Coordination Compounds 2 Question 55
57. Match each coordination compound in Column I with an appropriate pair of characteristics from Column II and select the correct answer using the codes given below the Columns (en $=H _2 NCH _2 CH _2 NH _2$; atomic numbers : $Ti=22$; $Cr=24 ; Co=27 ; Pt=78)$
(2014 Adv.)
| Column I | Column II | |
|---|---|---|
| (A) | $\left[\mathrm{Cr}\left(\mathrm{NH}_3\right)_4 \mathrm{Cl}_2\right] \mathrm{Cl}$ | 1. Paramagnetic and exhibits ionisation isomerism |
| (B) | $\left[\mathrm{Ti}\left(\mathrm{H}_2 \mathrm{O}\right)_5 \mathrm{Cl}\right]\left(\mathrm{NO}_3\right)_2$ | 2. Diamagnetic and exhibits cis-trans isomerism |
| (C) | $\left[\mathrm{Pt}(\mathrm{en})\left(\mathrm{NH}_3\right) \mathrm{Cl}\right] \mathrm{NO}_3$ | 3. Paramagnetic and exhibits cis-trans isomerism |
| (D) | $\left[\mathrm{Co}\left(\mathrm{NH}_3\right)_4\left(\mathrm{NO}_3\right)_2\right] \mathrm{NO}_3$ | 4. Diamagnetic and exhibits ionisation isomerism |
Codes
$\quad$ A $\hspace{2mm}$ B $\hspace{2mm}$ C $\hspace{2mm}$ D
(a) $4 \quad 2 \quad 3 \quad 1$
(b) $3 \quad 1 \quad 4 \quad 2$
(c) $2 \quad 1 \quad 3 \quad 4$
(d) $1 \quad 3 \quad 4 \quad 2$
Show Answer
Answer:
Correct Answer: 57. (b)
Solution:
- PLAN: This problem is based on concept of VBT and magnetic properties of coordination compound.
Draw VBT for each coordination compound.
If unpaired electron is present then coordination compound will be paramagnetic otherwise diamagnetic.
Coordination compounds of $\left[M A _4 B _2\right]$ type show geometrical isomerism. Molecular orbital electronic configuration (MOEC) for various coordination compound can be drawn using VBT as
A. $MO$ $EC$ for $\left[Cr\left(NH _3\right) _4 Cl _2\right] Cl$ is
Number of unpaired electrons $(n)=3$
Magnetic properties $=$ paramagnetic
Geometrical isomers of $\left[Cr\left(NH _3\right) _4 Cl _2\right]^{+}$are
B. $n=1$
Magnetic properties $=$ paramagnetic
Ionisation isomers of $\left[Ti\left(H _2 O\right) _5 Cl\right]\left(NO _3\right) _2$ are
$\left[Ti\left(H _2 O\right) _5 Cl\right]\left(NO _3\right) _2$ and $\left[Ti\left(H _2 O\right) _5\left(NO _3\right)\right] Cl\left(NO _3\right)$
C. $MO$ $EC$ of $\left[Pt(en)\left(NH _3\right) Cl\right] NO _3$ is
Magnetic property $=$ diamagnetic
Ionisation isomers are $\left[Pt(en) _2\left(NH _3\right) Cl\right] NO _3$
and $\left[Pt(en) _2 NH _3\left(NO _3\right)\right] Cl$
D. $MO$ $ EC$ of $\left[Co\left(NH _3\right) _4\left(NO _3\right) _2\right] NO _3$
Magnetic property $=$ Diamagnetic
Geometrical isomers are
Thus, magnetic property and isomerism in given coordination compound can be summarised as
(P) $\left[Cr\left(NH _3\right) _4 Cl _2\right] Cl \rightarrow$ Paramagnetic and exhibits cis-trans isomerism (3)
(Q) $\left[Ti\left(H _2 O\right) _5 Cl\right]\left(NO _3\right) _2 \rightarrow$ Paramagnetic and exhibits ionisation isomerism (1)
(R) $\left[Pt(en)\left(NH _3\right) Cl\right] NO _3 \rightarrow$ Diamagnetic and exhibits ionisation isomerism (4)
(S) $\left[Co\left(NH _3\right) _4\left(NO _3\right) _2\right] NO _3 \rightarrow$ Diamagnetic and exhibits cis-trans isomerism (2)
$\therefore P \rightarrow 3, Q \rightarrow 1, R \rightarrow 4, S \rightarrow 2$
Hence, (b) is the correct choice.
BETA
