JEE MAIN - Chemistry (2018 (Offline) - No. 12)
The oxidation states of Cr in [Cr(H2O)6]Cl3, [Cr(C6H6)2] and K2[Cr(CN)2(O)2(O2)(NH3)] respectively are :
+3, 0 and +4
+3, +4 and +6
+3, +2 and +4
+3, 0 and +6
Explanation
Assume oxidation state of Cr in all the compounds = x
(i)$$\,\,\,$$ In [Cr(H2O)6] Cl3 oxidation state of Cr is
x + 0 $$ \times $$ 6 + ($$-$$1 $$ \times $$ ) = O
$$ \Rightarrow \,\,\,\,\,$$ x + 0 $$-$$ 3 = O
$$ \Rightarrow \,\,\,\,$$ x = + 3
(ii)$$\,\,\,$$ [Cr (C6 H6)2] oxidation state of Cr is
x + 0 $$ \times $$ 2 = 0
$$ \Rightarrow \,\,\,\,$$ x = 0
(iii) $$\,\,\,$$ In K2 [ Cr(CN)2 (O)2 (O2) (NH3)] oxidation state of Cr is
1 $$ \times $$ 2 + x + ($$-$$ 1 $$ \times $$ 2) + ($$-$$2 $$ \times $$ 2) + ($$-$$2) + 0 = 0
$$ \Rightarrow \,\,\,\,$$ x = + 6
$$\therefore\,\,\,$$ + 3, 0 and + 6 is the correct answer.
Note :
O2 molecule can have 0, $$-$$ 1, $$-$$ 2 oxidation state but in K2 [ Cr (CN)2 (O)2 (O2) NH3 ] if we choose zero as the oxidation state of O2 then for Cr oxidation state will be $$+$$ 4. But + 4 oxidation state of Cr is unstable and $$+$$ 6 is most stable that is why we choose $$-$$ 2 oxidation state of O2.
(i)$$\,\,\,$$ In [Cr(H2O)6] Cl3 oxidation state of Cr is
x + 0 $$ \times $$ 6 + ($$-$$1 $$ \times $$ ) = O
$$ \Rightarrow \,\,\,\,\,$$ x + 0 $$-$$ 3 = O
$$ \Rightarrow \,\,\,\,$$ x = + 3
(ii)$$\,\,\,$$ [Cr (C6 H6)2] oxidation state of Cr is
x + 0 $$ \times $$ 2 = 0
$$ \Rightarrow \,\,\,\,$$ x = 0
(iii) $$\,\,\,$$ In K2 [ Cr(CN)2 (O)2 (O2) (NH3)] oxidation state of Cr is
1 $$ \times $$ 2 + x + ($$-$$ 1 $$ \times $$ 2) + ($$-$$2 $$ \times $$ 2) + ($$-$$2) + 0 = 0
$$ \Rightarrow \,\,\,\,$$ x = + 6
$$\therefore\,\,\,$$ + 3, 0 and + 6 is the correct answer.
Note :
O2 molecule can have 0, $$-$$ 1, $$-$$ 2 oxidation state but in K2 [ Cr (CN)2 (O)2 (O2) NH3 ] if we choose zero as the oxidation state of O2 then for Cr oxidation state will be $$+$$ 4. But + 4 oxidation state of Cr is unstable and $$+$$ 6 is most stable that is why we choose $$-$$ 2 oxidation state of O2.
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