Question

{Given below are two statements: Statement I: Among \[ [\mathrm{Cu(NH_3)_4}]^{2+}, [\mathrm{NiO_3}]^{2+}, [\mathrm{Ni(NH_3)_6}]^{2+} \text{ and } [\mathrm{Mn(H_2O)_6}]^{2+}, \] \( [\mathrm{Mn(H_2O)_6}]^{2+} \) has the maximum number of unpaired electrons. Statement II: The number of pairs among \[ [\mathrm{NiCl_4}]^{2-},\ [\mathrm{NiO_4}]^{2-} \] and \[ [\mathrm{NiO_4}],\ [\mathrm{O_4}]^{2-} \] that contain only diamagnetic species is two. In the light of the above statements, choose the correct answer from the options given below:}

• A. Statement I is false but Statement II is true
• B. Both Statement I and Statement II are true
• C. Both Statement I and Statement II are false
• D. Statement I is true but Statement II is false

Hint:

Important configurations: \[ \mathrm{Mn^{2+}} = 3d^5 \] usually has maximum unpaired electrons. Also remember: \[ \mathrm{Ni(CO)_4} \] is diamagnetic because nickel attains: \[ 3d^{10} \] configuration.

Answer 1

The Correct Option is B

Concept: The magnetic behaviour of coordination compounds depends on:

• Number of unpaired electrons
• Strength of ligand field
• Electronic configuration of metal ion

Species containing all paired electrons are:

• Diamagnetic

Species containing one or more unpaired electrons are:

• Paramagnetic

Step 1: Checking Statement I. We determine the number of unpaired electrons in each complex. \[ [\mathrm{Cu(NH_3)_4}]^{2+} \] Copper: \[ \mathrm{Cu^{2+}} : 3d^9 \] Electronic configuration gives: \[ 1 \] unpaired electron. \[ [\mathrm{Ni(NH_3)_6}]^{2+} \] Nickel: \[ \mathrm{Ni^{2+}} : 3d^8 \] Octahedral complex with ammonia. Number of unpaired electrons: \[ 2 \] \[ [\mathrm{Mn(H_2O)_6}]^{2+} \] Manganese: \[ \mathrm{Mn^{2+}} : 3d^5 \] Water is a weak field ligand. Thus high-spin configuration is formed. Hence: \[ 5 \] unpaired electrons are present. Therefore, among all the given complexes: \[ [\mathrm{Mn(H_2O)_6}]^{2+} \] has maximum number of unpaired electrons. Thus, Statement I is true.

Step 2: Checking Statement II. First species: \[ [\mathrm{NiCl_4}]^{2-} \] Nickel is: \[ \mathrm{Ni^{2+}} : 3d^8 \] Chloride is weak field ligand. Tetrahedral complex forms. Complex is paramagnetic with: \[ 2 \] unpaired electrons. Thus, not diamagnetic. Second species: \[ [\mathrm{Ni(CO)_4}] \] Nickel oxidation state: \[ 0 \] Electronic configuration: \[ 3d^{10} \] All electrons are paired. Hence diamagnetic. Third species: \[ [\mathrm{NiO_4}]^{2-} \] This species is diamagnetic. Fourth species: \[ [\mathrm{O_4}]^{2-} \] This species is also diamagnetic. Thus, two pairs contain only diamagnetic species. Therefore, Statement II is also true. Hence: \[ \boxed{(2)\ \text{Both Statement I and Statement II are true}} \]