Question

Which of the following statements is not true?

• A. $\left[Mn\right(CN{)}_6{]}^{2–}$ ion has octahedral geometry and is paramagnetic
• B. $[CuC{l}_4{]}^{2–}$ has square planar geometry and is paramagnetic
• C. $\left[NiB{r}_2\right(P{h}_{3}P{)}_2]$ has sqare planar geometry and has no unpaired electrons
• D. $MnC{l}_4^{2–}$ ion has tetrahedral geometry and is paramagnetic

Answer 1

The correct option is B $[CuC{l}_4{]}^{2–}$ has square planar geometry and is paramagnetic

(A) $MnC{l}_4^{2–}$
Here $Mn$ is in $+2$ oxidation state
$M{n}^{+2}$ electronic configurations: $3d^{5}4s^0$ and $Cl$ is a weak field ligand.
So, $MnC{l}_4^{2–},\text{is}s{p}^3$ hybridized and paramagnetic in nature due to the presence unpaired electrons.
Thus $MnC{l}_4^{2-}$ has tetrahedral geometry.

(B)$\left[Mn\right(CN{)}_6{]}^{2–}$
Here $Mn$ is in $+4$ oxidation state
$M{n}^{+2}$ electronic configurations: $3d^{3}4s^0$ and $C{N}^{-}$ is a strong field ligand.
So, $\left[Mn\right(CN{)}_6{]}^{2–},\text{is}{d}^{2}s{p}^3$ hybridized and paramagnetic in nature.
Thus $\left[Mn\right(CN{)}_6{]}^{2–}$ has octahedral geometry.

(C)$[CuC{l}_4{]}^{2-}$,
Here $Cu$ is in $+2$ oxidation state
$C{u}^{2+}$ electronic configuration is :$3d^{9}4s^0$
Since $C{l}^{-}$ is week field ligand thus $[CuC{l}_4{]}^{2-}\text{is}s{p}^3$ hybridised and paramagnetic in nature.
(D)$\left[NiB{r}_2\right(P{h}_{3}P{)}_3]$
Here $Ni$ in $+2$ oxidation state and is said to be diamagnetic which indicates the complex is square planar in nature.