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.
{$en=H_{2}NC{H}_{2}C{H}_{2}N{H}_2;$ atomic numbers:$Ti=22;Cr=24;Co=27;Pt=78$}

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

• A. P-4, Q-2, R-3, S-1
• B. P-3, Q-1, R-4, S-2
• C. P-2, Q-1, R-3, S-4
• D. P-1, Q-3, R-4, S-2

Answer 1

The correct option is B P-3, Q-1, R-4, S-2

P : 3
Compound cannot exhibit ionisation isomerism since only $Cl$ is there and thus cannot exhibit different ions in solution.
Checking the magnetism:
$C{r}^{3+}$ has $3d^3$ configuration, with 3 unpaired electrons.

Hence, it shows paramagnetic behaviour. Also complex of the type $M{a}_4{b}_2$ shows cis-trans isomerism.

Q :1
$T{i}^{3+}$ has $3d^1$ configuration, hence shows paramagnetic behaviour. complex gives $C{l}^{-}$ and $N{O}_3^{-}$ ions in solution hence, shows ionisation isomerism.


R : 4
$P{t}^{2+}$ has $5d^8$ configuration but strong field ligands forms square planar complex.

Thus, all electrons are paired and it also exhibits ionisation isomerism as it gives different ions in solution.


S : $C{o}^{3+}$ has $3d^6$ configuration. Here ligands present are strong enough to cause electron pairing. Hence, it Shows diamagnetic behaviour and exhibits cis-trans isomerism as it is a $M{a}_4{b}_2$ type complex.