Discussion Forum

1)

Match  each coordinates  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=H₂NCH₂CH₂NH₂. atomic numbers : Ti=22.: Cr=24: Co=27: Pt=78)

 

 

Answer:

Option B

Explanation:

Plan  This problem is based on concept of VBT  and magnetic properties of coordination compound.

 Draw VBT  for each coordination compound

 If an unpaired electron is present then the coordination compound will be paramagnetic otherwise diamagnetic.

coordination compounds of [MA₄B₂] type show geometrical isomerism.

Molecular orbital electrons  configuration (MOEC) for various coordination compound can be drawn using VBT as

 MOEC for [Cr(NH₃)₄Cl₂]Cl   is

                     

 Number of unpaired elecrtons (n)=3

 Magnetic properties = paramagnetic

 Geometrical isomers of (Cr(NH₃)₄Cl₂]⁺ are

 

MOEC of [Ti(H₂O)₅ Cl](NO₃)₂   is

   n=1

 Magnetic properties= paramagnetic

  Ionisation  isomers of [Ti(H₂O)₅Cl)(NO₃)₂ are     [Ti(H₂O)₅Cl(NO₃)₂    and [Ti(H₂O)₅(NO₃)]Cl(NO₃)

MOEC   of [Pt(en)(NH₃)Cl]NO₃   is 

 

 

n=0

 Magnetic property= diamgnetic

 Ionsiation isomers are [Pt(en)(NH₃)Cl)NO₃ and [Pt(en)(NH₃(NO₃)]Cl

 MOEC of [Co(NH₃)₄ (NO₃)₂]NO₃    is

 n=0

Magnetic property= Diamagnetic  Geometrical isomers are

  Thus, magnetic  property and isomerism in a given coordination compound can be summarised as

 (P)    [Cr(NH₃)₄ Cl₂]Cl  → Paramagnetic and exhibits cis-trans isomerism (3)

 (Q)    [Ti(H₂O)₅Cl](NO₃)₂   →  Paramagnetic and exhibits ionisation isomerism (1)

  (R)   [Pt(en) (NH₃)Cl]NO₃  →  Diamagnetic and exhibits  ionisation isomerism(4)

(S)   [Co(NH₃)₄ (NO₃)₂  ]NO₃   → Diamagnetic and exhibits cis-trans isomerism  (2)

     P → 3, Q→ 1, R→  4, S→ 2

 Hence, (b) is the correct choice