CBSE Class 12 Chemistry 2023: question 27

Hints

Open one hint at a time and try again before the next.

Step-by-step solution

Each step shows the marks it earns, as an examiner would award them. Checked by an independent re-solve from the question paper.

1. (a) With a strong field ligand (${\Delta }_0>P$), pairing in the lower $t_{2g}$ set is preferred, so the $d^5$ configuration is $t_{2g}^5{e}_g^0$ (low spin, one unpaired electron).

   1 mark

2. (b) $[\mathrm{Ni}(\mathrm{CO}{)}_4]$: Ni is in 0 state, $3d^{8}4s^2$. CO is a strong field ligand, so the 4s electrons are pushed into 3d, giving $3d^{10}$. The empty 4s and three 4p orbitals form **$s{p}^3$** hybrids → tetrahedral.

   ½ mark

3. $[\mathrm{Ni}(\mathrm{CN}{)}_4{]}^{2-}$: Ni is $\mathrm{N}{\mathrm{i}}^{2+}$, $3d^8$. Strong field $\mathrm{C}{\mathrm{N}}^{-}$ pairs the two unpaired 3d electrons, leaving one 3d orbital empty; one 3d, the 4s and two 4p orbitals form **$ds{p}^2$** hybrids → square planar.

   ½ mark

4. In both complexes all electrons are paired (no unpaired electrons), so both are diamagnetic despite their different geometries.

   1 mark

Answer: (a) $t_{2g}^5{e}_g^0$; (b) [Ni(CO)₄] sp³ (tetrahedral), [Ni(CN)₄]²⁻ dsp² (square planar); both have all electrons paired → diamagnetic

Revise the chapter

• Chapter: Coordination Compounds (notes, formulas, where marks are lost and every past question)
• Topic: Crystal Field Theory