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.
${en = H_{2}NCH_{2}CH_{2}NH_{2} ; At. Nos. : Ti = 22; Cr = 24 ; Co = 27; Pt = 78}$

List-I		List-II
$(P)$	$[Cr(NH_{3})_{4}]Cl_{2}]Cl$	$1$	Paramagnetic and exhibits ionisation isomerism
(Q)	$[Ti(H_{2}O)_{5}Cl](NO_{3})_{2}$	$2$	Diamagnetic and exhibits cis-trans isomerism
$(R)$	$[Pt(en)(NH_{3})Cl]NO_{3}$	$3$	Paramagnetic and exhibits cis-trans isomerism
$(S)$	$[Co (NH_{3})_{4}(NO_{3})_{2}]NO_{3}$	$4$	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)

$p : Cr^{3+}$ has $3d^{3}$ configuration, with $3$ unpaired electrons. Hence, it shows paramagnetic behaviour. Complex of the type $Ma_{4}b_{2}$ shows cis-trans isomerism. $Q : Ti^{3+}$ has $3d^{1}$ configuration, hence shows paramagnetic behaviour. Complex gives $Cl^{-}$ and $NO^{-}_{3}$ ions in solution hence, shows ionisation isomerism.
$R : Pt^{2+}$ 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.
$S : Co^{3+}$ 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 $Ma_{4}b_{2}$ type complex.