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(D) To determine the number of square planar species,we analyze the geometry of each:$1$. $XeF_4$: $sp^3d^2$ hybridization with $2$ lone pairs,resulti

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$4$

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(D) To determine the number of square planar species,we analyze the geometry of each:$1$. $XeF_4$: $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar shape.$2$. $SF_4$: $sp^3d$ hybridization with $1$ lone pair,resulting in a see-saw shape.$3$. $SiF_4$: $sp^3$ hybridization,resulting in a tetrahedral shape.$4$. $BF_4^-$: $sp^3$ hybridization,resulting in a tetrahedral shape.$5$. $BrF_4^-$: $sp^3d^2$ hybridization with $2$ lone pairs,resulting in a square planar shape.$6$. $[Cu(NH_3)_4]^{2+}$: $dsp^2$ hybridization,resulting in a square planar shape.$7$. $[FeCl_4]^{2-}$: $sp^3$ hybridization,resulting in a tetrahedral shape.$8$. $[CoCl_4]^{2-}$: $sp^3$ hybridization,resulting in a tetrahedral shape.$9$. $[PtCl_4]^{2-}$: $dsp^2$ hybridization,resulting in a square planar shape.The species with a square planar shape are $XeF_4, BrF_4^-, [Cu(NH_3)_4]^{2+},$ and $[PtCl_4]^{2-}$.Therefore,the total number of square planar species is $4$.

$A$. $[CoF_6]^{3-}$	$M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$	$N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$	$O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$	$P$. $d^2sp^3$,diamagnetic

$A$ list of species having the formula $XZ_4$ is given below.
$XeF_4, SF_4, SiF_4, BF_4^-, BrF_4^-, [Cu(NH_3)_4]^{2+}, [FeCl_4]^{2-}, [CoCl_4]^{2-}$ and $[PtCl_4]^{2-}$.
Defining shape on the basis of the location of $X$ and $Z$ atoms,the total number of species having a square planar shape is

Similar Questions

Chromium hexacarbonyl,$Cr(CO)_6$,is an octahedral compound involving which type of hybridization?

Which of the following statements are true about $[NiCl_4]^{2-}$?
$(a)$ The complex has tetrahedral geometry
$(b)$ Coordination number of $Ni$ is $2$ and oxidation state is $+4$
$(c)$ The complex is $sp^3$ hybridised
$(d)$ It is a high spin complex
$(e)$ The complex is paramagnetic

In the $Cu$-ammonia complex,the state of hybridization of $Cu^{2+}$ is:

Match the correct option.
$A$. $[CoF_6]^{3-}$ $M$. $sp^3d^2$,paramagnetic
$B$. $[Mn(CN)_6]^{3-}$ $N$. $sp^3d^2$,diamagnetic
$C$. $[Co(C_2O_4)_3]^{3-}$ $O$. $d^2sp^3$,paramagnetic
$D$. $[Zn(H_2O)_6]^{2+}$ $P$. $d^2sp^3$,diamagnetic

From the magnetic behaviour of $[NiCl_4]^{2-}$ (paramagnetic) and $[Ni(CO)_4]$ (diamagnetic),choose the correct geometry and oxidation state.

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$A$ list of species having the formula $XZ_4$ is given below.$XeF_4, SF_4, SiF_4, BF_4^-, BrF_4^-, [Cu(NH_3)_4]^{2+}, [FeCl_4]^{2-}, [CoCl_4]^{2-}$ and $[PtCl_4]^{2-}$.Defining shape on the basis of the location of $X$ and $Z$ atoms,the total number of species having a square planar shape is