Co - ordination Compounds MCQ Questions & Answers in Inorganic Chemistry | Chemistry

Hybridisation
$$\eqalign{ & \mathop {{{\left[ {Fe{{\left( {CN} \right)}_6}} \right]}^{4 - }}}\limits_{{d^2}s{p^3}} ,\mathop {{{\left[ {Mn{{\left( {CN} \right)}_6}} \right]}^{4 - }}}\limits_{{d^2}s{p^3}} , \cr & \mathop {{{\left[ {Co\left( {N{H_3}} \right.} \right]}^{3 + }}}\limits_{{d^2}s{p^3}} ,\mathop {{{\left[ {Ni{{\left( {N{H_3}} \right)}_6}} \right]}^{2 + }}}\limits_{s{p^3}{d^2}} \cr & {\text{Hence}}\,{\left[ {Ni{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }}{\text{is}}\,{\text{outer}}\,{\text{orbital}}\,{\text{complex}}. \cr} $$

Given complex compound is $$\left[ {M{{\left( {en} \right)}_2}\left( {{C_2}{O_4}} \right)} \right]Cl$$
Let oxidation number of $$M$$  is $$x.$$
$$\therefore x - 2 - 2 = - 1$$
$$or\,\,\,x = + 3$$
Now, as coordination number is defined as the total number of binding sites attached to the metal. Hence, in the given complex coordination number is 6.

A square planar complex is formed by hybridisation of $$s,{p_x},{p_y}$$  and $${d_{{x^2} - {y^2}}}$$  atomic orbitals

No explanation is given for this question. Let's discuss the answer together.

$${F^ - }$$ is a weak field ligand hence it does not cause pairing of electrons.

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$$\eqalign{ & \left( {\text{i}} \right)\mathop {{{\left[ {Cu{{\left( {CN} \right)}_4}} \right]}^{3 - }}}\limits_{{\text{(Diamagnetic)}}} \to \mu = 0 \cr & \left( {{\text{ii}}} \right)\mathop {{{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]}^{3 + }}}\limits_{{\text{(Diamagnetic)}}} \to \mu = 0 \cr & \left( {{\text{iii}}} \right)\mathop {{{\left[ {Ni{{\left( {N{H_3}} \right)}_6}} \right]}^{2 + }}}\limits_{{\text{(Paramagnetic)}}} \to \mu = \sqrt 8 BM \cr & \left( {{\text{iv}}} \right)\mathop {{{\left[ {Fe{{\left( {CN} \right)}_6}} \right]}^{3 - }}}\limits_{{\text{(Paramagnetic)}}} \to \mu = \sqrt 3 BM \cr} $$

$$\left[ {Co{{\left( {N{H_3}} \right)}_5}C{O_3}} \right]Cl{O_4}.$$     Six monodentate ligands are attached to $$Co$$  hence $$C. N.$$  of $$Co = 6;$$  $$O.N. = x + 5 \times \left( 0 \right) + 1 \times \left( { - 2} \right) + 1 \times \left( { - 1} \right) = 0$$       $$\therefore x = + 3;$$   electronic configuration of $$C{o^{3 + }}\left[ {Ar} \right]3{d^6}4{s^0}$$     hence number of $$d$$  electrons is 6. All $$d$$  electrons are paired due to strong ligand hence unpaired electron is zero.

Symmetrically filled $${t_{2g}}$$  and $${e_g}$$  are those, which contain equal distribution of electrons.

\[\underset{\begin{smallmatrix} \text{Lewis} \\ \text{acid} \end{smallmatrix}}{\mathop{C{{o}^{3+}}}}\,+\underset{\begin{smallmatrix} \text{Lewis} \\ \text{base} \end{smallmatrix}}{\mathop{6N{{H}_{3}}}}\,\to \underset{\text{adduct}}{\mathop{{{\left[ Co{{\left( N{{H}_{3}} \right)}_{6}} \right]}^{3+}}}}\,\]