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

As negative charge on metal carbonyl complex increases, back $$\pi $$-bonding increases and hence bond length of $$C—O$$  bond increases while bond length of metal-carbon bond decreases.
Hence, $${\left[ {Fe{{\left( {CO} \right)}_4}} \right]^{2 - }}$$   has longest $$C—O$$  bond length among the given complexes.
The correct order of bond length of the given complexes is
$${\left[ {Mn{{\left( {CO} \right)}_6}} \right]^ + } < \left[ {Ni{{\left( {CO} \right)}_4}} \right]$$      $$ < {\left[ {Co{{\left( {CO} \right)}_4}} \right]^ - } < {\left[ {Fe{{\left( {CO} \right)}_4}} \right]^{2 - }}$$

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

Oxidation state of $$Cr\,{\text{in}}\,{\left[ {Cr{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]^ + }.$$
Let it be $$x,\,1 \times x + 4 \times 0 + 2 \times \left( { - 1} \right) = 1\,{\text{Therefore}}\,x{\text{ = 3}}{\text{.}}$$

Tris - ( ethylenediamine ) cobalt (III) bromide $$\left[ {Co{{\left( {en} \right)}_3}} \right]B{r_3}$$    exhibits optical isomerism :

Positively charged ligands have suffix 'ium'.

$$X = {\left[ {Ni\left( {en} \right){{\left( {N{H_3}} \right)}_4}} \right]^{2 + }}$$     $${\text{neither }}G.I.{\text{ nor }}OI$$
$$Y = {\left[ {Ni{{\left( {en} \right)}_2}{{\left( {N{H_3}} \right)}_2}} \right]^{2 + }}$$     $$G.I{\text{ as well as }}OI$$
$$Z = {\left[ {Ni{{\left( {en} \right)}_3}} \right]^{2 + }}$$    $${\text{does not show }}G.I.{\text{ while show }}OI$$

In general, the ligands can be arranged in a series in the order of increasing field strength as
$${I^ - } < B{r^ - } < SC{N^ - } < C{l^ - } < {S^{2 - }}$$       $$ < {F^ - } < O{H^ - } < {C_2}O_4^{2 - } < {H_2}O$$       $$ < NC{S^ - } < EDT{A^{4 - }} < N{H_3}$$       $$ < en < NO_2^ - < C{N^ - } < CO$$

$$Cis$$  - platin is the isomer of $$\left[ {Pt{{\left( {N{H_3}} \right)}_2}C{l_2}} \right]$$    which is used as an anticancer drug for treating several types of malignant tumours.

Same magnetic moment = Same number of unpaired electrons.
$$\mu = \sqrt {n\left( {n + 2} \right)} $$
where $$n =$$  number of unpaired electrons
$$C{o^{2 + }} = 3{d^7},3$$    unpaired electrons;
$$C{r^{2 + }} = 3{d^4},4$$    unpaired electrons;
$$M{n^{2 + }} = 3{d^5},5$$    unpaired electrons;
$$F{e^{2 + }} = 3{d^6},4$$    unpaired electrons.