Question

Which of the following sequences of hybridisation, geometry and magnetic nature are correct for the given coordination compounds?}

• [A.] \([NiCl_4]^{2-}\) -- \(sp^3\), tetrahedral, paramagnetic
• [B.] \([Ni(NH_3)_6]^{2+}\) -- \(sp^3d^2\), octahedral, paramagnetic
• [C.] \([Ni(CO)_4]\) -- \(sp^3\), tetrahedral, paramagnetic
• [D.] \([Ni(CN)_4]^{2-}\) -- \(dsp^2\), square planar, diamagnetic

Choose the correct answer.

• A. A, B, C and D
• B. B, C and D only
• C. A, C and D only
• D. A, B and D only

Answer 1

The Correct Option is D

Step 1: Understanding the Question:
The question asks to identify the correct sets of hybridization, geometry, and magnetic properties for four different Nickel coordination complexes.
Step 2: Key Formula or Approach:
1. Determine the oxidation state of Nickel in each complex.
2. Write the electronic configuration of the Nickel ion.
3. Assess the nature of the ligand (Strong Field Ligand vs. Weak Field Ligand) using the Spectrochemical series.
4. Apply Valence Bond Theory (VBT) to find hybridization and magnetic behavior.
Step 3: Detailed Explanation:
A. \([NiCl_{4}]^{2-}\):
Ni is in +2 state (\(3d^{8} 4s^{0}\)). \(Cl^{-}\) is a weak field ligand (WFL), so no pairing occurs.
Electronic arrangement: \(t_{2g}^{6} e_{g}^{2}\) in CFT terms, but for VBT: Four \(sp^{3}\) orbitals are used.
Geometry: Tetrahedral. Magnetic nature: Paramagnetic (2 unpaired electrons). Correct.

B. \([Ni(NH_{3})_{6]^{2+}\):}
Ni is in +2 state (\(3d^{8}\)). \(NH_{3}\) is a moderately strong ligand. For octahedral Ni(II), it always forms outer orbital complexes because it cannot pair electrons to leave two empty 'd' orbitals.
Hybridization: \(sp^{3}d^{2}\). Geometry: Octahedral.
Magnetic nature: Paramagnetic (2 unpaired electrons). Correct.

C. \([Ni(CO)_{4]\):}
Ni is in 0 state (\(3d^{8} 4s^{2}\)). \(CO\) is a very strong field ligand (SFL). It causes all electrons to pair up in the 3d subshell (\(3d^{10}\)).
Hybridization: \(sp^{3}\). Geometry: Tetrahedral.
Magnetic nature: Diamagnetic (no unpaired electrons). The option says paramagnetic, so it is Incorrect.

D. \([Ni(CN)_{4]^{2-}\):}
Ni is in +2 state (\(3d^{8}\)). \(CN^{-}\) is a SFL. It causes pairing of the 3d electrons.
One 3d orbital becomes empty. Hybridization: \(dsp^{2}\).
Geometry: Square planar. Magnetic nature: Diamagnetic. Correct.
Step 4: Final Answer:
Since A, B, and D are correct, the right choice is option (D).