MCQ
Crystal field splitting energy for high spin $d^4$ octahedral complex is
- A$- 1.2 \Delta _o$
- ✓$- 0.6 \Delta _o$
- C$- 0.8 \Delta _o$
- D$- 1.6 \Delta _o$
For high spin $d^4$ octahedral complex,
therefore, Crystal field stabilisation energy
$=(-3 \times 0.4+1 \times 0.6) \Delta_{0}$
$=-(-1.2 \times 0.6) \Delta_{0}$
$=-0.6\Delta_{0}$
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$CH_3COOH + PCl_5 \longrightarrow (A)$ $\xrightarrow[(2){{H}_{2}}O]{(1)\,C{{H}_{3}}MgBr}\left( B \right)$ Product $B$ would be

| Peptide | $R _1$ | $R _2$ |
| $I$ | $H$ | $H$ |
| $II$ | $H$ | $CH _3$ |
| $III$ | $CH _2 COOH$ | $H$ |
| $IV$ | $CH _2 CONH _2$ | $( CH _2)_4 NH _2$ |
| $V$ | $CH _2 CONH _2$ | $CH _2 CONH _2$ |
| $VI$ | $( CH _2)_4 NH _2$ | $( CH _2)_4 NH _2$ |
| $VII$ | $CH _2 COOH$ | $CH _2 CONH _2$ |
| $VIII$ | $CH _2 OH$ | $( CH _2)_4 NH _2$ |
| $IX$ | $( CH _2)_4 NH _2$ | $CH _3$ |
