[Ni(PPh_3)_2Cl_2] is a paramagnetic complex. | vedclass

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(B) $[Ni(PPh_3)_2Cl_2]$ is a $d^8$ complex. Since it is paramagnetic, it must have a tetrahedral geometry.$(A)$ Tetrahedral complexes of the type $[MA

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$A, B$ and $D$ only

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(B) $[Ni(PPh_3)_2Cl_2]$ is a $d^8$ complex. Since it is paramagnetic, it must have a tetrahedral geometry.$(A)$ Tetrahedral complexes of the type $[MA_2B_2]$ do not show geometrical isomerism. Thus, statement $A$ is incorrect.$(B)$ The complex is blue in colour, not white. Thus, statement $B$ is incorrect.$(C)$ For $Ni^{2+}$ $(d^8)$ in a tetrahedral field, the configuration is $e^4 t_2^4$, which has $2$ unpaired electrons. The spin-only magnetic moment is $\mu = \sqrt{n(n+2)} = \sqrt{2(2+2)} = \sqrt{8} \approx 2.83 \ BM$. Thus, statement $C$ is correct.$(D)$ The $CFSE$ for a tetrahedral complex is calculated as $CFSE = (-0.6 	imes n_e + 0.4 	imes n_{t_2}) \Delta_t$. For $d^8$, $CFSE = (-0.6 	imes 4 + 0.4 	imes 4) \Delta_t = -0.8 \Delta_t$. The statement mentions $-0.8 \Delta_0$, which is incorrect.$(E)$ $Ni(CO)_4$ is tetrahedral, and $[Ni(PPh_3)_2Cl_2]$ is also tetrahedral. Thus, statement $E$ is correct.The incorrect statements are $A, B,$ and $D$.

Coordination number and type of hybridisation	Distribution of hybrid orbitals in space
$(a)$. $4, sp^3$	$(i)$. trigonal bipyramidal
$(b)$. $3, sp^2$	$(ii)$. octahedral
$(c)$. $5, sp^3 d$	$(iii)$. tetrahedral
$(d)$. $6, sp^3 d^2$	$(iv)$. trigonal planar

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