Question

Match each coordination compound in List-I with an appropriate pair of characteristics from List -II and select the correct the 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)	$\left[Cr\right(N{H}_3{)}_{4}C{l}_2]Cl$	1.	Paramagnetic and exhibits ionisation isomerism.
(Q)	$\left[Ti\right(H_{2}O{)}_{5}Cl\left]\right(N{O}_3{)}_2$	2.	Diamagnetic and exhibits cis-trans isomerism
(R)	$\left[Pt\right(en\left)\right(N{H}_3\left)Cl\right]N{O}_3$	3.	Paramagnetic and exhibits cis-trans isomerism
(Q)	$\left[Co\right(N{H}_3{)}_4\left(N{O}_3{)}_2\right]N{O}_3$	4.	Diamagnetic and exhibits 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

Answer: (B)

P-3 Q-1 R-4 S-2

P. $\left[Cr\right(N{H}_3{)}_{4}C{l}_2]Cl$ is Tetraamminedichlorochromium (III) Chloride.

Charge present on each ion is $Cl=-1$, $Cr=x$, $N{H}_3=0$.

Since chloride is outside the complex and it has a negative charge. So, the complex gets +1 charge.

Now, the addition of each ion inside = charge of overall complex

$x-2=1\Rightarrow x=3$

$Cr=Z=24$; Electronic Configuration = $3d^{5}4s^1$

$C{r}^{3+}=3d^3$

Therefore, hybridisation is $d^{2}s{p}^3$ and as there is no pairing of electron, it is left with 3 unpaired electrons.

So, it is paramagnetic. Also, it exhibits cis-trans isomerism (geometrical) as it is a octahedral complex of the form $M{A}_4{B}_2$

Q. $\left[Ti\right(H_{2}O{)}_{5}Cl\left]\right(N{O}_3{)}_2$

$Ti$ has an electronic configuration of $\left[Ar\right]3d^{2}4s^2$.

Thus, it has unpaired electrons, hence it is paramagnetic.

Also, here the counter ion, i.e., $N{O}_3$ can also act as a ligand. Hence, it also shows ionisation isomerism.

R. $\left[Pt\right(en\left)\right(N{H}_3\left)Cl\right]N{O}_3$

$Pt$ has an electronic configuration of $\left[Xe\right]4f^{14}5d^{9}6s^1$

It does not have any unpaired electrons, hence it is diamagnetic.

Also, here the counter ion, i.e., $N{O}_3$ can also act as a ligand. Hence, it also shows ionisation isomerism.

S. $\left[Co\right(N{H}_3{)}_4\left(N{O}_3{)}_2\right]N{O}_3$

$Co$ has electronic configuration $\left[Ar\right]3d^{7}4s^2$

It does not have any unpaired electrons, hence it is diamagnetic.

Also, it also shows cis-trans isomerism.