Q.The reaction of chromium metal with H2SO4 in the absence of air gives blue solution of chromium ion.
Cr(s) + 2H+(aq) -> Cr2+(aq) + H2(g)
Cr2+ forms octahedral complex with H2O ligands.

Step 1. (a) Formula of the complex. Chromium metal reacts with H2SO4 (air-free) to give Cr2+(aq): $\text{Cr(s)}+2\text{H}^{\oplus}(aq)\rightarrow\text{Cr}^{2\oplus}(aq)+\text{H}_2(g)$. Cr2+ forms an OCTAHEDRAL complex with six H2O ligands (given): the complex ion is [Cr(H2O)6]2+.

Step 2. (b) Free-ion configuration. Chromium (Z=24) has the anomalous ground-state configuration [Ar]3d5 4s1 (half-filled-shell stability). Cr2+ is formed by removing 2 electrons -- the one 4s electron plus one 3d electron -- giving a free-ion configuration of 3d4.

Step 3. CFT description. H2O sits toward the weaker-field end of the spectrochemical series (Table 9.6), so [Cr(H2O)6]2+ is HIGH SPIN. For a high-spin d4 ion (Table 9.7), the CFT diagram (Fig. 9.2 pattern) shows all four electrons occupying the t2g and eg sets singly, following Hund's rule: t2g3 eg1, i.e. three of the lower t2g orbitals and one of the upper eg orbitals each hold one electron -- 4 unpaired electrons total. (This is a genuine, well-known real-world case: high-spin d4 octahedral complexes like [Cr(H2O)6]2+ characteristically also show Jahn-Teller distortion, though that refinement is beyond this chapter's scope.)

Step 4. VBT description. Since this is a HIGH SPIN complex, no pairing of the 3d4 electrons happens before hybridisation; with no 3d orbital left vacant for hybridisation, the complex must instead use OUTER-orbital hybridisation: one 4s, three 4p and two 4d orbitals, giving sp3d2 hybridisation (matching the [CoF6]3- high-spin worked example of section 9.9.2). The six sp3d2 hybrid orbitals overlap with the six H2O lone pairs to form the Cr-O coordinate bonds, while the 3d4 electrons remain unpaired.

Step 5. Diagrams (described). Crystal field splitting diagram: t2g (lower, 3 orbitals, one electron each) and eg (upper, 2 orbitals, one electron in one, one empty), separated by Delta-o, matching Fig. 9.2's layout. Valence bond orbital diagram: 3d4 (4 singly-occupied orbitals, unaffected by bonding) shown alongside the six newly-formed sp3d2 hybrid orbitals (built from 4s, three 4p, two 4d) each overlapping with one H2O ligand.

Combining the balanced reaction (given) with Cr's anomalous ground-state configuration to get the free-ion d-count, then applying both CFT (spectrochemical series, high/low spin) and VBT (inner/outer orbital hybridisation) worked-example methods from sections 9.9.1-9.9.2 and 9.9.6

• Using chromium's 'expected' 3d4 4s2 configuration without accounting for the real, anomalous 3d5 4s1 ground state before removing electrons to form Cr2+.
• Assuming octahedral complexes always use d2sp3 (inner orbital) hybridisation, forgetting that a high-spin d4 ion with a weak-field ligand like H2O instead uses sp3d2 (outer orbital).