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BITSAT 2017
Exam
BITSAT
Year
2017
Subject
Chemistry
Chapter
Coordination Compounds

BITSAT 2017 · Chemistry · Coordination Compounds

Which of the following is a diamagnetic complex

Which of the following is a diamagnetic complex

Options

  1. A[Co(NH3)6]3+\left[Co\left({NH}_{3}\right)6\right]3 +
  2. B[NiCl4]2−\left[{NiCl}_{4}\right]2 -
  3. C[CuCl4]2−\left[{CuCl}_{4}\right]2 -
  4. D[Fe(H2O)6]3+\left[Fe\left({H}_{2}O\right)6\right]3 +

Correct answer

A. [Co(NH3)6]3+\left[Co\left({NH}_{3}\right)6\right]3 +

Step-by-step solution

Among the given option (a) is a diamagnetic complex.

(a) [Co(NH3)6]3+⇒Co(27)⇒Co3+=3d64s0\left[Co\left({NH}_{3}\right)6\right]3 + ⇒ Co\left(27\right) ⇒ {Co}^{3 +} = 3d^{6}4s^{0}.

Thus, has no unpaired electrons and is a diamagnetic complex.

(b) \left[{NiCl}_{4}\right]2 - ⇒ Ni\left(28\right) ⇒ {Ni}^{2 +} = 3d^{8}4s^{0} </p><p>Thus,hastwounpairedelectronsisparamagnetic.</p><p>(c)[CuCl4]2−⇒Cu(29)⇒Cu2+=3d94s0~</p><p>Thus, has two unpaired electrons is paramagnetic.</p><p>(c) \left[{CuCl}_{4}\right]2 - ⇒ Cu\left(29\right) ⇒ {Cu}^{2 +} = 3d^{9}4s^{0}

Thus, has one unpaired electrons and is paramagnetic.

(d) \left[Fe\left({H}_{2}O\right)6\right]3 + ⇒ Fe\left(26\right) ⇒ {Fe}^{3 +} = 3d^{5}4s^{0}

Thus, has five unpaired electrons and is paramagnetic.

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