Question

Give the correct of initials T or F for following statements. Use T if statement is true and F if it is false.
(I) $Co\left(III\right)$ is stabilised in presence of weak field ligands, while $Co\left(II\right)$ is stabilised in presence of strong field ligand.
(II) Four coordinated complexes of $Pd\left(II\right)$ and $Pt\left(II\right)$ are diamagnetic and square planar.
(III) $\left[Ni\right(CN{)}_4{]}^{4-}$ ion and $\left[Ni\right(CO{)}_4]$ are diamagnetic tetrahedral and square planar respectively.
(IV) $N{i}^{2+}$ ion does not form inner orbital octahedral complexes.

• A. TFTF
• B. TTTF
• C. TTFT
• D. FTFT

Answer 1

The correct option is D FTFT

(i) Both Co(II) and Co(III) are stabilized in presence of strong field ligand as the values of CSFE are high. Hence, the statement (i) is false.
(ii) Four coordinated complexes of Pd(II) and Pt(II) are diamagnetic and square planar. This is because the valence shells of $P{d}^{2+}$ and $P{t}^{2+}$ have high values for $Z_{eff}$. They always form inner orbital complex..
Hence, the statement (ii) is true.
(iii) Both $\left[Ni\right(CN{)}_4{]}^{2-}and\left[Ni\right(CO{)}_4]$ are diamagnetic with zero unpaired electrons.
$\left[Ni\right(CN{)}_4{]}^{4-}$ involves $ds{p}^2$ hybridization and is square planar complex.
$\left[Ni\right(CO{)}_4]$ involves $s{p}^3$ hybridization and has tetrahedral geometry.
Hence, the statement (iii) is false.
(iv) $N{i}^{2+}$ ion has outer electronic configuration $3d^8,4s^0$ or $t_{2g}^6{e}_g^2$. It remains unaffected in presence of strong and weak field ligands.
Hence, it does not form inner orbital octahedral complexes.
Hence, the statement (iv) is true.