4]^{+}, [Cu(CN)_4]^{3-}$,and $P_4$. The total count is $5$.

Among $[Co(CN)_4]^{4-}, [Co(CO)_3(NO)], XeF_4, [PCl_4]^{+}, [PdCl_4]^{2-}, [ICl_4]^{-}, [Cu(CN)_4]^{3-}$ and $P_4$,the total number of species with tetrahedral geometry is:

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