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Question

# Which has highest paramagnetism?

A
[Cr(H2O)6]3+
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B
[Fe(H2O)6]2+
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C
[Cu(H2O)6]2+
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D
[Zn(H2O)6]2+
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Solution

## The correct option is A [Fe(H2O)6]2+Paramagnetism is directly proportional to the number of unpaired electrons. The complex having the highest number of unpaired electrons will show the highest paramagnetism. (A) [Cr(H2O)6]3+ Oxidation state of Cr is +3 (i.e. x−0=+3), hence Cr3+ have 3d3 electron in outermost orbit, so it have three unpaired electrons (n=3). (B) [Fe(H2O)6]2+ oxidation state of Fe is +2 (i.e. x−0=+2), hence Fe2+ have 3d6 electron in outermost orbit, so it have four unpaired electron (n=4). (C) [Cu(H2O)6]2+ oxidation state of Cu is +2 (i.e. x−0=+2), hence Cu2+ have 3d9 electron in outermost orbit, so it have one unpaired electron (n=1). (D) [Zn(H2O)6]2+ oxidation state of Zn is +2 (i.e. x−0=+2), hence Zn2+ have 3d10 electron in outermost orbit, so all electrons are paired (n=0). So as explained above, the highest number of the unpaired electron is (n=4) for [Fe(H2O)6]2+, hence it shows the highest paramagnetism.

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