Chemistry MCQs for NEET — Practice Questions with Answers

(c) ${\mathrm{Fe}}^{2+}\left(\mathrm{aq}\right)+\mathrm{NO}+{\mathrm{SO}}_4^{2-}\to \left[\mathrm{Fe}\right({\mathrm{H}}_2\mathrm{O}{)}_{\mathrm{s}}\left(\mathrm{NO}\right){]}^{2+}+{\mathrm{SO}}_4^{2-}$

${\mathrm{\mu }}_{\mathrm{eff}}=3.89 \mathrm{BM}$

Hence, no. of unpaired electrons = 3 

(b) As ${\mathrm{NH}}_3$ is stronger ligand than ${\mathrm{H}}_2\mathrm{O}$, hence CFSE value for $\left[\mathrm{M}\right({\mathrm{NH}}_3{)}_6{]}^{2+}$> CFSE of$\left[\mathrm{M}\right({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+}$ therefore, absorption shifts to smaller wavelength . Also difference between splitting power of  ${\mathrm{NH}}_3$ and ${\mathrm{H}}_2\mathrm{O}$ is not very high.

(c) $$\begin{gathered} {∆}_{\mathrm{t}}=\frac{4}{9}{∆}_0 \\ \therefore {∆}_{\mathrm{t}} \mathrm{for} [{\mathrm{CoCl}}_4{]}^{2-}=\frac{4}{9}\times 18000=8000 {\mathrm{cm}}^{-1} \end{gathered}$$

(c) Octahedral complexes having metal cation with ${\mathrm{d}}^3$ and ${\mathrm{d}}^8$ configuration can not be defined in terms of high and low spin complexes .

(a) $\left[\stackrel{\mathrm{III}}{\mathrm{Co}}\right({\mathrm{NH}}_3{)}_4{\mathrm{CO}}_3{]}^{+}{\mathrm{ClO}}_4^{-},{\mathrm{CO}}_3^{-}$ is working as bidentate ligand ,have coordination number of $\stackrel{\mathrm{III}}{\mathrm{Co}}=6$.

(b) Vaccant non-axial d- orbital participate in $\mathrm{M}\stackrel{\mathrm{\pi }}{\to }{\mathrm{PR}}_3$ back bonding in its complexes.

(a) In the metal carbonyls:

Higher the negative oxidation state of central metal : ∝ Bond length of C-O

                                                                           : ∝ $\frac{1}{\mathrm{Bond} \mathrm{order} \mathrm{of} \mathrm{CO}}$

                                                                           : ∝ Bond order of M-C bond 

(b) In ${\mathrm{PR}}_3$ there is vacant atomic d-orbital on P-atom , which can be involved in Synergic bonding .

  CO bond order ∝ $\frac{1}{\mathrm{Extent} \mathrm{of} \mathrm{back} \mathrm{bonding} (\mathrm{M}\to \mathrm{CO})}$

Correct sequence of CO bond order :

Mn⁺= 3d⁵4s¹; in presence of CO, effective configuration =3d⁶4s⁰. 

$\underset{(\mathrm{CO} \mathrm{bond} \mathrm{order} >3.0)}{{\mathrm{H}}_3\mathrm{B}\stackrel{}{\leftarrow \mathrm{CO}}}>\underset{\mathrm{B}.\mathrm{O}.=3.0}{\mathrm{CO}}>\underset{(\mathrm{CO} \mathrm{bond} \mathrm{order}<3.0)}{\left[\mathrm{Mn}\right(\mathrm{CO}{)}_5{]}^{+}} >\mathrm{Fe}(\mathrm{CO}{)}_5$

(d) $\stackrel{\mathrm{II}}{\left[\mathrm{Mn}\right(\mathrm{CN}{)}_6{]}^{4-}}$ : Hyb.  :${\mathrm{d}}^2{\mathrm{sp}}^3$,${\mathrm{\mu }}_{\mathrm{eff}}=\sqrt{3}\mathrm{BM}$

 $\stackrel{\mathrm{III}}{\left[\mathrm{Fe}\right({\mathrm{NH}}_3{)}_6{]}^{3+}}: \mathrm{Hyb}.: {\mathrm{d}}^2{\mathrm{sp}}^3, {\mathrm{\mu }}_{\mathrm{eff}}=\sqrt{3}\mathrm{BM}$

$\left[\mathrm{Co}\right(\mathrm{CO}{)}_4]:\mathrm{Hyb}: {\mathrm{dsp}}^2, {\mathrm{\mu }}_{\mathrm{eff}}=\sqrt{3}\mathrm{BM}$

$\left[\mathrm{Cu}\right({\mathrm{H}}_2\mathrm{O}{)}_6{]}^{2+} : \mathrm{Hyb}. :{\mathrm{sp}}^3{\mathrm{d}}^2, {\mathrm{\mu }}_{\mathrm{eff}}=\sqrt{3}\mathrm{BM}$