321.
Consider the following complex ions, $$P,Q$$ and $$R.$$
$$P = {\left[ {Fe{F_6}} \right]^{3 - }},Q = {\left[ {V{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}{\text{and}}\,R = {\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }}$$
The correct order of the complex ions, according to their spin-only magnetic moment values $$\left( {{\text{in}}\,{\text{B}}{\text{.M}}.} \right)$$ is
The electronic configuration of central metal ion in complex ions $$P,Q\,{\text{and}}\,R$$ are
$$P = {\left[ {Fe{F_6}} \right]^{3 - }};F{e^{3 + }}:$$
$$Q = {\left[ {V{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }};{V^{2 + }}$$
$$R = {\left[ {Fe{{\left( {{H_2}O} \right)}_6}} \right]^{2 + }};F{e^{2 + }}$$
Higher the no. of unpaired electron(s), higher will be magnetic moment. Thus the correct order of spin only magnetic moment is $$Q < R < P$$
322.
Which of the following has a square planar geometry?
$$\left( {{\text{At}}{\text{.}}\,{\text{nos}}.:Fe = 26,Co = 27,Ni = 28,Pt = 78} \right)$$
No explanation is given for this question. Let's discuss the answer together.
326.
As per IUPAC nomenclature, the name of the complex $$\left[ {Co{{\left( {{H_2}O} \right)}_4}{{\left( {N{H_3}} \right)}_2}} \right]C{l_3}\,{\text{is}}:$$
$$Ni{\left( {CO} \right)_4}.$$ The $$O.S.$$ of $$Ni$$ is zero. Electronic configuration is $$\left[ {Ar} \right]3{d^8}4{s^2}4{p^0}.$$ In presence of strong ligand $$CO$$ the paring of electrons take place and electronic configuration will be $$\left[ {Ar} \right]3{d^{10}}4{s^0}4{p^0}.$$ Hence unpaired electrons is zero .
328.
A solution containing $$2.675g$$ of a cobalt (III) chloride ammonia complex ( molar mass $$ = 267.5\,g\,mo{l^{ - 1}}$$ ) is passed through a cation exchanger. The chloride ions obtained in solution were treated with excess of $$AgN{O_3}$$ to give $$4.78\,g$$ of $$AgCl$$ ( molar mass = $$143.5\,g\,mo{l^{ - 1}}$$ ). The formula of the complex is (At. mass of $$Ag = 108\,u$$ )
A
$$\left[ {CoCl{{\left( {N{H_3}} \right)}_5}} \right]C{l_2}$$
B
$$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}$$
C
$$\left[ {CoC{l_2}{{\left( {N{H_3}} \right)}_4}} \right]Cl$$
D
$$\left[ {CoC{l_3}{{\left( {N{H_3}} \right)}_3}} \right]$$
No. of moles of $$CoC{l_3} \cdot 6N{H_3} = \frac{{2.675}}{{267.5}} = 0.01$$
No. of moles of $$AgCl = \frac{{4.78}}{{143.5}} = 0.03$$
Since 0.01 moles of the complex gives 0.03 moles of $$AgCl$$ on treatment with $$AgN{O_3},$$ it implies that 3 chloride ions are ionisable, in the complex. Thus, the formula of the complex is $$\left[ {Co{{\left( {N{H_3}} \right)}_6}} \right]C{l_3}.$$
329.
One mole of the complex compound $$Co{\left( {N{H_3}} \right)_5}C{l_3},$$ gives $$3$$ moles ofions on dissolution in water. One mole of the same complex reacts with two moles of $$AgN{O_3}$$ solution to yield two moles of $$AgCl\left( s \right).$$ The structure of the complex is
A
$$\left[ {Co{{\left( {N{H_3}} \right)}_3}C{l_3}} \right].2\,N{H_3}$$
B
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}C{l_2}} \right]Cl.\,N{H_3}$$
C
$$\left[ {Co{{\left( {N{H_3}} \right)}_4}Cl} \right]C{l_2}.\,N{H_3}$$
D
$$\left[ {Co{{\left( {N{H_3}} \right)}_5}Cl} \right]C{l_2}$$