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Coordination Compounds — Chemistry STD 12 Science — Question

Gujarat BoardEnglish MediumSTD 12 ScienceChemistryCoordination Compounds4 Marks
Question
Read the passage given below and answer the following questions:
Iron forms many complexes in its $+2$ and $+3$ oxidation states such as $[Fe(H_2O)_6]^{2+} (A); [Fe(CN)_6]^{4-} (B); [Fe(H_2O)_6]^{3+} (C); [Fe(CN)_6]^{3-} (D),$ etc., They exhibit, different magnetic properties and undergo different hybridisation of iron.
The following questions are multiple choice questions. Choose the most appropriate answer:
  1. Which of the following statements is correct?
  1. $(B)$ is paramagnetic while $(C)$ is diamagnetic.
  2. Both $(B)$ and $(D)$ are outer orbital complexe.
  3. Both $(A)$ and $(C)$ are paramagnetic.
  4. $(A)$ is outer orbital complex and $(C)$ is inner orbital complex.
  1. The complex having maximum magnetic moment is:
  1. $(A)$
  2. $(B)$
  3. $(C)$
  4. $(D)$
  1. Which of the following does not represent correct configuration of the d-orbitals in the given complexes?
  1. $\text{(A)}:\text{t}^4_{2\text{g}}\text{e}^2_\text{g}$
  2. $\text{(B)}:\text{t}^6_{2\text{g}}\text{e}^0_\text{g}$
  3. $\text{(C)}:\text{t}^4_{2\text{g}}\text{e}^1_\text{g}$
  4. $\text{(D)}:\text{t}^5_{2\text{g}}\text{e}^0_\text{g}$
  1. The spin only magnetic moment of complexes $(A), (B), (C)$ and $(D)$ are respectively $($in $BM).$
  1. $2\sqrt{6},0,\sqrt{35},\sqrt{3}$
  2. $0,2\sqrt{6},\sqrt{35},\sqrt{3}$
  3. $\sqrt{15,}2\sqrt{6},\sqrt{3},0$
  4. $\sqrt{3},\sqrt{8},0,\sqrt{15}$
  1. Which of the given complexes are outer orbital complexes?
  1. $(A)$ and $(B)$ only
  2. $(B)$ and $(C)$ only
  3. $(A)$ and $(C)$ only
  4. $(B)$ and $(D)$ only
✓

Answer

  1. $(c)$ Both $(A)$ and $(C)$ are paramagnetic.
Explanation:

(A): $sp^3d^2$ hybridisation (outer orbital)

No. of unpaired electrons $= 4$

(B): $d^2sp^3$ hybridisation (inner orbital)

No. of unpaired electrons $= 0$

(C): $sp^3d^2$ hybridisation (outer orbital)

No. of unpaired electrons $= 5$

(D): $d^2sp^3$ hybridisation (inner orbital)

No. of unpaired electron $= 1$
  1. (c) $(C)$
Explanation:

It has $5$ unpaired electrons.
  1. (c) $\text{(C)}:\text{t}^4_{2\text{g}}\text{e}^1_\text{g}$
Explanation:

As $H_2​​​​​​​O$ is a weak ligand so, it should be $\text{t}^4_{2\text{g}}\text{e}^1_\text{g}.$
  1. (a) $2\sqrt{6},0,\sqrt{35},\sqrt{3}$
Explanation:

Magnetic moments of $(A), (B), (C)$ and $(D)$ are respectively.

$\sqrt{4(4+2)},0,\sqrt{5(5+2)},\sqrt{1(1+2)}$
  1. (c) $(A)$ and $(C)$ only

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Read the passage given below and answer the following questions:
Dehydration of alcohols can lead to the formation of either alkenes or ethers. This dehydration can be carried out either with protonic acids such as cone.$H_2SO_4, H_3PO_4$ or catalysts such as anhydrous $ZnCl_2$ or $Al_2O_3$. When primary alcohols are heated with cone. $H_2SO_4$ at $433-44 K$, they undergo intramolecular dehydration to form alkenes. Secondary and tertiary alcohols undergo dehydration under milder conditions. The ease of dehydration of alcohols follows the order: $3^\circ > 2^\circ > 1^\circ$.
The dehydration of alcohols always occurs in accordance with the Saytzeff's rule. Primary alcohols when heated with protic acid at 413K, gives dialkyl ether.
$\text{CH}_3\text{CH}_2\text{OH}\xrightarrow[433.433\text{K}]{\text{conc. H}_2\text{SO}_4}\text{CH}_2=\text{CH}_2+\text{H}_2\text{O}$
$2\text{CH}_3\text{CH}_2\text{OH}\xrightarrow[433\text{K}]{\text{conc. H}_2\text{SO}_4}\text{CH}_3\text{CH}_2-\text{O}-\text{CH}_2\text{CH}_3+\text{H}_2\text{O}$
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The following questions are multiple choice questions. Choose the most appropriate answer:
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  1. N-alkyl benzene sulphonamide soluble in KOH solution.
  2. N-alkyl benzene sulphonamide insoluble in KOH solution.
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  3. 2-Methyl-1-propanol
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