MCQ
Maximum oxidation state of $Cr$ is
- A$3$
- B$4$
- ✓$6$
- D$7$
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Reason : Oxygen forms $p\pi \,-\, p\pi $ multiple bond due to small size and small bond length but $p\pi \,-\,p\pi $ bonding is not possible in sulphur.
| $LIST-I$ | $LIST-II$ |
| $(I)$ $\left[ Cr ( CN )_6\right]^{4-}$ | $(P)$ $t_{2 g }$ orbitals contain 4 electrons |
| $(II)$ $\left[ RuCl _6\right]^{2-}$ | $(Q)$ $\mu$ (spin-only) $=4.9 BM$ |
| $(III)$ $\left[ Cr \left( H _2 O \right)_6\right]^{2+}$ | $(R)$ low spin complex ion |
| $(IV)$ $\left[ Fe \left( H _2 O \right)_6\right]^{2+}$ | $(S)$ metal ion in $4$+ oxidation state |
| $(T)$ $d^4$ species |
[Given : Atomic number of $Cr =24, Ru =44, Fe =26$ ] Metal each metal species in $LIST-I$ with their properties in $LIST-II$, and choose the correct option