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Coordination Compounds question

2020 · 5 Sep · Shift 2 · Q4
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  5. /2020 · 5 Sep · Shift 2 · Q4

Coordination Compounds question

2020 · 5 Sep · Shift 2 · Q4

JEE MainChemistryCoordination CompoundsNumerical+4 / −1
Considering that Δ\DeltaΔ 0 > P, the magnetic moment (in BM) of [Ru(H2O)6]2+[Ru(H_2O)_6]^{2+}[Ru(H2​O)6​]2+ would be ‾\underline{\hspace{2cm}}​.
Numerical answer
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Correct answer: 0

  1. Find the oxidation state of Ru in [Ru(H2O)6]2+[Ru(H_2O)_6]^{2+}[Ru(H2​O)6​]2+.

Since H2OH_2OH2​O is a neutral ligand,

extOxidationstateofRu=+2 ext{Oxidation state of Ru} = +2extOxidationstateofRu=+2

So the metal ion is Ru2+Ru^{2+}Ru2+.

  1. Write the electronic configuration of Ru and Ru2+Ru^{2+}Ru2+.

Atomic number of Ru is 444444. Ground-state configuration of Ru:

Ru:[Kr]4d75s1Ru: [Kr]4d^7 5s^1Ru:[Kr]4d75s1

Removing two electrons for Ru2+Ru^{2+}Ru2+:

Ru2+:[Kr]4d6Ru^{2+}: [Kr]4d^6Ru2+:[Kr]4d6

So it is a d6d^6d6 system.

  1. Use the given condition Δo>P\Delta_o > PΔo​>P.

This means the octahedral crystal field splitting is greater than pairing energy, so the complex is low spin.

For an octahedral low-spin d6d^6d6 configuration:

t2g6eg0t_{2g}^6 e_g^0t2g6​eg0​

All six electrons are paired.

  1. Count unpaired electrons.

Number of unpaired electrons,

n=0n = 0n=0
  1. Calculate magnetic moment.

Using spin-only formula:

μ=n(n+2)\mu = \sqrt{n(n+2)}μ=n(n+2)​

Thus,

μ=0(0+2)=0 BM\mu = \sqrt{0(0+2)} = 0\ \text{BM}μ=0(0+2)​=0 BM
  1. Final answer

The magnetic moment of [Ru(H2O)6]2+[Ru(H_2O)_6]^{2+}[Ru(H2​O)6​]2+ is:

0 BM0\ \text{BM}0 BM
  1. Comparison with stored correct answer

Stored correct answer = 000

My derived answer also is 000, so they agree.

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