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

Outer orbital complex utilises $$\left( {n - 1} \right)$$  $$d$$ - orbitals for bonding and exhibit paramagnetic behaviour, only if there present unpaired electrons.
$$\left( {\text{A}} \right){\text{In}}{\left[ {Ni{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }}:$$
$$N{i^{2 + }} = \left[ {Ar} \right]3{d^8}4{s^0}$$

$${\left[ {Ni{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }} = $$


So, this is an outer orbital complex as it involve $$4d$$ - orbitals for bonding, but having paramagnetic character.
$$\left( {\text{B}} \right){\text{In}}{\left[ {Zn{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }}:$$
$$Z{n^{2 + }} = \left[ {Ar} \right]3{d^{10}}$$

$${\left[ {Zn{{\left( {N{H_3}} \right)}_6}} \right]^{2 + }} = $$

Thus, it is also an outer orbital complex as it involve $$4d$$ - orbitals for bonding but it is diamagnetic as all the electrons are paired.
$$\left( {\text{C}} \right){\text{In}}{\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }},C{r^{3 + }} = \left[ {Ar} \right]3{d^3}$$
$${\left[ {Cr{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }} = $$

Because of the involvement of $$\left( {n - 1} \right)d,$$   i.e. $$3d$$ - orbital in hybridisation, it is an inner orbital complex. Its nature is paramagnetic because of the presence of three unpaired electrons.
$$\left( {\text{D}} \right){\text{In}}{\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }}:$$
$$C{o^{3 + }} = \left[ {Ar} \right]3{d^6}$$

$${\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]^{3 + }} = $$

Because of the involvement of $$(n-1)d$$   orbital in hybridisation, it is an inner orbital complex. As all the electrons are paired, it is a diamagnetic complex.

With increase in synergic bonding, $$C–C$$  bond length increases in Zeise’s salt.

In the given complex $${\left[ {E{{\left( {en} \right)}_2}\left( {{C_2}{O_4}} \right)} \right]^ + }NO_2^ - $$     ethylenediamine is a bidentate ligand and oxalate ion $$\left( {{C_2}O_4^{2 - }} \right)$$  is also bidentate ligand. Therefore, coordination number of the complex is 6 i.e., it is an octahedral complex.
Oxidation number of E in the given complex is
$$x + 2 \times 0 + 1 \times \left( { - 2} \right) = + 1\,\,\therefore x = 3$$

Magnetic moment indicates that there are three unpaired electrons present in chromium. These must be present in lower energy orbitals which are $$3{d_{xy}},3{d_{yz}}$$   and $$3{d_{xz}}.$$

$$C{u^{2 + }} + 2O{H^ - } \to Cu{\left( {OH} \right)_2}$$
$$Cu{\left( {OH} \right)_2} + 4N{H_3} \to $$     $$\mathop {{{\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]}^{2 + }}}\limits_{{\text{deep blue}}} + 2O{H^ - }$$
In $${\left[ {Cu{{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$$   oxidation state of $$Cu = + 2$$

Hybridisation is $$ds{p^2}$$  hence geometry is square planar and paramagnetic due to presence of one unpaired electron.

In $$Fe{\left( {CO} \right)_5},$$   $$Fe$$  is $$s{p^3}d$$  hybridised and thus trigonal bipyramidal in shape.

Complex is not superimposable on its mirror image hence optically active i.e., rotate plane polarized light.

In octahedral complex the magnitude of $${\vartriangle _o}$$ will be highest in a complex having strongest ligand. Of the given ligands $$C{N^ - }$$  is strongest so $${\vartriangle _o}$$ will be highest for $${\left( {Co{{\left( {CN} \right)}_6}} \right)^{3 - }}.$$   Thus option (A) is correct.

Crystal field stabilisation energy for tetrahedral complexes is less than pairing energy hence they do not pair up to form low spin complexes.

$${S_2}O_3^{2 - }$$   is monodentate ligand where as other ligands are bidentate.