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Coordination Compounds

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Question
CBSEENCH11008501
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[Co(NH3)4(NO2)2]Cl exhibits:

  • linkage isomerism, ionisation isomerism and optical isomerism

  • Linkage isomerism, ionisation isomerism and geometrical isomerism

  • ionization isomerism, geometrical isomerism and optical isomerism

  • linkage isomerism, geometrical isomerism and optical isomerism

Solution

B.

Linkage isomerism, ionisation isomerism and geometrical isomerism

The compound [Co(NH3)4((NO2)2]Cl exhibits linkage, ionisation and geometrical isomerism.
Hence, its linkage isomers are
(i) [Co(NH3)2(NO2)2]Cl and [Co(NH3)4(ONO)2]Cl
(ii) Its ionisation isomers are 
[Co(NH3)4 (NO2)Cl]NO2 and [Co(NH3)4(NO2)2]Cl
(iii) Its geometrical isomers are

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Question
CBSEENCH12011402
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[Cr(H2O)6]Cl3 (at. no. of Cr = 24) has a magnetic moment of 3.83 BM, the correct distribution of 3d electrons in the chromium of the complex is:

  • 3 straight d subscript straight x squared minus straight y squared end subscript superscript 1 comma space 3 straight d subscript straight z squared end subscript superscript 1 comma space 3 straight d subscript xz superscript 1
  • 3 straight d subscript xy superscript 1 comma space 3 straight d subscript x squared minus y squared end subscript superscript 1 comma space 3 straight d subscript yz superscript 1
  • 3 straight d subscript xy superscript 1 comma space 3 straight d subscript zy superscript 1 comma space 3 straight d subscript zx superscript 1
  • 3 straight d subscript xy superscript 1 comma space 3 straight d subscript yz superscript 1 comma space 3 straight d subscript straight z squared end subscript superscript 1

Solution

C.

3 straight d subscript xy superscript 1 comma space 3 straight d subscript zy superscript 1 comma space 3 straight d subscript zx superscript 1

Magnetic moment,
straight mu space equals space square root of straight n left parenthesis straight n plus 2 right parenthesis end root space BM 3.83 space equals space square root of straight n space left parenthesis straight n plus 2 right parenthesis end root 3.83 space straight x space 3.83 space equals space straight n squared space plus 2 straight n 14.6689 space equals space straight n squared space plus 2 straight n On space solving space this comma space we space get space straight n space equals 3
Hence, a number of unpaired electrons in d- subshell of the penultimate shell of chromium (Cr = 24).
so, the configuration of chromium ion is 
Cr3+ = 1s2, 2s2, 2p6, 3s2, 3p6 3d3
In [Cr(H2O)6]Cl2 oxidation state of Cr is +3
Hence in 3d3 the distribution of electrons
3 straight d subscript xy superscript 1 comma space 3 straight d subscript yz superscript 1 comma space 3 straight d subscript zx superscript 1 comma space 3 straight d subscript straight x squared minus straight y squared end subscript superscript 0 comma space 3 straight d subscript straight z squared end subscript superscript 0

Question
CBSEENCH12011121
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A magnetic moment of 1.73 BM will be shown by one among the following

  • [Cu(NH3)4]2+

  • [Ni(CN)4]2-

  • TiCl4

  • [CoCl6]4-

Solution

A.

[Cu(NH3)4]2+

Magneitc space moment comma space straight mu space is space related space with space number space of space unpaired space electrons space as straight mu space equals space square root of straight n left parenthesis straight n plus 2 right parenthesis end root space BM left parenthesis 1.73 right parenthesis to the power of 2 space end exponent equals straight n left parenthesis straight n plus 2 right parenthesis On space solving space straight n equals 1 Thus comma space the space complex divided by compound space having space one space unpaired space electron exhibit space straight a space magnetic space moment space of space 1.73 space BM straight a right parenthesis space In space left square bracket Cu left parenthesis NH subscript 3 right parenthesis subscript 4 right square bracket to the power of 2 plus end exponent Cu to the power of 2 plus end exponent space equals space left square bracket Ar right square bracket 3 straight d to the power of 9 straight b right parenthesis space In space left square bracket Ni left parenthesis CN right parenthesis subscript 4 right square bracket to the power of 2 minus end exponent Ni to the power of 2 plus end exponent space equals space left square bracket Ar right square bracket 3 straight d to the power of 8 straight c right parenthesis space In space left square bracket TiCl subscript 4 right square bracket Ti to the power of 4 plus end exponent space equals space left square bracket Ar right square bracket 3 straight d to the power of 0 straight d right parenthesis space In space left square bracket CoCl subscript 6 right square bracket to the power of 4 minus end exponent Co to the power of 2 plus end exponent space equals space left square bracket Ar right square bracket 3 straight d to the power of 7 In space above space all space configuration space only space space left square bracket Cu left parenthesis NH subscript 3 right parenthesis subscript 4 right square bracket to the power of 2 plus end exponent has space one space unpaired space electron.

Question
CBSEENCH12011100
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Among the following complexes, the one which shows zero crystal field stabilisation energy (CFSE) is

  • [Mn(H2O)6]3+

  • [Fe(H2O)6]3+

  • [Co(H2O)6]3+

  • [Co(H2O)]3+

Solution

B.

[Fe(H2O)6]3+

The CFSE for octahedral complex is given by
CFSE = [-0.4 t2g e- +0.6 ege-]
For Mn3+, [3d4]--> t2g3e1g
therefore,
CFSE = [-0.4 x 3 + 0.6 x1]

=-0.6
For Fe3+, [3d] --> t2g3e2g
CFSE = [(-0.4 x 3) + (0.6 x2)] = 0
For Co2+, [3d7] --> t2g5e2g 
CFSE = [(-0.4 x 5) + (0.6 x2)] =-0.8
For Co3+, [3d6] ---> t2g4e2g 

CFSE = [(-0.4 x4) + (0.6 x2)] = -0.4

Question
CBSEENCH12011422
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An example of a sigma bonded organometallic compound is :

  • Ruthenocene

  • Grignard's reagent

  • Ferrocene

  • Cobaltocene

Solution

B.

Grignard's reagent

Grignard's reagent i.e., RMgX is σ-bonded organometallic compound.

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