A) +3, 3d and 4
B) +3, 4d and 6
C) +3, 3d and 6
D) +2, 3d and 6
Correct Answer: C
Solution :
Let the oxidation number of Cr is\[x\]. \[cis[Cr{{(en)}_{2}}C{{l}_{2}}]Cl\] \[x+(0)\times 2+(-1)\times 2-1=0\] \[x-3=0\] \[x=+3\] In\[[Cr{{(en)}_{2}}C{{l}_{2}}]Cl,\]the Cr atom is present as\[C{{r}^{3+}}\]. \[C{{r}^{3+}}=[Ar]\,3{{d}^{3}}4{{s}^{0}}\] \[cis[Cr{{(en)}_{2}}C{{l}_{2}}]=[Ar]\] (en being a strong field ligand, can cause pairing.) Thus, 3d orbitals are occupied. Since, two\[Cl\]atoms (monodentate) and two ethylene diammine, ie, en molecules (didentate) are linked two the central atom, the coordination number of central atomYou need to login to perform this action.
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