NEET Practice Questions (MCQs) with Answers & Solutions

(a) $$\begin{gathered} \mathrm{Mn}(\mathrm{CO}{)}_{\mathrm{s}}+\stackrel{-}{\mathrm{e} }\to {\left[\mathrm{Mn}(\mathrm{CO}{)}_{\mathrm{s}}\right]}^{-} \\ \mathrm{less} \mathrm{stable} \mathrm{more} \mathrm{stable},\mathrm{as} \mathrm{EAN} \mathrm{of} \mathrm{Mn}=36(\mathrm{Kr}\left) \\ \right(\mathrm{O}.\mathrm{A}.) \end{gathered}$$

(d) In complexes ${\left[\mathrm{Rh}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+} \mathrm{and} {\left[\mathrm{Co}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}$, central metal cations have same oxidation state as well as same ligands and they fall in same group, but ${∆}_0$ of ${\left[\mathrm{Rh}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}>{∆}_0 \mathrm{of} {\left[\mathrm{Co}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}$ because ${\mathrm{Rh}}^{3+}$ has high ${\mathrm{Z}}_{\mathrm{eff}}$ value than ${\mathrm{CO}}^{3+}$.

(c) Number of unpaired electrons =3

${\mathrm{\mu }}_{\mathrm{eff}}=3.9 \mathrm{BM}$

type of hybridisation = ${\mathrm{sp}}^3{\mathrm{d}}^2$

(b) Increasing stability order :

  (a) ${\left[\mathrm{Cu}({\mathrm{NH}}_3{)}_4\right]}^{2+}<{\left[\mathrm{Cu}(\mathrm{en}{)}_2\right]}^{2+}<{\left[\mathrm{Cu}\left(\mathrm{trien}\right)\right]}_{}^{2+}$

 Their formation entropy increases in the same order, because denticity of ligand increases 

(b) ${\left[\mathrm{Fe}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}<{\left[\mathrm{Fe}({\mathrm{NO}}_2{)}_6\right]}^{3-}<{\left[\mathrm{Fe}({\mathrm{NH}}_3{)}_6\right]}^{3+}$

${\mathrm{NO}}_2^{-}$ is stronger ligand than ${\mathrm{NH}}_3$

(c) ${\left[\mathrm{Co}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}<{\left[\mathrm{Rh}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}<{\left[\mathrm{Ir}({\mathrm{H}}_2\mathrm{O}{)}_6\right]}^{3+}$

${\mathrm{Z}}_{\mathrm{eff}}$ value increases from ${\mathrm{Co}}^{3+ }\mathrm{to} {\mathrm{Ir}}^{3+}$

(d) ${\left[\mathrm{Cr}({\mathrm{NH}}_3{)}_6\right]}^{1+}<{\left[\mathrm{Cr}({\mathrm{NH}}_3{)}_6\right]}^{2+}<{\left[\mathrm{Cr}({\mathrm{NH}}_3{)}_6\right]}^{3+}$

oxidation state of Cr atom increases from +1 to +3

$\left[{\mathrm{PtCl}}_2\right({\mathrm{NH}}_3\left)\right(\mathrm{OH}{)}_2]$ ; Pt-oxidation state +2;

Valence configuration = $5{\mathrm{d}}^{8^{}}$

Complex is diamagnetic and exhibits geometrical isomerism.

(c) Due to cationic part of the complex, it is green , because there are three unpaired electrons in ${\mathrm{t}}_{2\mathrm{g}}$ orbitals of $C{r}^{\mathit{3}\mathit{+}}$ cation, both cationic and anionic parts are complex ions , hence, coordination isomerism will be exhibited by it.

(b) $\left[\mathrm{Ti}\right({\mathrm{H}}_2\mathrm{O}{)}_6]{\mathrm{Cl}}_4$

Coordination number$\Rightarrow$Octahedral complex

Ti is in +4 oxidation state $\Rightarrow$no. of unpaired electrons

$\Rightarrow$magnetic moment = 0 B.M.

(d) $\left[\mathrm{CoBr}\right({\mathrm{NH}}_3{)}_5]{\mathrm{SO}}_4 \rightleftharpoons [\mathrm{CoBr}({\mathrm{NH}}_3{)}_5{]}^{2+}+{\mathrm{SO}}_4^{2-}$ ;        Total charges = 4 

$\left[\mathrm{Co}\right({\mathrm{SO}}_4\left)\right({\mathrm{NH}}_3{)}_5]\mathrm{Br} \rightleftharpoons [\mathrm{Co}\left({\mathrm{SO}}_4\right)({\mathrm{NH}}_3{)}_5{]}^{+} + {\mathrm{Br}}^{-}$ ;      Total Charges = 2

hence, both have different molar conductivities.

(d) Complexes $\left[\mathrm{Co}\right({\mathrm{SO}}_4\left)\right({\mathrm{NH}}_3{)}_5]\mathrm{Br} \mathrm{and} [\mathrm{Co}\left({\mathrm{SO}}_4\right)\left({\mathrm{NH}}_3{)}_5\right]\mathrm{Cl}$  are not any isomers to each other.