Question

Match each coordination compound in List-I with an appropriate pair of characteristics from List-II and select the correct answer using the code given below the lists.
$en=H_{2}NC{H}_{2}C{H}_{2}N{H}_2;At.Nos.:Ti=22;Cr=24;Co=27;Pt=78$

List-I		List-II
$(P)$	$[Cr(N{H}_3{)}_4]C{l}_2]Cl$	$1$	Paramagnetic and exhibits ionisation isomerism
(Q)	$[Ti(H_{2}O{)}_{5}Cl](N{O}_3{)}_2$	$2$	Diamagnetic and exhibits cis-trans isomerism
$(R)$	$[Pt(en)(N{H}_3)Cl]N{O}_3$	$3$	Paramagnetic and exhibits cis-trans isomerism
$(S)$	$[Co(N{H}_3{)}_4(N{O}_3{)}_2]N{O}_3$	$4$	Diamagnetic and exhibits ionisation isomerism

	$P$	$Q$	$R$	$S$
(a)	$4$	$2$	$3$	$1$
(b)	$3$	$1$	$4$	$2$
(c)	$2$	$1$	$3$	$4$
(d)	$1$	$3$	$4$	$2$

• A. a
• B. b
• C. c
• D. d

Answer 1

Answer: (B)

$p:C{r}^{3+}$ has $3d^3$ configuration, with $3$ unpaired electrons. Hence, it shows paramagnetic behaviour. Complex of the type $M{a}_4{b}_2$ shows cis-trans isomerism. $Q:T{i}^{3+}$ has $3d^1$ configuration, hence shows paramagnetic behaviour. Complex gives $C{l}^{-}$ and $N{O}_3^{-}$ ions in solution hence, shows ionisation isomerism.
$R:P{t}^{2+}$ has $3d^8$ configuration but ligands are strong field ligands hence, it forms square planar complex. Thus, all electrons are paired and it also exhibits ionisation isomerism.
$S:C{o}^{3+}$ has $3d^6$ configuration. But, ligands present are strong enough to cause electron pairing, hence, it shows diamagnetic behaviour and exhibits cis-trans isomerism as it is $M{a}_4{b}_2$ type complex.